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Title: INTERMEDIATE ALGEBRA (English)
Description: It's very helpful for new intermediate students.
Description: It's very helpful for new intermediate students.
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Beginning and Intermediate Algebra
An open source (CC-BY) textbook
Available for free download at: http://wallace
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Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
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Special thanks to: My beautiful wife, Nicole Wallace
who spent countless hours typing problems and
my two wonderful kids for their patience and
support during this project
Another thanks goes to the faculty reviewers who reviewed this text: Donna Brown, Michelle
Sherwood, Ron Wallace, and Barbara Whitney
One last thanks to the student reviewers of the text: Eloisa Butler, Norma Cabanas, Irene
Chavez, Anna Dahlke, Kelly Diguilio, Camden Eckhart, Brad Evers, Lisa Garza, Nickie Hampshire, Melissa Hanson, Adriana Hernandez, Tiffany Isaacson, Maria Martinez, Brandon Platt,
Tim Ries, Lorissa Smith, Nadine Svopa, Cayleen Trautman, and Erin White
3
Table of Contents
Chapter 0: Pre-Algebra
Chapter 3: Inequalities
0
...
7
3
...
118
0
...
12
3
...
124
0
...
18
3
...
128
0
...
22
Chapter 4: Systems of Equations
Chapter 1: Solving Linear Equations
4
...
134
1
...
28
4
...
139
1
...
33
4
...
146
1
...
37
4
...
151
1
...
43
4
...
158
1
...
47
4
...
167
1
...
52
Chapter 5: Polynomials
1
...
57
5
...
177
1
...
64
5
...
183
1
...
72
5
...
188
1
...
79
5
...
192
Chapter 2: Graphing
5
...
196
2
...
89
5
...
201
2
...
95
5
...
205
2
...
102
2
...
107
2
...
112
4
Chapter 6: Factoring
Chapter 9: Quadratics
6
...
212
9
...
326
6
...
216
9
...
332
6
...
221
9
...
337
6
...
226
9
...
343
6
...
229
9
...
348
6
...
234
9
...
352
6
...
237
9
...
357
Chapter 7: Rational Expressions
9
...
364
7
...
243
9
...
370
7
...
248
9
...
373
7
...
253
9
...
380
7
...
257
Chapter 10: Functions
7
...
262
10
...
386
7
...
268
10
...
393
7
...
274
10
...
401
7
...
279
10
...
406
Chapter 8: Radicals
10
...
410
8
...
288
10
...
414
8
...
292
10
...
420
8
...
295
10
...
428
8
...
298
Answers
...
5 Rationalize Denominators
...
6 Rational Exponents
...
7 Radicals of Mixed Index
...
8 Complex Numbers
...
1 Integers
...
2 Fractions
...
3 Order of Operations
...
4 Properties of Algebra
...
1
Pre-Algebra - Integers
Objective: Add, Subtract, Multiply and Divide Positive and Negative
Numbers
...
For this reason we will do a quick review of adding, subtracting, multiplying and dividing of integers
...
As this is intended to be a review of integers, the
descriptions and examples will not be as detailed as a normal lesson
...
When adding integers we have two cases to consider
...
If the signs match we will add the numbers together and keep the sign
...
− 5 + ( − 3)
−8
Same sign, add 5 + 3, keep the negative
Our Solution
− 7 + ( − 5)
− 12
Same sign, add 7 + 5, keep the negative
Our Solution
Example 2
...
This means if the larger number is positive, the answer is positive
...
This is shown in the following
examples
...
−7+2
−5
Different signs, subtract 7 − 2, use sign from bigger number, negative
Our Solution
Example 4
...
4 + ( − 3)
1
Different signs, subtract 4 − 3, use sign from bigger number, positive
Our Solution
Example 6
...
The way we change a subtraction to an addition is to add the opposite of the number after the subtraction
sign
...
” This is illustrated in
the following examples
...
8−3
8 + ( − 3)
5
Add the opposite of 3
Different signs, subtract 8 − 3, use sign from bigger number, positive
Our Solution
Example 8
...
9 − ( − 4)
9+4
13
Add the opposite of − 4
Same sign, add 9 + 4, keep the positive
Our Solution
Example 10
...
The
short description of the process is we multiply and divide like normal, if the signs
match (both positive or both negative) the answer is positive
...
This is shown
in the following examples
Example 11
...
− 36
−9
4
Signs match, answer is positive
Our Solution
Example 13
...
15
−3
Signs do not match, answer is negative
−5
Our Solution
A few things to be careful of when working with integers
...
The second problem is a multiplication problem because there is nothing between the 3 and the parenthesis
...
The − 3 − 8 problem, is subtraction because the subtraction separates the 3 from what comes after it
...
They can look very similar, for
example if the signs match on addition, the we keep the negative, − 3 + ( − 7) = −
10, but if the signs match on multiplication, the answer is positive, ( − 3)( − 7) =
21
...
1 Practice - Integers
Evaluate each expression
...
31) (4)( − 1)
32) (7)( − 5)
33) (10)( − 8)
34) ( − 7)( − 2)
35) ( − 4)( − 2)
36) ( − 6)( − 1)
37) ( − 7)(8)
38) (6)( − 1)
39) (9)( − 4)
40) ( − 9)( − 7)
41) ( − 5)(2)
42) ( − 2)( − 2)
43) ( − 5)(4)
44) ( − 3)( − 9)
10
45) (4)( − 6)
Find each quotient
...
2
Pre-Algebra - Fractions
Objective: Reduce, add, subtract, multiply, and divide with fractions
...
Here we will
briefly review reducing, multiplying, dividing, adding, and subtracting fractions
...
World View Note: The earliest known use of fraction comes from the Middle
Kingdom of Egypt around 2000 BC!
We always like our final answers when working with fractions to be reduced
...
This is shown in the following example
Example 15
...
We also could have divided by 2 twice and then
divided by 3 once (in any order)
...
The easiest operation with fractions is multiplication
...
Example 16
...
We can either
reduce vertically with a single fraction, or diagonally with several fractions, as
long as we use one number from the numerator and one number from the denominator
...
25 32
·
24 55
5 4
·
3 11
20
33
Reduce 25 and 55 by dividing by 5
...
Dividing
fractions requires us to first take the reciprocal of the second fraction and multiply
...
13
Example 18
...
Reduce 6 and 16 by dividing by 2
Multiply numerators across and denominators across
Our Soultion
To add and subtract fractions we will first have to find the least common denominator (LCD)
...
One way is to find the
smallest multiple of the largest denominator that you can also divide the small
denomiator by
...
Find the LCD of 8 and 12
12
12?
8
24
24?
=3
8
24
Test multiples of 12
Can ′t divide 12 by 8
Yes! We can divide 24 by 8!
Our Soultion
Adding and subtracting fractions is identical in process
...
Example 20
...
For this reason we will always use the improper fraction, not
the mixed number
...
13 9
−
6
6
Same denominator, subtract numerators 13 − 9
4
6
Reduce answer, dividing by 2
2
3
Our Solution
If the denominators do not match we will first have to identify the LCD and build
up each fraction by multiplying the numerators and denominators by the same
number so the denominator is built up to the LCD
...
5 4
+
6 9
3·5 4·2
+
3·6 9·2
15
8
+
18 18
23
18
LCD is 18
...
2 1
−
3 6
2·2 1
−
2·3 6
4 1
−
6 6
LCD is 6
Multiply first fraction by 2, the second already has a denominator of 6
Same denominator, subtract numerators, 4 − 1
3
6
Reduce answer, dividing by 3
1
2
Our Solution
15
0
...
Leave your answer as an improper fraction
...
5
8
22) ( − 2)( − 6 )
21) (9)( 9 )
2
24) ( − 2)( 3 )
1
13
26) ( 2 )( 2 )
23) (2)( − 9 )
25) ( − 2)( 8 )
6
27) ( − 5 )( −
11
)
8
3
1
3
28) ( − 7 )( −
11
)
8
1
30) ( − 2)( − 7 )
9
3
32) ( −
29) (8)( 2 )
2
31) ( 3 )( 4 )
3
3
17
33 (2)( 2 )
1
3
17
)( − 5 )
9
34) ( 9 )( − 5 )
7
35) ( 2 )( − 5 )
1
5
36) ( 2 )( 7 )
16
Find each quotient
...
53)
1
3
+ ( − 3)
55)
3
7
−7
57)
11
6
+6
59)
3
5
+4
5
60) ( − 1) − 3
61)
2
5
+4
5
62)
63)
9
8
+ ( − 7)
4
1
7
11
)
7
5
58) ( − 2) + ( −
2
12
7
9
−7
5
2
64) ( − 2) + 6
1
66)
1
2
68)
11
8
−2
70)
6
5
8
65) 1 + ( − 3 )
1
3
67) ( − 2 ) + 2
69)
1
5
3
+4
5
71) ( − 7 ) −
73) 6 −
75)
3
2
−
77) ( −
11
6
1
−5
8
72) ( − 3 ) + ( − 5 )
5
8
7
74) ( − 6) + ( − 3 )
15
8
76) ( − 1) − ( − 3 )
15
5
)+ 3
8
78)
1
1
5
3
−
1
15
8
79) ( − 1) − ( − 6 )
81)
15
)
8
1
− ( − 3)
3
2
9
+7
1
3
80) ( − 2 ) − ( − 5 )
82)
17
9
7
5
− ( − 3)
0
...
When simplifying expressions it is important that we simplify them in the correct
order
...
2 + 5 · 3 Add First
7·3
Multiply
21
Solution
2 + 5 · 3 Multiply
2 + 15
Add
17
Solution
The previous example illustrates that if the same problem is done two different
ways we will arrive at two different solutions
...
It turns out the second method, 17, is the correct method
...
Before addition is completed we must do repeated addition or multiplication (or
division)
...
When we want to do something out of order and make it come first
we will put it in parenthesis (or grouping symbols)
...
Order of Operations:
Parenthesis (Grouping)
Exponents
Multiply and Divide (Left to Right)
Add and Subtract (Left to Right)
Multiply and Divide are on the same level because they are the same operation
(division is just multiplying by the reciprocal)
...
The
same is true for adding and subtracting (subtracting is just adding the opposite)
...
However, it is the
P
E
author’s suggestion to think about PEMDAS as a vertical word written as:
MD
AS
so we don’t forget that multiplication and division are done left to right (same
with addition and subtraction)
...
World View Note: The first use of grouping symbols are found in 1646 in the
Dutch mathematician, Franciscus van Schooten’s text, Vieta
...
So problems like 2(3 + 5) were written
as 2 · 3 + 5
...
2 + 3(9 − 4)2
2 + 3(5)2
2 + 3(25)
2 + 75
77
Parenthesis first
Exponents
Multiply
Add
Our Solution
It is very important to remember to multiply and divide from from left to right!
Example 26
...
If there are several parenthesis in a problem we will start with the inner most
parenthesis and work our way out
...
To make it easier to know which parenthesis goes with
which parenthesis, different types of parenthesis will be used such as { } and [ ]
and ( ), these parenthesis all mean the same thing, they are parenthesis and must
be evaluated first
...
2{82 − 7[32 − 4(32 + 1)]( − 1)}
˜
2{82 − 7[32 − 4(9 + 1)]( − 1)}
2{82 − 7[32− 4(10)]( − 1)}
2{82 − 7[32 − 40]( − 1)}
2{82 − 7[ − 8]( − 1)}
˜
2{64− 7[ − 8]( − 1)}
2{64 + 56( − 1)}
2{64 − 56}
2{8}
16
Inner most parenthesis, exponents first
Add inside those parenthesis
Multiply inside inner most parenthesis
Subtract inside those parenthesis
Exponents next
Multiply left to right, sign with the number
Finish multiplying
Subtract inside parenthesis
Multiply
Our Solution
As the above example illustrates, it can take several steps to complete a problem
...
This will reduce the chance of making
a mistake along the way
...
One type is a fraction bar
...
In these
cases we can simplify in both the numerator and denominator at the same time
...
˜
24 − ( − 8) · 3
15 ÷ 5 − 1
Exponent in the numerator, divide in denominator
16 − ( − 8) · 3
3−1
Multiply in the numerator, subtract in denominator
16 − ( − 24)
2
Add the opposite to simplify numerator, denominator is done
...
When we have absolute value we will evaluate everything inside the absolute value, just as if it were a normal parenthesis
...
Example 29
...
Exponents only are considered to be on the number they are attached to
...
But when the
negative is in parentheses, such as ( − 5)2 the negative is part of the number and
is also squared giving us a positive solution, 25
...
3 Practice - Order of Operation
Solve
...
4
Pre-Algebra - Properties of Algebra
Objective: Simplify algebraic expressions by substituting given values,
distributing, and combining like terms
In algebra we will often need to simplify an expression to make it easier to use
...
World View Note: The term “Algebra” comes from the Arabic word al-jabr
which means “reunion”
...
The first form of simplifying expressions is used when we know what number each
variable in the expression represents
...
Example 30
...
Notice the 3 and 5 in the previous example are in
parenthesis
...
Sometimes the parenthesis won’t make a difference, but it is a good
habbit to always use them to prevent problems later
...
x
when x = − 6 and z = − 2
3
( − 6)
( − 6) + ( − 2)( − 6)(3 − ( − 2))
3
− 6 + ( − 2)( − 6)(5)( − 2)
− 6 + 12(5)( − 2)
− 6 + 60( − 2)
− 6 − 120
− 126
x + zx(3 − z)
22
Replace all x ′s with 6 and z ′s with 2
Evaluate parenthesis
Multiply left to right
Multiply left to right
Multiply
Subtract
Our Solution
It will be more common in our study of algebra that we do not know the value of
the variables
...
One way we can simplify expressions is to combine
like terms
...
Examples of like terms would be 3xy and − 7xy or 3a2b and 8a2b or −
3 and 5
...
This is shown in the following examples
Example 32
...
8x2 − 3x + 7 − 2x2 + 4x − 3
6x2 + x + 4
Combine like terms 8x2 − 2x2 and − 3x + 4x and 7 − 3
Our Solution
As we combine like terms we need to interpret subtraction signs as part of the following term
...
A final method to simplify is known as distributing
...
To get rid of these unwanted parenthesis we have the distributive
property
...
Distributive Property: a(b + c) = ab + ac
Several examples of using the distributive property are given below
...
4(2x − 7)
8x − 28
Multiply each term by 4
Our Solution
Example 35
...
The most common error in distributing is a sign error, be very
careful with your signs!
23
It is possible to distribute just a negative through parenthesis
...
This is shown in the following example
...
− (4x − 5y + 6)
− 1(4x − 5y + 6)
− 4x + 5y − 6
Negative can be thought of as − 1
Multiply each term by − 1
Our Solution
Distributing through parenthesis and combining like terms can be combined into
one problem
...
Thus we do each problem in two steps, distribute then combine
...
5 + 3(2x − 4)
5 + 6x − 12
− 7 + 6x
Distribute 3, multipling each term
Combine like terms 5 − 12
Our Solution
Example 38
...
This is because we will
always treat subtraction like a negative sign that goes with the number after it
...
Following are more involved examples of distributing and combining like terms
...
2(5x − 8) − 6(4x + 3)
10x − 16 − 24x − 18
− 14x − 34
Distribute 2 into first parenthesis and − 6 into second
Combine like terms 10x − 24x and − 16 − 18
Our Solution
Example 40
...
4 Practice - Properties of Algebra
Evaluate each using the values given
...
53) 9(b + 10) + 5b
54) 4v − 7(1 − 8v)
55) − 3x(1 − 4x) − 4x2
56) − 8x + 9( − 9x + 9)
57) − 4k 2 − 8k (8k + 1)
58) − 9 − 10(1 + 9a)
59) 1 − 7(5 + 7p)
60) − 10(x − 2) − 3
61) − 10 − 4(n − 5)
62) − 6(5 − m) + 3m
63) 4(x + 7) + 8(x + 4)
64) − 2r(1 + 4r) + 8r( − r + 4)
65) − 8(n + 6) − 8n(n + 8)
66) 9(6b + 5) − 4b(b + 3)
67) 7(7 + 3v) + 10(3 − 10v)
68) − 7(4x − 6) + 2(10x − 10)
69) 2n( − 10n + 5) − 7(6 − 10n)
70) − 3(4 + a) + 6a(9a + 10)
71) 5(1 − 6k) + 10(k − 8)
72) − 7(4x + 3) − 10(10x + 10)
73) (8n2 − 3n) − (5 + 4n2)
74) (7x2 − 3) − (5x2 + 6x)
75) (5p − 6) + (1 − p)
76) (3x2 − x) − (7 − 8x)
77) (2 − 4v 2) + (3v 2 + 2v)
78) (2b − 8) + (b − 7b2)
79) (4 − 2k 2) + (8 − 2k 2)
80) (7a2 + 7a) − (6a2 + 4a)
81) (x2 − 8) + (2x2 − 7)
82) (3 − 7n2) + (6n2 + 3)
26
Chapter 1 : Solving Linear Equations
1
...
28
1
...
33
1
...
37
1
...
43
1
...
47
1
...
52
1
...
57
1
...
64
1
...
72
1
...
79
27
1
...
In
algebra, we are often presented with a problem where the answer is known, but
part of the problem is missing
...
An example of such a problem is shown below
...
4x + 16 = − 4
Notice the above problem has a missing part, or unknown, that is marked by x
...
This is shown in Example 2
...
4( − 5) + 16 = − 4
− 20 + 16 = − 4
−4=−4
Multiply 4( − 5)
Add − 20 + 16
True!
Now the equation comes out to a true statement! Notice also that if another
number, for example, 3, was plugged in, we would not get a true statement as
seen in Example 3
...
4(3) + 16 = − 4
12 + 16 = − 4
28 − 4
Multiply 4(3)
Add 12 + 16
False!
Due to the fact that this is not a true statement, this demonstates that 3 is not
the solution
...
Thus, we take a more algebraic approach
to solving equations
...
While these equations often seem
very fundamental, it is important to master the pattern for solving these problems
so we can solve more complex problems
...
For example, consider
the following example
...
x+7=−5
−7 −7
x = − 12
The 7 is added to the x
Subtract 7 from both sides to get rid of it
Our solution!
Then we get our solution, x = − 12
...
Example 45
...
Addition Examples
Subtraction Problems
In a subtraction problem, we get rid of negative numbers by adding them to both
sides of the equation
...
Example 46
...
The same process is used in each of the following
examples
...
29
Example 47
...
Subtraction Examples
Multiplication Problems
With a multiplication problem, we get rid of the number by dividing on both
sides
...
Example 48
...
If
x is multiplied by a negative then we will divide by a negative
...
Example 49
...
Notice how negative
and positive numbers are handled as each problem is solved
...
30
8x = − 24
8
8
x=−3
− 4x = − 20
−4 −4
x=5
42 = 7x
7
7
6=x
Table 3
...
For example consider our next example
...
x
=−3
5
x
(5) = − 3(5)
5
x = − 15
Variable is divided by 5
Multiply both sides by 5
Our Solution!
Then we get our solution x = − 15
...
Example 52
...
Division Examples
The process described above is fundamental to solving equations
...
These problems may seem more complex, but the process and patterns used will remain the
same
...
31
1
...
1) v + 9 = 16
2) 14 = b + 3
3) x − 11 = − 16
4) − 14 = x − 18
5) 30 = a + 20
6) − 1 + k = 5
7) x − 7 = − 26
8) − 13 + p = − 19
9) 13 = n − 5
10) 22 = 16 + m
11) 340 = − 17x
12) 4r = − 28
n
5
9
13) − 9 = 12
14)
15) 20v = − 160
16) − 20x = − 80
17) 340 = 20n
19) 16x = 320
21) − 16 + n = − 13
23) p − 8 = − 21
25) 180 = 12x
b
=9
18)
1
2
20)
k
13
a
=8
= − 16
22) 21 = x + 5
24) m − 4 = − 13
26) 3n = 24
27) 20b = − 200
29)
r
14
=
5
14
31) − 7 = a + 4
33) 10 = x − 4
35) 13a = − 143
37)
p
20
= − 12
39) 9 + m = − 7
x
28) − 17 = 12
30) n + 8 = 10
32) v − 16 = − 30
34) − 15 = x − 16
36) − 8k = 120
x
38) − 15 = 9
n
40) − 19 = 20
32
1
...
After mastering the technique for solving equations that are simple one-step equations, we are ready to consider two-step equations
...
We learned that to
clear “divided by seven” we multiply by seven on both sides
...
When solving for our variable x, we use order
of operations backwards as well
...
So to solve the equation in the first
example,
Example 53
...
We need to move the 4 and the
20 to the other side
...
If order of operations is done backwards, we will add or subtract first
...
Once we are done with that, we will divide both sides by 4
...
4x − 20 = − 8
+ 20 + 20
4x
= 12
4
4
x=3
Start by focusing on the subtract 20
Add 20 to both sides
Now we focus on the 4 multiplied by x
Divide both sides by 4
Our Solution!
Notice in our next example when we replace the x with 3 we get a true statement
...
Add or subtract first,
then multiply or divide
...
Example 54
...
With this in mind the process is almost identical to our first example
...
Remember the sign always stays with the number
...
Example 55
...
Consider
the next example
...
8−x=2
−8
−8
−x=−6
− 1x = − 6
Start by focusing on the positive 8
Subtract 8 from both sides
Negative (subtraction) stays on the x
Remember, no number in front of variable means 1
34
−1 −1
x=6
Divide both sides by − 1
Our Solution!
Solving two-step equations is a very important skill to master, as we study
algebra
...
This pattern is seen in each of the following examples
...
− 3x + 7 = − 8
−7 −7
− 3x = − 15
−3 −3
x=5
7 − 5x = 17
−7
−7
− 5x = 10
−5 −5
x=−2
− 2 + 9x = 7
+2
+2
9x = 9
9
9
x=1
− 5 − 3x = − 5
+5
+5
− 3x = 0
−3 −3
x=0
8 = 2x + 10
− 10
− 10
− 2 = 2x
2
2
−1=x
x
−3= 5 −4
+4
+4
x
(5)(1) = 5 (5)
5=x
Table 5
...
In fact, as we solve problems like those in the next example,
each one of them will have several steps to solve, but the last two steps are a twostep equation like we are solving here
...
3x2 + 4 − x + 6
1
1 1
+ =
x−8 x 3
√
5x − 5 + 1 = x
log5(2x − 4) = 1
World View Note: Persian mathematician Omar Khayyam would solve algebraic problems geometrically by intersecting graphs rather than solving them
algebraically
...
2 Practice - Two-Step Problems
Solve each equation
...
3
Solving Linear Equations - General Equations
Objective: Solve general linear equations with variables on both sides
...
This section will focus on manipulating an equation we are asked to solve in such a way that we can use our pattern for solving two-step equations to ultimately arrive at the solution
...
Often the parenthesis
can get in the way of solving an otherwise easy problem
...
This is
shown in the following example
...
Example 59
...
Example 2 shows distributing to clear the parenthesis and then combining
like terms next
...
Once we have done this, our next example solves just like any other two-step
equation
...
3(2x − 4) + 9 = 15
6x − 12 + 9 = 15
6x − 3 = 15
+3 +3
6x = 18
Distribute the 3 through the parenthesis
Combine like terms, − 12 + 9
Focus on the subtraction first
Add 3 to both sides
Now focus on multiply by 6
37
6
6
x=3
Divide both sides by 6
Our Solution
A second type of problem that becomes a two-step equation after a bit of work is
one where we see the variable on both sides
...
Example 61
...
This can
make it difficult to decide which side to work with
...
It
doesn’t matter which term gets moved, 4x or 2x, however, it would be the
author’s suggestion to move the smaller term (to avoid negative coefficients)
...
4x − 6 = 2x + 10
− 2x − 2x
2x − 6 = 10
+6 +6
2x = 16
2 2
x=8
Notice the variable on both sides
Subtract 2x from both sides
Focus on the subtraction first
Add 6 to both sides
Focus on the multiplication by 2
Divide both sides by 2
Our Solution!
The previous example shows the check on this solution
...
Example 62
...
Notice
first the smaller term with the variable is moved to the other side, this time by
adding because the coefficient is negative
...
− 3x + 9 = 6x − 27
+ 3x + 3x
9 = 9x − 27
+ 27
+ 27
36 = 9x
9 9
4=x
Notice the variable on both sides, − 3x is smaller
Add 3x to both sides
Focus on the subtraction by 27
Add 27 to both sides
Focus on the mutiplication by 9
Divide both sides by 9
Our Solution
Linear equations can become particularly intersting when the two processes are
combined
...
Notice in each of the following examples we distribute, then combine like
terms, then move the variable to one side of the equation
...
2(x − 5) + 3x = x + 18
2x − 10 + 3x = x + 18
5x − 10 = x + 18
−x
−x
4x − 10 = 18
+ 10 + 10
4x = 28
4
4
x=7
Distribute the 2 through parenthesis
Combine like terms 2x + 3x
Notice the variable is on both sides
Subtract x from both sides
Focus on the subtraction of 10
Add 10 to both sides
Focus on multiplication by 4
Divide both sides by 4
Our Solution
Sometimes we may have to distribute more than once to clear several parenthesis
...
3(4x − 5) − 4(2x + 1) = 5
12x − 15 − 8x − 4 = 5
4x − 19 = 5
+ 19 + 19
4x = 24
Distribute 3 and − 4 through parenthesis
Combine like terms 12x − 8x and − 15 − 4
Focus on subtraction of 19
Add 19 to both sides
Focus on multiplication by 4
39
4
4
x=6
Divide both sides by 4
Our Solution
This leads to a 5-step process to solve any linear equation
...
1
...
2
...
3
...
Solve the remaining 2-step equation (add or subtract then multiply or
divide)
5
...
The order of these steps is very important
...
Example 66
...
Multiply inside parenthesis
Finish parentesis on left, multiply on right
40
4[ − 2] + 9 = − 15 + 8(2)
− 8 + 9 = − 15 + 16
1=1
Finish multiplication on both sides
Add
True!
When we check our solution of x = 2 we found a true statement, 1 = 1
...
There are two special cases that can come up as we are solving these linear equations
...
Notice we start by distributing and moving the variables all to the same side
...
3(2x − 5) = 6x − 15
6x − 15 = 6x − 15
− 6x
− 6x
− 15 = − 15
Distribute 3 through parenthesis
Notice the variable on both sides
Subtract 6x from both sides
Variable is gone! True!
Here the variable subtracted out completely! We are left with a true statement,
− 15 = − 15
...
Thus, for our solution we say all real numbers or R
...
2(3x − 5) − 4x = 2x + 7
6x − 10 − 4x = 2x + 7
2x − 10 = 2x + 7
− 2x
− 2x
− 10 7
Distribute 2 through parenthesis
Combine like terms 6x − 4x
Notice the variable is on both sides
Subtract 2x from both sides
Variable is gone! False!
Again, the variable subtracted out completely! However, this time we are left with
a false statement, this indicates that the equation is never true, no matter what x
is
...
41
1
...
1) 2 − ( − 3a − 8) = 1
2) 2( − 3n + 8) = − 20
5) 66 = 6(6 + 5x)
6) 32 = 2 − 5( − 4n + 6)
3) − 5( − 4 + 2v) = − 50
7) 0 = − 8(p − 5)
4) 2 − 8( − 4 + 3x) = 34
8) − 55 = 8 + 7(k − 5)
9) − 2 + 2(8x − 7) = − 16
10) − (3 − 5n) = 12
13) − 1 − 7m = − 8m + 7
14) 56p − 48 = 6p + 2
11) − 21x + 12 = − 6 − 3x
12) − 3n − 27 = − 27 − 3n
15) 1 − 12r = 29 − 8r
16) 4 + 3x = − 12x + 4
19) − 32 − 24v = 34 − 2v
20) 17 − 2x = 35 − 8x
17) 20 − 7b = − 12b + 30
18) − 16n + 12 = 39 − 7n
21) − 2 − 5(2 − 4m) = 33 + 5m
22) − 25 − 7x = 6(2x − 1)
25) − 6v − 29 = − 4v − 5(v + 1)
26) − 8(8r − 2) = 3r + 16
23) − 4n + 11 = 2(1 − 8n) + 3n
27) 2(4x − 4) = − 20 − 4x
29) − a − 5(8a − 1) = 39 − 7a
24) − 7(1 + b) = − 5 − 5b
28) − 8n − 19 = − 2(8n − 3) + 3n
30) − 4 + 4k = 4(8k − 8)
31) − 57 = − ( − p + 1) + 2(6 + 8p)
32) 16 = − 5(1 − 6x) + 3(6x + 7)
35) 50 = 8 (7 + 7r) − (4r + 6)
36) − 8(6 + 6x) + 4( − 3 + 6x) = − 12
33) − 2(m − 2) + 7(m − 8) = − 67
37) − 8(n − 7) + 3(3n − 3) = 41
39) − 61 = − 5(5r − 4) + 4(3r − 4)
34) 7 = 4(n − 7) + 5(7n + 7)
38) − 76 = 5(1 + 3b) + 3(3b − 3)
40) − 6(x − 8) − 4(x − 2) = − 4
41) − 2(8n − 4) = 8(1 − n)
42) − 4(1 + a) = 2a − 8(5 + 3a)
45) − 7(x − 2) = − 4 − 6(x − 1)
46) − (n + 8) + n = − 8n + 2(4n − 4)
49) − 2(1 − 7p) = 8(p − 7)
50) 8( − 8n + 4) = 4( − 7n + 8)
43) − 3( − 7v + 3) + 8v = 5v − 4(1 − 6v)
47) − 6(8k + 4) = − 8(6k + 3) − 2
44) − 6(x − 3) + 5 = − 2 − 5(x − 5)
48) − 5(x + 7) = 4( − 8x − 2)
42
1
...
Often when solving linear equations we will need to work with an equation with
fraction coefficients
...
This is
demonstrated in our next example
...
3
7 5
x− =
4
2 6
+
Focus on subtraction
7
7
+
2
2
Add
7
to both sides
2
Notice we will need to get a common denominator to add
5
6
7
+ 2
...
So we build up the denominator,
can now add the fractions:
3
21 5
x−
=
4
6
6
+
21 21
+
6
6
7
2
3
3
=
21
,
6
and we
Same problem, with common denominator 6
Add
21
to both sides
6
26
3
x=
4
6
Reduce
26 13
to
6
3
3
13
x=
4
3
Focus on multiplication by
3
3
4
3
We can get rid of 4 by dividing both sides by 4
...
13 4
4 3
x=
3 3
3 4
52
x=
9
Multiply by reciprocal
Our solution!
While this process does help us arrive at the correct solution, the fractions can
make the process quite difficult
...
Clearing fractions is nice as it gets rid
of the fractions for the majority of the problem
...
This is shown in the
next example, the same problem as our first example, but this time we will solve
by clearing fractions
...
3
7 5
x− =
4
2 6
(12)3
(12)7 (12)5
x−
=
4
2
6
(3)3x − (6)7 = (2)5
9x − 42 = 10
+ 42 + 42
9x = 52
9
9
52
x=
9
LCD = 12, multiply each term by 12
Reduce each 12 with denominators
Multiply out each term
Focus on subtraction by 42
Add 42 to both sides
Focus on multiplication by 9
Divide both sides by 9
Our Solution
The next example illustrates this as well
...
Example 71
...
We will first
distribute the coefficient in front of the parenthesis, then clear the fractions
...
44
Example 72
...
15 15
10
x=
3
Distribute
3
through parenthesis, reducing if possible
2
LCD = 18, multiply each term by 18
Reduce 18 with each denominator
Multiply out each term
Focus on addition of 4
Subtract 4 from both sides
Focus on multiplication by 15
Divide both sides by 15
...
Our Solution
While the problem can take many different forms, the pattern to clear the fraction is the same, after distributing through any parentheses we multiply each term
by the LCD and reduce
...
The following example again illustrates this process
...
3
1 1 3
7
x − = ( x + 6) −
4
2 3 4
2
1 1
7
3
x− = x+2−
2 4
2
4
(4)1 (4)1
(4)2 (4)7
(4)3
x−
=
x+
−
2
4
1
2
4
(1)3x − (2)1 = (1)1x + (4)2 − (2)7
3x − 2 = x + 8 − 14
3x − 2 = x − 6
−x
−x
2x − 2 = − 6
+2 +2
2x = − 4
2
2
x=−2
1
Distribute , reduce if possible
3
LCD = 4, multiply each term by 4
...
The most famous mathematical document from Ancient Egypt
is the Rhind Papyrus where the unknown variable was called “heap”
45
1
...
3
1
21
1) 5 (1 + p) = 20
5
3
4
7)
635
72
5
− 4m =
5
9
9) 2b + 5 = −
11)
8
11
4
+ 4r =
163
32
16
9
4 5
6)
11
4
8) −
+ x)
3
10)
3
2
12)
11
5
3 7
3
( n + 1) = 2
2 3
41
9
= − 3 ( 3 + n)
7
17)
16
9
=−
5 3
5
( p − 3)
2 2
=−
4
4
4
( − 3n − 3)
3
5
15)
55
6
5
5
11
3
3
5
7
3
5
3
23) − ( − 2 x − 2 ) = − 2 + x
25)
45
16
3
7
19
+ 2 n = 4 n − 16
3
3
7
27) 2 (v + 2 ) = − 4 v −
29)
47
9
3
1 2
3
9
18) 3 (m + 4 ) −
20)
1
12
=−
7
13
8
83
7
6
4
10
3
53
= − 18
5
7
= 3 x + 3 (x − 4 )
4
3
3
− 3 n = − 2 n + 2(n + 2 )
24) −
149
16
−
7 5
11
7
r= − 4 r
3
1
26) − 2 ( 3 a + 3 ) =
8
19
6
10
k
3
16) − 2 ( 3 x − 4 ) − 2 x = − 24
5
+ 2 b = 2 (b − 3 )
1
14) 3 ( − 4 k + 1) −
22)
3
2
= 2 (x + 3 ) − 3 x
2
19) − 8 = 4 (r − 2 )
21) −
9
− 4v = − 8
1
5
8
19
( − 3 a + 1) = − 4
4
13) − a −
29
4) 2 n − 3 = − 12
113
24
= − 2( −
3
3
6
3) 0 = − 4 (x − 5 )
5)
3
2) − 2 = 2 k + 2
1
5
11
25
a+ 8
4
4
2
28) − 3 − 2 x = − 3 x − 3 ( −
1
30) 3 n +
5 5
+ 2 x = 3 ( 2 x + 1)
46
29
6
4
4
− 4 ( − 3 r + 1)
2
= 2( 3 n + 3 )
13
x + 1)
4
1
...
Solving formulas is much like solving general linear equations
...
For example, we may have a formula such as A = πr 2 +
πrs (formula for surface area of a right circular cone) and we may be interested in
solving for the varaible s
...
So a solution might look like s =
A − πr 2
...
In this section we will discuss how we can
move from the first equation to the second
...
This is shown in
the following example
...
3x = 12
3 3
x=4
wx = z In both problems, x is multiplied by something
w w To isolate the x we divide by 3 or w
...
Because we are solving for x we treat all the other variables the same way we
would treat numbers
...
This
same idea is seen in the following example
...
m + n = p for n Solving for n, treat all other variables like numbers
−m −m
Subtract m from both sides
n= p−m
Our Solution
As p and m are not like terms, they cannot be combined
...
This same one-step process can be used with grouping
symbols
...
a(x − y) = b
for a Solving for a, treat (x − y) like a number
(x − y) (x − y)
Divide both sides by (x − y)
a=
b
x−y
Our Solution
Because (x − y) is in parenthesis, if we are not searching for what is inside the
parenthesis, we can keep them together as a group and divide by that group
...
The following example is the same formula, but this time we will solve for x
...
a(x − y) = b for x
ax − ay = b
+ ay + ay
ax = b + ay
a
a
b + ay
x=
a
Solving for x, we need to distribute to clear parenthesis
This is a two − step equation, ay is subtracted from our x term
Add ay to both sides
The x is multipied by a
Divide both sides by a
Our Solution
Be very careful as we isolate x that we do not try and cancel the a on top and
bottom of the fraction
...
There is no reducing possible in this problem, so our final reduced
b + ay
answer remains x = a
...
y = mx + b for m
−b
−b
y − b = mx
x
x
y −b
=m
x
Solving for m, focus on addition first
Subtract b from both sides
m is multipied by x
...
The next example is also a two-step equation, it is the problem we started with at
the beginning of the lesson
...
A = πr 2 + πrs for s
− πr 2 − πr 2
A − πr 2 = πrs
πr
πr
2
A − πr
=s
πr
Solving for s, focus on what is added to the term with s
Subtract πr2 from both sides
s is multipied by πr
Divide both sides by πr
Our Solution
Again, we cannot reduce the πr in the numerator and denominator because of the
subtraction in the problem
...
First identify the LCD and then multiply each
term by the LCD
...
Example 80
...
Example 81
...
These represent different values and we
must be careful not to combine a capital variable with a lower case variable
...
a=
A
for b Use LCD (2 − b) as a group
2−b
49
(2 − b)a =
(2 − b)A
2−b
(2 − b)a = A
2a − ab = A
− 2a
− 2a
− ab = A − 2a
−a
−a
A − 2a
b=
−a
Multiply each term by (2 − b)
reduce (2 − b) with denominator
Distribute through parenthesis
Subtract 2a from both sides
The b is multipied by − a
Divide both sides by − a
Our Solution
Notice the A and a were not combined as like terms
...
Often with formulas there is more than one way to solve for a variable
...
After clearing the
denominator, we divide by a to move it to the other side, rather than distributing
...
a=
A
for b Use LCD = (2 − b) as a group
2−b
(2 − b)A
2−b
(2 − b)a = A
a
a
A
2−b=
a
−2
−2
A
−b= −2
a
A
( − 1)( − b) = ( − 1) − 2( − 1)
a
A
b=− +2
a
(2 − b)a =
Multiply each term by (2 − b)
Reduce (2 − b) with denominator
Divide both sides by a
Focus on the positive 2
Subtract 2 from both sides
Still need to clear the negative
Multiply (or divide) each term by − 1
Our Solution
Both answers to the last two examples are correct, they are just written in a different form because we solved them in different ways
...
World View Note: The father of algebra, Persian mathematician Muhammad
ibn Musa Khwarizmi, introduced the fundamental idea of blancing by subtracting
the same term to the other side of the equation
...
50
1
...
2) g =
f
3) g x = b for x
h
i
for h
4) p =
1) ab = c for b
3y
q
for y
a
5) 3x = b for x
6)
7) E = mc2 for m
8) DS = ds for D
9) V =
4
πr3
3
for π
13) c =
4y
m+n
for y
15) V =
πDn
12
for D
17) P = n(p − c) for n
D−d
L
c
= d for y
10) E =
11) a + c = b for c
19) T =
ym
b
for D
mv2
2
for m
12) x − f = g for x
14)
rs
a−3
= k for r
16) F = k(R − L) for k
18) S = L + 2B for L
20) I =
Ea − E q
R
for Ea
21) L = Lo(1 + at) for Lo
22) ax + b = c for x
23) 2m + p = 4m + q for m
24) q = 6(L − p) for L
25)
k−m
r
26) R = aT + b for T
= q for k
27) h = vt − 16t2 for v
29) Q1 = P (Q2 − Q1) for Q2
31) R =
kA(T1 + T2)
d
for T1
28) S = πrh + πr 2 for h
30) L = π(r1 + r2) + 2d for r1
32) P =
V1(V2 − V1)
g
33) ax + b = c for a
34) rt = d for r
35) lwh = V for w
36) V =
for V2
37)
1
a
c
+ b = a for a
39) at − bw = s for t
38)
1
a
πr 2h
3
for h
c
+ b = a for b
40) at − bw = s for w
41) ax + bx = c for a
42) x + 5y = 3 for x
43) x + 5y = 3 for y
44) 3x + 2y = 7 for x
45) 3x + 2y = 7 for y
46) 5a − 7b = 4 for a
47) 5a − 7b = 4 for b
48) 4x − 5y = 8 for x
49) 4x − 5y = 8 for y
50) C = 9 (F − 32) for F
5
51
1
...
When solving equations with absolute value we can end up with more than one
possible answer
...
This is illustrated in the following example
...
|x| = 7
x = 7 or x = − 7
Absolute value can be positive or negative
Our Solution
Notice that we have considered two possibilities, both the positive and negative
...
World View Note: The first set of rules for working with negatives came from
7th century India
...
”
When we have absolute values in our problem it is important to first isolate the
absolute value, then remove the absolute value by considering both the positive
and negative solutions
...
Example 85
...
− 4|x| = − 20
−4
−4
Notice absolute value is not alone
Divide both sides by − 4
52
|x| = 5
x = 5 or x = − 5
Absolute value can be positive or negative
Our Solution
Notice we never combine what is inside the absolute value with what is outside
the absolute value
...
The next example requires two steps to isolate the absolute value
...
Example 87
...
Often the absolute value will have more than
just a variable in it
...
This is shown in the next
example
...
|2x − 1| = 7
2x − 1 = 7 or 2x − 1 = − 7
Absolute value can be positive or negative
Two equations to solve
Now notice we have two equations to solve, each equation will give us a different
solution
...
2x − 1 = 7
+1+1
2x = 8
2
2
x=4
2x − 1 = − 7
+1 +1
2x = − 6
2
2
x=−3
or
53
Thus, from our previous example we have two solutions, x = 4 or x = − 3
...
This is illustrated in
below
...
2 − 4|2x + 3| = − 18
To get the absolute value alone we first need to get rid of the 2 by subtracting,
then divide by − 4
...
Also notice we do
not distribute the − 4 into the absolute value
...
Thus we get
the absolute value alone in the following way:
2 − 4|2x + 3| = − 18
−2
−2
− 4|2x + 3| = − 20
−4
−4
|2x + 3| = 5
2x + 3 = 5 or 2x + 3 = − 5
Notice absolute value is not alone
Subtract 2 from both sides
Absolute value still not alone
Divide both sides by − 4
Absoloute value can be positive or negative
Two equations to solve
Now we just solve these two remaining equations to find our solutions
...
As we are solving absolute value equations it is important to be aware of special
cases
...
Notice
what happens in the next example
...
7 + |2x − 5| = 4
−7
−7
|2x − 5| = − 3
Notice absolute value is not alone
Subtract 7 from both sides
Result of absolute value is negative!
Notice the absolute value equals a negative number! This is impossible with absolute value
...
One other type of absolute value problem is when two absolute values are equal to
eachother
...
Example 91
...
The second equation considers
the negative possibility
...
So
we solve both these equations as follows:
2x − 7 = 4x + 6
− 2x − 2x
− 7 = 2x + 6
−6
−6
− 13 = 2x
2
2
− 13
=x
2
This gives us our two solutions, x =
2x − 7 = − (4x + 6)
2x − 7 = − 4x − 6
+ 4x + 4x
6x − 7 = − 6
+7 +7
6x = 1
6
6
1
x=
6
or
− 13
2
1
or x = 6
...
6 Practice - Absolute Value Equations
Solve each equation
...
7
Solving Linear Equations - Variation
Objective: Solve variation problems by creating variation equations and
finding the variation constant
...
Often different events are
related by what is called the constant of variation
...
The faster you
travel, the less time it take to get there
...
Variation problems have two or three
variables and a constant in them
...
There are two ways to set up a variation problem, the first solves for one of the
variables, a second method is to solve for the constant
...
The greek letter pi (π) is used to represent the ratio of the circumference of a
circle to its diameter
...
1415926)
...
14159
...
This relationship
variation or directly proportional
...
by the diameter
a bigger circumis called direct
the problem we
Example 92
...
If
you multiply the force required to break a board by the length of the board you
will also get a constant
...
This relationship is called
indirect variation or inversly proportional
...
Example 93
...
If we divide the
1
area by the base times the height we will also get a constant, 2
...
If we see this phrase in the
problem we know to divide the first variable by the product of the other two to
find the constant of variation
...
′′
Jointly ′′ tells us to divide by the product
A varies jointly as x and y
A
=k
xy
Our formula for the relationship
Once we have our formula for the relationship in a variation problem, we use
given or known information to calculate the constant of variation
...
Example 95
...
c varies indirectly as d and c = 4
...
5)(6) = k
27 = k
′′
indirectly ′′ tells us to multiply
Substitute known values
Evaluate to find our constant
Example 97
...
Each of these problems we solve will have three
important steps, none of which should be skipped
...
Find the formula for the relationship using the type of variation
2
...
Answer the question using the constant of variation
The next three examples show how this process is worked out for each type of
variation
...
The price of an item varies directly with the sales tax
...
5 = k
40
= 12
...
5(t)
t
40 = 12
...
5 12
...
2 = t
Multiply by LCD = t to clear fraction
Reduce the t with the denominator
Divide by 12
...
20
Example 99
...
If he travels 35 miles per hour it will take him 2
...
How long will it take him if he travels 55 miles per hour?
rt = k
(35)(2
...
5 = k
55t = 87
...
59
′′
Inversely ′′ tells us to multiply the rate and time
Substitute known values for rate and time
Evaluate to find our constant
Using our constant, substitute 55 for rate to find the time
Divide both sides by 55
Our solution: It takes him 1
...
59
The amount of simple interest earned on an investment varies jointly as the principle (amount invested) and the time it is invested
...
How much interest would be earned on a S220
investment for 3 years?
I
=k
Pt
(12)
=k
(150)(2)
0
...
04
(220)(3)
I
= 0
...
04(660)
660
I = 26
...
40 in interest
Sometimes a variation problem will ask us to do something to a variable as we set
up the formula for the relationship
...
This is still direct variation, we say the area
varies directly as the radius square and thus our variable is squared in our formula
...
Example 101
...
A circle
with a radius of 10 has an area of 314
...
14 = k
A
= 3
...
14
16
(16)A
= 3
...
24
′′
Direct ′′ tells us to divide, be sure we use r 2 for the denominator
Substitute known values into our formula
Exponents first
Divide to find our constant
Using the constant, use 4 for r, don ′t forget the squared!
Evaluate the exponent
Multiply both sides by 16
Our Solution: Area is 50
...
60
1
...
c varies directly as a
2
...
w varies inversely as x
4
...
f varies jointly as x and y
6
...
h is directly proportional to b
8
...
a is inversely proportional to b
Find the constant of variation and write the formula to express the
relationship using that constant
10
...
p is jointly proportional to q and r and p = 12 when q = 8 and r = 3
12
...
t varies directly as the square of u and t = 6 when u = 3
14
...
w is inversely proportional to the cube of x and w is 54 when x = 3
16
...
a is jointly proportional with the square of x and the square root of y and
a = 25 when x = 5 and y = 9
18
...
8 when n = 2
...
19
...
When 15 volts are applied, the current is 5 amperes
...
The current in an electrical conductor varies inversely as the resistance of the
conductor
...
Hooke’s law states that the distance that a spring is stretched by hanging
object varies directly as the mass of the object
...
The volume of a gas varies inversely as the pressure upon it
...
What will be its volume under
a pressure of 40 kg/cm2?
23
...
If 250 people use 60,000 cans in one year, how many
cans are used each year in Dallas, which has a population of 1,008,000?
24
...
It takes 5hr for 7 bricklayers to build a park well
...
According to Fidelity Investment Vision Magazine, the average weekly
allowance of children varies directly as their grade level
...
66 dollars per week
...
The wavelength of a radio wave varies inversely as its frequency
...
What is the length
of a wave with a frequency of 800 kilohertz?
27
...
A 96-kg person contains 64 kg of water
...
The time required to drive a fixed distance varies inversely as the speed
...
How long will it
take to drive the same distance at a speed of 70 km/h?
29
...
A
person weighs 95lb on Earth weighs 38 lb on Mars
...
At a constant temperature, the volume of a gas varies inversely as the pres62
sure
...
The time required to empty a tank varies inversely as the rate of pumping
...
The weight of an object varies inversely as the square of the distance from the
center of the earth
...
How far above the earth must the astronaut be in
order to weigh 64 lb?
33
...
If a car, traveling 60 mph can stop in 200
ft, how fast can a car go and still stop in 72 ft?
34
...
If a boat going 6
...
2 ft2, how fast must a boat
with 28
...
The intensity of a light from a light bulb varies inversely as the square of the
distance from the bulb
...
How much further would it be to a point
where the intesity is 40 W/m2?
36
...
If a cone with a height of 8 centimeters and a radius of 2 centimeters has a
volume of 33
...
The intensity of a television signal varies inversely as the square of the distance from the transmitter
...
56 W/m2?
38
...
The unit for measuring the intesity of illumination is usually the footcandle
...
8
Linear Equations - Number and Geometry
Objective: Solve number and geometry problems by creating and
solving a linear equation
...
Often it takes a bit of practice to convert the
English sentence into a mathematical sentence
...
A few important phrases are described below that can give us clues for how to set
up a problem
...
Example 102
...
What is the number?
5x − 28
5x − 28 = 232
+ 28 + 28
5x = 260
5
5
x = 52
Subtraction is built backwards, multiply the unknown by 5
Is translates to equals
Add 28 to both sides
The variable is multiplied by 5
Divide both sides by 5
The number is 52
...
Example 103
...
What is the number
3x + 15
6x − 10
3x + 15 = 6x − 10
− 3x
− 3x
15 = 3x − 10
+ 10
+ 10
25 = 3x
3
3
25
=x
3
First, addition is built backwards
Then, subtraction is also built backwards
Is between the parts tells us they must be equal
Subtract 3x so variable is all on one side
Now we have a two − step equation
Add 10 to both sides
The variable is multiplied by 3
Divide both sides by 3
25
Our number is
3
Another type of number problem involves consecutive numbers
...
If we are
looking for several consecutive numbers it is important to first identify what they
look like with variables before we set up the equation
...
Example 104
...
What are the integers?
First x
Second x + 1
Third x + 2
F + S + T = 93
(x) + (x + 1) + (x + 2) = 93
x + x + 1 + x + 2 = 93
3x + 3 = 93
−3 −3
3x = 90
3 3
x = 30
First 30
Second (30) + 1 = 31
Third (30) + 2 = 32
Make the first number x
To get the next number we go up one or + 1
Add another 1(2 total) to get the third
First (F ) plus Second (S) plus Third (T ) equals 93
Replace F with x, S with x + 1, and T with x + 2
Here the parenthesis aren ′t needed
...
When we had consecutive integers, we only had to add 1 to get to
the next number so we had x, x + 1, and x + 2 for our first, second, and third
number respectively
...
So
if we want three consecutive even numbers, if the first is x, the next number
would be x + 2, then finally add two more to get the third, x + 4
...
It is important to note that we are still adding 2 and 4 even
when the numbers are odd
...
Consider the next two examples
...
The sum of three consecutive even integers is 246
...
The numbers are 80, 82, and 84
...
Find three consecutive odd integers so that the sum of twice the first, the second
and three times the third is 152
...
The numbers added do not change with odd or even,
it is our answer for x that will be odd or even
...
A well known property of triangles is that all three angles
will always add to 180
...
If you add these together, 50 + 30 + 100 =
180
...
World View Note: German mathematician Bernhart Thibaut in 1809 tried to
prove that the angles of a triangle add to 180 without using Euclid’s parallel postulate (a point of much debate in math history)
...
Example 107
...
The third angle is 40 less than
the first
...
First x
Second 2x
Third x − 40
F + S + T = 180
(x) + (2x) + (x − 40) = 180
x + 2x + x − 40 = 180
4x − 40 = 180
+ 40 + 40
4x = 220
4
4
x = 55
First 55
Second 2(55) = 110
Third (55) − 40 = 15
With nothing given about the first we make that x
The second is double the first,
The third is 40 less than the first
All three angles add to 180
Replace F , S , and T with the labeled values
...
Combine like terms, x + 2x + x
Add 40 to both sides
The variable is multiplied by 4
Divide both sides by 4
Our solution for x
Replace x with 55 in the original list of angles
Our angles are 55, 110, and 15
Another geometry problem involves perimeter or the distance around an object
...
There are
two lengths and two widths in a rectangle (opposite sides) so we add 8 + 8 + 3 +
3 = 22
...
So for the
rectangle of length 8 and width 3 the formula would give, P = 2(8) + 2(3) = 16 +
6 = 22
...
Example 108
...
The length is 5 less than double the width
...
Length x
Width 2x − 5
P = 2L + 2W
(44) = 2(x) + 2(2x − 5)
44 = 2x + 4x − 10
44 = 6x − 10
+ 10
+ 10
54 = 6x
6 6
9=x
Length 9
Width 2(9) − 5 = 13
We will make the length x
Width is five less than two times the length
The formula for perimeter of a rectangle
Replace P , L, and W with labeled values
Distribute through parenthesis
Combine like terms 2x + 4x
Add 10 to both sides
The variable is multiplied by 6
Divide both sides by 6
Our solution for x
Replace x with 9 in the origional list of sides
The dimensions of the rectangle are 9 by 13
...
This is important in all word
problems involving variables, not just consective numbers or geometry problems
...
Example 109
...
The sofa costs double the love seat
...
Replace S and L with labeled values
Parenthesis are not needed, combine like terms x + 2x
Divide both sides by 3
Our solution for x
Replace x with 148 in the origional list
The love seat costs S148 and the sofa costs S296
...
Many students see the phrase “double” and
believe that means we only have to divide the 444 by 2 and get S222 for one or
both of the prices
...
By clearly labeling the variables in the original list we know exactly how to set up and solve these problems
...
8 Practice - Number and Geometry Problems
Solve
...
When five is added to three more than a certain number, the result is 19
...
If five is subtracted from three times a certain number, the result is 10
...
When 18 is subtracted from six times a certain number, the result is − 42
...
A certain number added twice to itself equals 96
...
A number plus itself, plus twice itself, plus 4 times itself, is equal to − 104
...
Sixty more than nine times a number is the same as two less than ten times
the number
...
Eleven less than seven times a number is five more than six times the number
...
8
...
What is the number?
9
...
What are the integers?
10
...
What are the integers?
11
...
12
...
What are the integers?
13
...
What are the integers?
69
14
...
What are the integers?
15
...
16
...
The third
angle is 12 degrees larger than the first angle
...
Two angles of a triangle are the same size
...
Find the measure the angles
...
Two angles of a triangle are the same size
...
How large are the angles?
19
...
The second angle is
4 times the third
...
20
...
The third
angle is 30 degrees more than the first angle
...
21
...
The measure of
the third angle is 20 degrees greater than the first
...
The second angle of a triangle is three times as large as the first
...
How large
are the three angles?
23
...
The measure
of the third angle is 12 degrees greater than that of the first angle
...
The second angle of a triangle is three times the first, and the third is 12
degrees less than twice the first
...
25
...
Find the measures of the angles
...
The perimeter of a rectangle is 150 cm
...
Find the dimensions
...
The perimeter of a rectangle is 304 cm
...
Find the length and width
...
The perimeter of a rectangle is 152 meters
...
Find the length and width
...
The perimeter of a rectangle is 280 meters
...
Find the length and width
...
The perimeter of a college basketball court is 96 meters and the length is 14
meters more than the width
...
A mountain cabin on 1 acre of land costs S30,000
...
A horse and a saddle cost S5000
...
A bicycle and a bicycle helmet cost S240
...
Of 240 stamps that Harry and his sister collected, Harry collected 3 times as
many as his sisters
...
If Mr
...
Brown had 10 times as
much as his son, how much money had each?
36
...
How
many of each were there?
37
...
How
many sheep had each?
38
...
What was the cost of each?
39
...
If Jamal
furnished half as much capital as Moshe, how much did each furnish?
40
...
41
...
How long
are the pieces?
42
...
One piece is 2 ft longer than the
other
...
An electrician cuts a 30 ft piece of wire into two pieces
...
How long are the pieces?
44
...
Tuition costs S704 more than room and board
...
The cost of a private pilot course is S1,275
...
What is the cost of each?
71
1
...
An application of linear equations is what are called age problems
...
Using the clues given in the problem we will be
working to find their current age
...
To help us organize and
solve our problem we will fill out a three by three table for each problem
...
Structure of Age Table
Normally where we see “Person 1” and “Person 2” we will use the name of the
person we are talking about
...
Example 110
...
In two years Brian will be twice as old as
Adam
...
We are given information about
Adam, not Brian
...
To show Adam
is 20 years younger we subtract 20, Adam is x − 20
...
This is done by adding
2 to both Adam ′s and Brian ′s now column as shown
in the table
...
72
B = 2A
(x + 2) = 2(x − 18)
x + 2 = 2x − 36
−x
−x
2 = x − 36
+ 36 + 36
38 = x
Age now
Adam 38 − 20 = 18
Brian
38
Our equation comes from the future statement:
Brian will be twice as old as Adam
...
Replace B and A with the information in their future
cells, Adam (A) is replaced with x − 18 and Brian (B)
is replaced with (x + 2) This is the equation to solve!
Distribute through parenthesis
Subtract x from both sides to get variable on one side
Need to clear the − 36
Add 36 to both sides
Our solution for x
The first column will help us answer the question
...
Adam is 18 and Brian is 38
Solving age problems can be summarized in the following five steps
...
1
...
The person we know nothing about is x
...
Fill in the future/past collumn by adding/subtracting the change to the
now column
...
Make an equation for the relationship in the future
...
4
...
Solve the equation for x, use the solution to answer the question
These five steps can be seen illustrated in the following example
...
Carmen is 12 years older than David
...
How old are they now?
Carmen
David
Carmen
David
Age Now − 5
Age Now − 5
x + 12
x
Age Now
−5
Carmen x + 12 x + 12 − 5
David
x
x−5
Five years ago is − 5 in the change column
...
We don ′t
know about David so he is x, Carmen then is x + 12
Subtract 5 from now column to get the change
73
Age Now − 5
Carmen x + 12 x + 7
David
x
x−5
C + D = 28
(x + 7) + (x − 5) = 28
x + 7 + x − 5 = 28
2x + 2 = 28
−2 −2
2x = 26
2 2
x = 13
Age Now
Caremen 13 + 12 = 25
David
13
Simplify by combining like terms 12 − 5
Our table is ready!
The sum of their ages will be 29
...
Remove parenthesis
Combine like terms x + x and 7 − 5
Subtract 2 from both sides
Notice x is multiplied by 2
Divide both sides by 2
Our solution for x
Replace x with 13 to answer the question
Carmen is 25 and David is 13
Sometimes we are given the sum of their ages right now
...
In this case we will write the sum above the now column and make the
first person’s age now x
...
This is shown in the next example
...
The sum of the ages of Nicole and Kristin is 32
...
How old are they now?
32
Age Now + 2
Nicole
x
Kristen 32 − x
Nicole
Kristen
Age Now
+2
x
x+2
32 − x 32 − x + 2
Age Now + 2
Nicole
x
x+2
Kristen 32 − x 34 − x
N = 3K
(x + 2) = 3(34 − x)
x + 2 = 102 − 3x
+ 3x
+ 3x
The change is + 2 for two years in the future
The total is placed above Age Now
The first person is x
...
Replace variables with information in change cells
Distribute through parenthesis
Add 3x to both sides so variable is only on one side
74
4x + 2 = 102
−2 −2
4x = 100
4
4
x = 25
Age Now
Nicole
25
Kristen 32 − 25 = 7
Solve the two − step equation
Subtract 2 from both sides
The variable is multiplied by 4
Divide both sides by 4
Our solution for x
Plug 25 in for x in the now column
Nicole is 25 and Kristin is 7
A slight variation on age problems is to ask not how old the people are, but
rather ask how long until we have some relationship about their ages
...
In the change column because we don’t know the time
to add or subtract we will use a variable, t, and add or subtract this from the now
column
...
Example 113
...
Her daughter is 4 years old
...
The change is unknown, so we write + t for
the change
Age Now + t
Louis
26
26 + t
Daughter
4
4+t
Fill in the change column by adding t to each person ′s
age
...
L = 2D
(26 + t) = 2(4 + t)
26 + t = 8 + 2t
−t
−t
26 = 8 + t
−8−8
18 = t
Louis will be double her daughter
Replace variables with information in change cells
Distribute through parenthesis
Subtract t from both sides
Now we have an 8 added to the t
Subtract 8 from both sides
In 18 years she will be double her daughter ′s age
Age problems have several steps to them
...
World View Note: The oldest man in the world was Shigechiyo Izumi from
Japan who lived to be 120 years, 237 days
...
75
1
...
A boy is 10 years older than his brother
...
Find the present age of each
...
A father is 4 times as old as his son
...
Find the present age of each
...
Pat is 20 years older than his son James
...
How old are they now?
4
...
In 6 years Diane will be twice
as old as Amy
...
Fred is 4 years older than Barney
...
How old are they now?
6
...
Five years ago the sum of their ages was
50
...
Tim is 5 years older than JoAnn
...
How old are they now?
8
...
In three years the sum of their ages will be 54
...
The sum of the ages of John and Mary is 32
...
Find the present age of each
...
The sum of the ages of a father and son is 56
...
Find the present age of each
...
The sum of the ages of a china plate and a glass plate is 16 years
...
Find the
present age of each plate
...
The sum of the ages of a wood plaque and a bronze plaque is 20 years
...
Find
the present age of each plaque
...
A is now 34 years old, and B is 4 years old
...
A man’s age is 36 and that of his daughter is 3 years
...
An Oriental rug is 52 years old and a Persian rug is 16 years old
...
A log cabin quilt is 24 years old and a friendship quilt is 6 years old
...
The age of the older of two boys is twice that of the younger; 5 years ago it
was three times that of the younger
...
18
...
How many years ago was
the pitcher twice as old as the vase?
19
...
The sum of their ages seven years ago was
13
...
The sum of Jason and Mandy’s age is 35
...
How old are they now?
21
...
In 6 years, the silver coin
will be twice as old as the bronze coin
...
22
...
In how many years will the
table be twice as old as the sofa?
23
...
In 12 years, the
limestone will be three times as old as the marble statue
...
24
...
In how many
years will the silver bowl be twice the age of the pewter bowl?
25
...
In four years the sum of their ages will
be 91
...
A kerosene lamp is 95 years old, and an electric lamp is 55 years old
...
A father is three times as old as his son, and his daughter is 3 years younger
77
than the son
...
28
...
In four years, Wendy will be three
times as old as Clyde
...
The sum of the ages of two ships is 12 years
...
Find the present age of
each ship
...
Chelsea’s age is double Daniel’s age
...
How old are they now?
31
...
One year ago, she was three times
as old as her son
...
The sum of the ages of Kristen and Ben is 32
...
How old are they both now?
33
...
Thirty years ago, the mosaic
was three times as old as the engraving
...
34
...
Four years ago Elli was 3 times as
old as Dan
...
A wool tapestry is 32 years older than a linen tapestry
...
Find the present age of
each
...
Carolyn’s age is triple her daughter’s age
...
How old are they now?
37
...
Emma is 2 years old
...
The sum of the ages of two children is 16 years
...
Find the present age
of each child
...
Mike is 4 years older than Ron
...
How old are they now?
40
...
In how
many years will the terra-cotta bust be three times as old as the marble bust?
78
1
...
An application of linear equations can be found in distance problems
...
For example, if a person were to travel 30 mph for 4 hours
...
This person travel a distance of 120 miles
...
So to keep the information in the
problem organized we will use a table
...
Structure of Distance Problem
The third column, distance, will always be filled in by multiplying the rate and
time columns together
...
We will now use this table to
set up and solve the following example
79
Example 114
...
One jogger is running at a rate of 4 mph, and the other is running at a
rate of 6 mph
...
These are added to the table
We only know they both start and end at the
same time
...
This same process can be seen in the following example
Example 115
...
Bob
walks 2 miles per hour faster than Fred
...
How fast did each walk?
Bob
Fred
Rate Time Distance
3
3
The basic table with given times filled in
Both traveled 3 hours
80
Rate Time Distance
Bob r + 2
3
Fred
r
3
Rate Time Distance
Bob r + 2
3
3r + 6
Fred
r
3
3r
30
3r + 6 + 3r = 30
6r + 6 = 30
−6 −6
6r = 24
6 6
r=4
Rate
Bob 4 + 2 = 6
Fred
4
Bob walks 2 mph faster than Fred
We know nothing about Fred, so use r for his rate
Bob is r + 2, showing 2 mph faster
Distance column is filled in by multiplying rate by
Time
...
Total distance is put under distance
The distance columns is our equation, by adding
Combine like terms 3r + 3r
Subtract 6 from both sides
The variable is multiplied by 6
Divide both sides by 6
Our solution for r
To answer the question completely we plug 4 in for
r in the table
...
One example of this is if we are given a total time, rather than the individual times like we had in the previous example
...
This is shown in
the next example
Example 116
...
They turned around and paddled back upstream at an average
rate of 4 mph
...
After how much time did the campers
turn around downstream?
Rate Time Distance
Down 12
Up
4
Basic table for down and upstream
Given rates are filled in
1
Rate Time Distance
Down 12
t
Up
4 1−t
Total time is put above time column
As we have the total time, in the first time we have
t, the second time becomes the subtraction,
total − t
81
Rate Time Distance
=
Down 12
t
12t
Up
4
1 − t 4 − 4t
12t = 4 − 4t
+ 4t
+ 4t
16t = 4
16 16
1
t=
4
Distance column is found by multiplying rate
by time
...
As they cover the same distance,
= is put after the down distance
With equal sign, distance colum is equation
Add 4t to both sides so variable is only on one side
Variable is multiplied by 16
Divide both sides by 16
1
Our solution, turn around after hr (15 min )
4
Another type of a distance problem where we do some work is when one person
catches up with another
...
Our startegy for this problem will be to use t
for the faster person’s time, and add amount of time the head start was to get the
slower person’s time
...
Example 117
...
Joy leaves 6 hours later to catch
up with him traveling 8 miles per hour
...
Be sure to distribute the 2(t + 6) for Mike
As they cover the same distance, = is put after
Mike ′s distance
Now the distance column is the equation
Subtract 2t from both sides
The variable is multiplied by 6
Divide both sides by 6
82
2=t
Our solution for t, she catches him after 2 hours
World View Note: The 10,000 race is the longest standard track event
...
2 miles
...
53 second
...
7 miles per hour!
As these example have shown, using the table can help keep all the given information organized, help fill in the cells, and help find the equation we will solve
...
Example 118
...
The trip took 2
...
For
how long did the car travel 40 mph?
Rate Time Distance
Fast 55
Slow 40
2
...
5 − t
2
...
5 − t 100 − 40t
130
55t + 100 − 40t = 130
15t + 100 = 130
− 100 − 100
15t = 30
15 15
t=2
Time
Fast
2
Slow 2
...
5
Basic table for fast and slow speeds
The given rates are filled in
Total time is put above the time column
As we have total time, the first time we have t
The second time is the subtraction problem
2
...
Be sure to distribute 40(2
...
To answer the question we plug 2 in for t
The car traveled 40 mph for 0
...
10 Practice - Distance, Rate, and Time Problems
1
...
An automobile at A starts for B at the rate of 20 miles
an hour at the same time that an automobile at B starts for A at the rate of
25 miles an hour
...
Two automobiles are 276 miles apart and start at the same time to travel
toward each other
...
If they
meet after 6 hours, find the rate of each
...
Two trains travel toward each other from points which are 195 miles apart
...
If they start at the
same time, how soon will they meet?
4
...
If A went at the rate of 20 miles an hour, at what rate must B travel if they
meet in 5 hours?
5
...
If the rate of the passenger train exceeds the rate
of the freight train by 15 miles per hour, and they meet after 4 hours, what
must the rate of each be?
6
...
Their rates were 25 and 35 miles per hour respectively
...
A man having ten hours at his disposal made an excursion, riding out at the
rate of 10 miles an hour and returning on foot, at the rate of 3 miles an hour
...
8
...
How far can he walk into the
country and ride back on a trolley that travels at the rate of 20 miles per hour,
if he must be back home 3 hours from the time he started?
9
...
The round trip requires 2 hours
...
A motorboat leaves a harbor and travels at an average speed of 15 mph
toward an island
...
How far
was the island from the harbor if the total trip took 5 hours?
11
...
Find the distance to the
resort if the total driving time was 8 hours
...
As part of his flight trainging, a student pilot was required to fly to an airport
and then return
...
Find the distance between the two
airports if the total flying time was 7 hours
...
A, who travels 4 miles an hour starts from a certain place 2 hours in advance
of B, who travels 5 miles an hour in the same direction
...
A man travels 5 miles an hour
...
When will the
second man overtake the first?
15
...
Two hours later a cabin cruiser leaves the same harbor and
travels at an average speed of 16 mph toward the same island
...
A long distance runner started on a course running at an average speed of 6
mph
...
How long after the second runner started will the second
runner overtake the first runner?
17
...
How far from the starting point does the car overtake
the cyclist?
18
...
The propeller-driven plane is traveling at 200 mph
...
Two men are traveling in opposite directions at the rate of 20 and 30 miles an
hour at the same time and from the same place
...
Running at an average rate of 8 m/s, a sprinter ran to the end of a track and
then jogged back to the starting point at an average rate of 3 m/s
...
Find the
length of the track
...
A motorboat leaves a harbor and travels at an average speed of 18 mph to an
island
...
How far was the
island from the harbor if the total trip took 5 h?
22
...
Two hours later a cabin cruiser leaves the same harbor and
travels at an average speed of 18 mph toward the same island
...
A jet plane traveling at 570 mph overtakes a propeller-driven plane that has
had a 2 h head start
...
How far from the starting point does the jet overtake the propeller-driven
plane?
24
...
If the rate of one is 6 miles per hour more than the rate of the
other and they are 168 miles apart at the end of 4 hours, what is the rate of
each?
25
...
The average speed on the way to the airport was 100 mph, and
the average speed returning was 150 mph
...
26
...
One
cyclist rides twice as fast as the other
...
Find the rate of each cyclist
...
A car traveling at 56 mph overtakes a cyclist who, riding at 14 mph, has had
a 3 h head start
...
Two small planes start from the same point and fly in opposite directions
...
In two hours
the planes are 430 miles apart
...
29
...
If the car had a 1 h head start, how far from the starting point does
the bus overtake the car?
30
...
The first plane is flying 25 mph slower than the second plane
...
Find the rate of each plane
...
A truck leaves a depot at 11 A
...
and travels at a speed of 45 mph
...
At what time does the van overtake the truck?
32
...
Find the distance to the
resort if the total driving time was 13 h
...
Three campers left their campsite by canoe and paddled downstream at an
average rate of 10 mph
...
How long
did it take the campers to canoe downstream if the total trip took 1 hr?
34
...
The motorcycle was being driven at 45 mph, and the rider walks at a
speed of 6 mph
...
How far did the motorcycle go before if broke
down?
35
...
The student averages 5 km/hr
walking and 9 km/hr jogging
...
How far does the student jog?
36
...
The trip took a
total of 2
...
For how long did the car travel at 40 mph?
37
...
The trip
took a total of 5 h
...
An executive drove from home at an average speed of 40 mph to an airport
where a helicopter was waiting
...
The entire
distance was 150 mi
...
Find the distance from the
airport to the corporate offices
...
1 Points and Lines
...
2 Slope
...
3 Slope-Intercept Form
...
4 Point-Slope Form
...
5 Parallel and Perpendicular Lines
...
1
Graphing - Points and Lines
Objective: Graph points and lines using xy coordinates
...
A graph is simply a picture of the solutions to an equation
...
Following
is an example of what is called the coordinate plane
...
Where the two lines meet in the center
is called the origin
...
As we move
to the right the numbers count up
from zero, representing x = 1, 2, 3
...
Similarly, as we move up the number count up from zero, y = 1, 2, 3
...
We can put dots on the
graph which we will call points
...
The first number will be the value on the x − axis or horizontal number line
...
The second
number will represent the value on the y − axis or vertical number line
...
The points are given as an
ordered pair (x, y)
...
This “origin is just off the western coast of Africa
...
Example 119
...
A
B
C
89
Tracing from the origin, point A is
right 1, up 4
...
Point B is left 5, up 3
...
C is straight down 2 units
...
This means we go
right zero so the point is C(0, − 2)
...
Example 120
...
Following these instructions, starting
from the origin, we get our point
...
This is also illustrated on the graph
...
A
B
Left 2 Right 3
Down 3
Down 4
D
C
The fourth point, D ( − 2, − 3) is left
2, down 3 (both negative, both move
backwards)
The last three points have zeros in them
...
If there is a zero there is just no movement
...
This is left 3 (negative is backwards), and up zero, right
on the x − axis
...
This is right zero,
and up two, right on the y − axis
...
This point has no
movement
...
90
B
A
F
Our Solution
G
E
D
C
The main purpose of graphs is not to plot random points, but rather to give a
picture of the solutions to an equation
...
We may be interested in what type of solution are possible in this equation
...
We will have to start by
finding possible x and y combinations
...
Example 121
...
Any three can be used
Evaluate each by replacing x with the given value
x = − 1; y = 2( − 1) − 3 = − 2 − 3 = − 5
x = 0; y = 2(0) − 3 = 0 − 3 = − 3
x = 1; y = 2(1) − 3 = 2 − 3 = − 1
These then become the points to graph on our equation
91
nect the dots to make a line
...
Once the point are on the graph, con-
The graph is our solution
What this line tells us is that any point on the line will work in the equation y =
2x − 3
...
If we use
x = 2, we should get y = 1
...
Thus we have the line is a picture of all the solutions for y = 2x − 3
...
Example 122
...
Any three can be used
...
Table becomes points to graph
Graph points and connect dots
Our Solution
93
2
...
D
K
J
G
C
E
1)
I
H
F
B
Plot each point
...
1
3) y = − 4 x − 3
5) y = −
5
x−4
4
7) y = − 4x + 2
3
9) y = 2 x − 5
4
11) y = − 5 x − 3
13) x + 5y = − 15
15) 4x + y = 5
4) y = x − 1
3
6) y = − 5 x + 1
5
8) y = 3 x + 4
10) y = − x − 2
1
12) y = 2 x
14) 8x − y = 5
16) 3x + 4y = 16
18) 7x + 3y = − 12
17) 2x − y = 2
20) 3x + 4y = 8
19) x + y = − 1
22) 9x − y = − 4
21) x − y = − 3
94
2
...
As we graph lines, we will want to be able to identify different properties of the
lines we graph
...
Slope
is a measure of steepness
...
A
1
line with a small slope, such as 10 is very flat
...
A line that goes up from left to right will have a positive
slope and a line that goes down from left to right will have a negative slope
...
For this reason we will describe slope as the fraction run
...
Run would be a horizontal change, or a change in the x-values
...
It turns out that if we have a graph we can draw
vertical and horiztonal lines from one point to another to make what is called a
slope triangle
...
The following
examples show graphs that we find the slope of using this idea
...
To find the slope of this line we will
consider the rise, or verticle change
and the run or horizontal change
...
This is rise − 4, run
−4
6
...
2
Reduce the fraction to get − 3
...
95
Run 3
Rise 6
To find the slope of this line, the rise
is up 6, the run is right 3
...
This fraction reduces to 2
...
2 Our Solution
There are two special lines that have unique slopes that we need to be aware of
...
Example 125
...
This slope becomes
0
3
=
0
...
This line has a rise of 5, but no run
...
This
line,
and all vertical lines, have no slope
...
Remember, slope is a measure of steepness
...
Therefore it has a zero slope
...
It is so steep that there is no number large
enough to express how steep it is
...
We can find the slope of a line through two points without seeing the points on a
graph
...
If the rise is the change in y values,
we can calculate this by subtracting the y values of a point
...
In this way we get the following equation for slope
...
For this reason we often represent the slope with the variable m
...
Slope = m =
rise change in y
y2 − y1
=
=
run change in x x2 − x1
As we subtract the y values and the x values when calculating slope it is important we subtract them in the same order
...
Example 126
...
Find the slope between (4,
6) and (2, − 1)
(x1, y1) and (x2, y2)
−1−6
2−4
−7
m=
−2
7
m=
2
m=
Identify x1, y1, x2, y2
Use slope formula, m =
Simplify
y2 − y1
x2 − x1
Reduce, dividing by − 1
Our Solution
We may come up against a problem that has a zero slope (horiztonal line) or no
slope (vertical line) just as with using the graphs
...
Find the slope between ( − 4, − 1) and ( − 4, − 5)
97
Identify x1, y1, x2, y2
(x1 , y1) and (x2, y2)
− 5 − ( − 1)
− 4 − ( − 4)
−4
m=
0
m = no slope
m=
Use slope formula, m =
y2 − y1
x2 − x1
Simplify
Can ′t divide by zero, undefined
Our Solution
Example 129
...
Zero is an
integer and it has a value, the slope of a flat horizontal line
...
Using the slope formula we can also find missing points if we know what the slope
is
...
Example 130
...
Find the value of x between the points ( − 3, 2) and (x, 6) with slope
y2 − y1
x2 − x1
6−2
2
=
5 x − ( − 3)
2
4
=
5 x+3
2
(x + 3) = 4
5
2
(5) (x + 3) = 4(5)
5
2(x + 3) = 20
2x + 6 = 20
−6 −6
2x = 14
2 2
x=7
m=
2
5
We will plug values into slope formula
Simplify
Multiply both sides by (x + 3)
Multiply by 5 to clear fraction
Simplify
Distribute
Solve
...
2 Practice - Slope
Find the slope of each line
...
11) ( − 2, 10), ( − 2, − 15)
12) (1, 2), ( − 6, − 14)
13) ( − 15, 10), (16, − 7)
14) (13, − 2), (7, 7)
15) (10, 18), ( − 11, − 10)
16) ( − 3, 6), ( − 20, 13)
17) ( − 16, − 14), (11, − 14)
18) (13, 15), (2, 10)
19) ( − 4, 14), ( − 16, 8)
20) (9, − 6), ( − 7, − 7)
21) (12, − 19), (6, 14)
22) ( − 16, 2), (15, − 10)
23) ( − 5, − 10), ( − 5, 20)
24) (8, 11), ( − 3, − 13)
25) ( − 17, 19), (10, − 7)
26) (11, − 2), (1, 17)
27) (7, − 14), ( − 8, − 9)
28) ( − 18, − 5), (14, − 3)
29) ( − 5, 7), ( − 18, 14)
30) (19, 15), (5, 11)
Find the value of x or y so that the line through the points has the
given slope
...
3
Graphing - Slope-Intercept Form
Objective: Give the equation of a line with a known slope and y-intercept
...
However, if we can identify some properties of the line, we may be able to
make a graph much quicker and easier
...
The slope can be represented by m and the yintercept, where it crosses the axis and x = 0, can be represented by (0, b) where b
is the value where the graph crosses the vertical y-axis
...
Using this information we will look at the slope
formula and solve the formula for y
...
m, (0, b), (x, y)
y −b
=m
x−0
y −b
=m
x
y − b = mx
+b +b
y = mx + b
Using the slope formula gives:
Simplify
Multiply both sides by x
Add b to both sides
Our Solution
This equation, y = mx + b can be thought of as the equation of any line that as a
slope of m and a y-intercept of b
...
Slope − Intercept Equation: y = m x + b
If we know the slope and the y-intercept we can easily find the equation that represents the line
...
3
Slope = , y − intercept = − 3
4
y = mx + b
3
y= x−3
4
Use the slope − intercept equation
m is the slope, b is the y − intercept
Our Solution
We can also find the equation by looking at a graph and finding the slope and yintercept
...
102
Identify the point where the graph
crosses the y-axis (0,3)
...
Idenfity one other point and draw a
slope triangle to find the slope
...
However, it will be
important for the equation to first be in slope intercept form
...
Example 135
...
Example 136
...
Once we have both points, connect the
dots to get our graph
...
Graph 3x + 4y = 12
− 3x
− 3x
4y = − 3x + 12
4
4
4
3
y=− x+3
4
y = mx + b
3
m=− ,b=3
4
Not in slope intercept form
Subtract 3x from both sides
Put the x term first
Divide each term by 4
Recall slope − intercept equation
Idenfity m and b
Make the graph
Starting with a point at the y-intercept of 3,
rise
Then use the slope run , but its negative so it will go downhill, so we will
drop 3 units and run 4 units to find
the next point
...
We want to be very careful not to confuse using slope to find the next point with
use a coordinate such as (4, − 2) to find an individule point
...
Slope starts from a point on the line that could be anywhere on the graph
...
Lines with zero slope or no slope can make a problem seem very different
...
So the equation simply becomes y = b or y is equal to the y-coordinate of the graph
...
There is no y in
these equations
...
Example 138
...
Because we have a vertical line and no
slope there is no slope-intercept equation we can use
...
3 Practice - Slope-Intercept
Write the slope-intercept form of the equation of each line given the
slope and the y-intercept
...
9)
11)
10)
12)
13)
14)
105
15) x + 10y = − 37
16) x − 10y = 3
17) 2x + y = − 1
18) 6x − 11y = − 70
19) 7x − 3y = 24
20) 4x + 7y = 28
21) x = − 8
22) x − 7y = − 42
23) y − 4 = − (x + 5)
24) y − 5 = 2 (x − 2)
25) y − 4 = 4(x − 1)
26) y − 3 = − 3 (x + 3)
27) y + 5 = − 4(x − 2)
28) 0 = x − 4
1
29) y + 1 = − 2 (x − 4)
5
2
6
30) y + 2 = 5 (x + 5)
Sketch the graph of each line
...
4
Graphing - Point-Slope Form
Objective: Give the equation of a line with a known slope and point
...
In these cases we can’t use the slope intercept equation, so we will use a different more flexible formula
...
We can use the slope formula to make a second equation
...
m, (x1, y1), (x, y)
y2 − y1
=m
x2 − x1
y − y1
=m
x − x1
y − y1 = m(x − x1)
Recall slope formula
Plug in values
Multiply both sides by (x − x1)
Our Solution
If we know the slope, m of an equation and any point on the line (x1, y1) we can
easily plug these values into the equation above which will be called the pointslope formula
...
3
Write the equation of the line through the point (3, − 4) with a slope of 5
...
If the direc107
tions ask for the answer in slope-intercept form we will simply distribute the
slope, then solve for y
...
Write the equation of the line through the point ( − 6, 2) with a slope of −
slope-intercept form
...
This means when you simplify the signs
you will have the opposite of the numbers in the point
...
In order to find the equation of a line we will always need to know the slope
...
Example 142
...
y2 − y1
x2 − x1
−8
4
−3−5
=
=−
m=
6
3
4 − ( − 2)
y − y1 = m(x − x1)
4
y − 5 = − (x − ( − 2))
3
4
y − 5 = − (x + 2)
3
m=
First we must find the slope
Plug values in slope formula and evaluate
With slope and either point, use point − slope formula
Simplify signs
Our Solution
Example 143
...
y2 − y1
x2 − x1
−6
−2−4
=
=−3
m=
2
− 1 − ( − 3)
y − y1 = m(x − x1)
y − 4 = − 3(x − ( − 3))
y − 4 = − 3(x + 3)
y − 4 = − 3x − 9
+4
+4
y = − 3x − 5
m=
First we must find the slope
Plug values in slope formula and evaluate
With slope and either point, point − slope formula
Simplify signs
Distribute slope
Solve for y
Add 4 to both sides
Our Solution
Example 144
...
y2 − y1
x2 − x1
3
3
1 − ( − 2)
=
=−
m=
− 10
10
−4−6
y − y1 = m(x − x1)
3
y − ( − 2) = − (x − 6)
10
3
y + 2 = − (x − 6)
10
9
3
y+2=− x+
5
10
10
−2
−
5
3
1
y=− x−
10
5
m=
First we must find the slope
Plug values into slope formula and evaluate
Use slope and either point, use point − slope formula
Simplify signs
Distribute slope
Solve for y
...
There were 7
bridges that connected the parts of the city
...
It turned out that this problem was impossible, but the work laid the foundation of what would become graph theory
...
4 Practice - Point-Slope Form
Write the point-slope form of the equation of the line through the
given point with the given slope
...
17) through: ( − 1, − 5), slope = 2
18) through: (2, − 2), slope = − 2
2
3
20) through: ( − 2, − 2), slope = − 3
19) through: (5, − 1), slope = − 5
7
22) through: (4, − 3), slope = − 4
1
21) through: ( − 4, 1), slope = 2
5
24) through: ( − 2, 0), slope = − 2
3
23) through: (4, − 2), slope = − 2
2
7
25) through: ( − 5, − 3), slope = − 5
26) through: (3, 3), slope = 3
27) through: (2, − 2), slope = 1
28) through: ( − 4, − 3), slope = 0
29) through:( − 3, 4), slope=undefined
30) through: ( − 2, − 5), slope = 2
1
31) through: ( − 4, 2), slope = − 2
6
32) through: (5, 3), slope = 5
110
Write the point-slope form of the equation of the line through the
given points
...
43) through: ( − 5, 1) and ( − 1, − 2)
44) through: ( − 5, − 1) and (5, − 2)
45) through: ( − 5, 5) and (2, − 3)
46) through: (1, − 1) and ( − 5, − 4)
47) through: (4, 1) and (1, 4)
48) through: (0, 1) and ( − 3, 0)
49) through: (0, 2) and (5, − 3)
50) through: (0, 2) and (2, 4)
51) through: (0, 3) and ( − 1, − 1)
52) through: ( − 2, 0) and (5, 3)
111
2
...
There is an interesting connection between the slope of lines that are parallel and
the slope of lines that are perpendicular (meet at a right angle)
...
Example 145
...
The slope of the top line is
2
down 2, run 3, or − 3
...
The above graph has two perpendicular lines
...
The slope of the steeper line is down 3,
3
3
run 2 or − 2
...
In it is the famous parallel postulate which
mathematicians have tried for years to drop from the list of postulates
...
We can use these properties to make conclusions about parallel and perpendicular lines
...
Find the slope of a line parallel to 5y − 2x = 7
...
Parallel lines have the same slope
m=
2
5
Our Solution
Example 147
...
Perpendicular lines have opposite reciprocal slopes
m=−
4
3
Our Solution
Once we have a slope, it is possible to find the complete equation of the second
line if we know one point on the second line
...
Find the equation of a line through (4, − 5) and parallel to 2x − 3y = 6
...
3
Find the equation of the line through (6, − 9) perpendicular to y = − 5 x + 4 in
slope-intercept form
...
Because a horizontal line is perpendicular to a vertical line we can
say that no slope and zero slope are actually perpendicular slopes!
Example 150
...
114
2
...
2
1) y = 2x + 4
2) y = − 3 x + 5
3) y = 4x − 5
4) y = −
5) x − y = 4
6) 6x − 5y = 20
7) 7x + y = − 2
8) 3x + 4y = − 8
10
x−5
3
Find the slope of a line perpendicular to each given line
...
17) through: (2, 5), parallel to x = 0
7
18) through: (5, 2), parallel to y = 5 x + 4
9
19) through: (3, 4), parallel to y = 2 x − 5
3
20) through: (1, − 1), parallel to y = − 4 x + 3
7
21) through: (2, 3), parallel to y = 5 x + 4
22) through: ( − 1, 3), parallel to y = − 3x − 1
23) through: (4, 2), parallel to x = 0
7
24) through: (1, 4), parallel to y = 5 x + 2
25) through: (1, − 5), perpendicular to − x + y = 1
26) through: (1, − 2), perpendicular to − x + 2y = 2
115
27) through: (5, 2), perpendicular to 5x + y = − 3
28) through: (1, 3), perpendicular to − x + y = 1
29) through: (4, 2), perpendicular to − 4x + y = 0
30) through: ( − 3, − 5), perpendicular to 3x + 7y = 0
31) through: (2, − 2) perpendicular to 3y − x = 0
32) through: ( − 2, 5)
...
33) through: (4, − 3), parallel to y = − 2x
3
34) through: ( − 5, 2), parallel to y = 5 x
4
35) through: ( − 3, 1), parallel to y = − 3 x − 1
5
36) through: ( − 4, 0), parallel to y = − 4 x + 4
1
37) through: ( − 4, − 1), parallel to y = − 2 x + 1
5
38) through: (2, 3), parallel to y = 2 x − 1
1
39) through: ( − 2, − 1), parallel to y = − 2 x − 2
3
40) through: ( − 5, − 4), parallel to y = 5 x − 2
41) through: (4, 3), perpendicular to x + y = − 1
42) through: ( − 3, − 5), perpendicular to x + 2y = − 4
43) through: (5, 2), perpendicular to x = 0
44) through: (5, − 1), perpendicular to − 5x + 2y = 10
45) through: ( − 2, 5), perpendicular to − x + y = − 2
46) through: (2, − 3), perpendicular to − 2x + 5y = − 10
47) through: (4, − 3), perpendicular to − x + 2y = − 6
48) through: ( − 4, 1), perpendicular to 4x + 3y = − 9
116
Chapter 3 : Inequalities
3
...
118
3
...
124
3
...
128
117
3
...
When we have an equation such as x = 4 we have a specific value for our variable
...
To do this we
will not use equals, but one of the following symbols:
>
<
Greater than
Greater than or equal to
Less than
Less than or equal to
World View Note: English mathematician Thomas Harriot first used the above
symbols in 1631
...
If we have an expression such as x < 4, this means our variable can be any number
smaller than 4 such as − 2, 0, 3, 3
...
999999999 as long as it is smaller
118
than 4
...
9999, or even − 2
...
We will start from the
value in the problem and bold the lower part of the number line if the variable is
smaller than the number, and bold the upper part of the number line if the variable is larger
...
Once the graph is drawn we can quickly convert the graph into what is called
interval notation
...
If there is no largest value, we can use ∞
(infinity)
...
If we
use either positive or negative infinity we will always use a curved bracket for that
value
...
Graph the inequality and give the interval notation
x<2
( − ∞, 2)
Start at 2 and shade below
Use ) for less than
Our Graph
Interval Notation
Example 152
...
119
Example 153
...
Curved bracket means just greater than
x>3
Our Solution
Example 154
...
Square bracket means less than or
equal to
x
−4
Our Solution
Generally when we are graphing and giving interval notation for an inequality we
will have to first solve the inequality for our variable
...
Consider the following inequality
and what happens when various operations are done to it
...
5>1
8>4
6>2
12 > 6
6>3
5>2
9>6
− 18 < − 12
3>2
Add 3 to both sides
Subtract 2 from both sides
Multiply both sides by 3
Divide both sides by 2
Add − 1 to both sides
Subtract − 4 from both sides
Multiply both sides by − 2
Divide both sides by − 6
Symbol flipped when we multiply or divide by a negative!
As the above problem illustrates, we can add, subtract, multiply, or divide on
both sides of the inequality
...
We will keep that in mind as we solve
inequalities
...
Solve and give interval notation
5 − 2x
11
Subtract 5 from both sides
120
−5
−5
− 2x 6
−2 −2
x −3
( − ∞, − 3]
Divide both sides by − 2
Divide by a negative − flip symbol!
Graph, starting at − 3, going down with ] for less than or equal to
Interval Notation
The inequality we solve can get as complex as the linear equations we solved
...
Just remember that any time we multiply or divide by a negative the
symbol switches directions (multiplying or dividing by a positive does not change
the symbol!)
Example 156
...
Often students draw their graphs the
wrong way when this is the case
...
So we must shade above the 4
...
1 Practice - Solve and Graph Inequalities
Draw a graph for each inequality and give interval notation
...
7)
8)
9)
10)
11)
12)
122
Solve each inequality, graph each solution, and give interval notation
...
2
Inequalities - Compound Inequalities
Objective: Solve, graph and give interval notation to the solution of
compound inequalities
...
There are three types of compound inequalities which we will investigate in this lesson
...
For this type of
inequality we want a true statment from either one inequality OR the other
inequality OR both
...
When we give interval notation for our solution, if there are two different parts to
the graph we will put a ∪ (union) symbol between two sets of interval notation,
one for each part
...
Solve each inequality, graph the solution, and give interval notation of solution
2x − 5 > 3 or 4 − x 6
−4
+5+5 −4
2x > 8 or − x 2
2 2
−1 −1
x > 4 or x − 2
Solve each inequality
Add or subtract first
Divide
Dividing by negative flips sign
Graph the inequalities separatly above number line
( − ∞, − 2] ∪ (4, ∞) Interval Notation
World View Note: The symbol for infinity was first used by the Romans,
although at the time the number was used for 1000
...
There are several different results that could result from an OR statement
...
Notice how interval
notation works for each of these cases
...
When the graphs are combined they
cover the entire number line
...
AND inequalities
require both statements to be true
...
When we
graph these inequalities we can follow a similar process, first graph both inequalities above the number line, but this time only where they overlap will be drawn
onto the number line for our final graph
...
Example 158
...
2x + 8 5x − 7 and 5x − 3 > 3x + 1
− 3x
− 3x
− 2x
− 2x
8 3x − 7 and 2x − 3 > 1
+7
+7
+3+3
15 3x and 2x > 4
3
2 2
3
5 x and x > 2
(2, 5]
Move variables to one side
Add 7 or 3 to both sides
Divide
Graph, x is smaller (or equal) than 5,
greater than 2
Interval Notation
Again, as we graph AND inequalities, only the overlapping parts of the individual
graphs makes it to the final number line
...
The first is shown in the above
125
example
...
The third is if the arrows point opposite ways but don’t
overlap, this is shown below on the right
...
In this graph, the overlap is only the
smaller graph, so this is what makes it
to the final number line
...
Because their is no overlap, no
values make it to the final number
line
...
When
our variable (or expression containing the variable) is between two numbers, we
can write it as a single math sentence with three parts, such as 5 < x 8, to show
x is between 5 and 8 (or equal to 8)
...
The graph then is simply the values between the numbers with appropriate brackets on the ends
...
Solve the inequality, graph the solution, and give interval notation
...
2 Practice - Compound Inequalities
Solve each compound inequality, graph its solution, and give interval
notation
...
3
Inequalities - Absolute Value Inequalities
Objective: Solve, graph and give interval notation for the solution to
inequalities with absolute values
...
The way we remove
the absolute value depends on the direction of the inequality symbol
...
Absolute value is defined as distance from zero
...
So on a number line we
will shade all points that are less than 2 units away from zero
...
So |x| < 2 becomes − 2 < x < 2, as the
graph above illustrates
...
Absolute value is defined as distance from zero
...
So on the number
line we shade all points that are more than 2 units away from zero
...
So |x| > 2 becomes x > 2 or x < − 2, as the graph
above illustrates
...
The first known use of the symbol for integers comes from a 1939
128
edition of a college algebra text!
For all absolute value inequalities we can also express our answers in interval
notation which is done the same way it is done for standard compound inequalities
...
Our first step will be to isolate the absolute value
...
Then we will solve these inequalites
...
Solve, graph, and give interval notation for the solution
4x − 5 6
+5+5
4x
4
x
|4x − 5| 6
OR 4x − 5 − 6
+5 +5
11 OR 4x − 1
4
4
4
11
1
OR x −
4
4
− ∞, −
Absolute value is greater, use OR
Solve
Add 5 to both sides
Divide both sides by 4
Graph
11
1
,∞
∪
4
4
Interval notation
Example 161
...
So we must first clear the − 4 by
adding 4, then divide by − 3
...
Example 162
...
We can never distribute or combine things outside the absolute value with what is
inside the absolute value
...
130
It is important to remember as we are solving these equations, the absolute value
is always positive
...
Similarly, if absolute value is greater than a negative, this will always happen
...
Example 163
...
Solve, graph, and give interval notation for the solution
5 − 6|x + 7| 17
−5
−5
− 6|x + 7| 12
−6
−6
|x + 7| − 2
Subtract 5 from both sides
Divide both sides by − 6
Dividing by a negative flips the symbol
Absolute value always greater than negative
All Real Numbers or R
131
3
...
1) |x| < 3
2) |x|
3) |2x| < 6
4) |x + 3| < 4
5) |x − 2| < 6
6) x − 8| < 12
7) |x − 7| < 3
8) |x + 3|
9) |3x − 2| < 9
10) |2x + 5| < 9
11) 1 + 2|x − 1|
9
8
4
12) 10 − 3|x − 2|
13) 6 − |2x − 5| > = 3
14) |x| > 5
15) |3x| > 5
16) |x − 4| > 5
17) |x = 3| > = 3
4
18) |2x − 4| > 6
19) |3x − 5| >
3
20) 3 − |2 − x| < 1
21) 4 + 3|x − 1| > = 10
22) 3 − 2|3x − 1|
23) 3 − 2|x − 5|
24) 4 − 6| − 6 − 3x|
−5
26) − 3 − 2|4x − 5|
1
− 15
25) − 2 − 3|4 − 2x|
−8
−7
27) 4 − 5| − 2x − 7| < − 1
28) − 2 + 3|5 − x|
29) 3 − 2|4x − 5|
30) − 2 − 3| − 3x − 5
1
4
31) − 5 − 2|3x − 6| < − 8
32) 6 − 3|1 − 4x| < − 3
33) 4 − 4| − 2x + 6| > − 4
34) − 3 − 4| − 2x − 5|
−5
35) | − 10 + x|
8
132
−7
Chapter 4 : Systems of Equations
4
...
134
4
...
139
4
...
146
4
...
151
4
...
158
4
...
167
133
4
...
We have solved problems like 3x − 4 = 11 by adding 4 to both sides and then
dividing by 3 (solution is x = 5)
...
It turns out that to solve for more than one variable we will need the same number of equations as variables
...
When we have
several equations we are using to solve, we call the equations a system of equations
...
This solution is usually given as an ordered pair (x, y)
...
Show (2,1) is the solution to the system
(2, 1)
x = 2, y = 1
3(2) − (1) = 5
6−1=5
5=5
(2) + (1) = 3
3=3
3x − y = 5
x+y=3
Identify x and y from the orderd pair
Plug these values into each equation
First equation
Evaluate
True
Second equation, evaluate
True
As we found a true statement for both equations we know (2,1) is the solution to
the system
...
In this lesson we will be working to find this point given the equations
...
If the graph of a line is a picture of all the solutions, we can graph two lines on
the same coordinate plane to see the solutions of both equations
...
Example 166
...
(4,1)
To graph each equation, we start at
rise
the y-intercept and use the slope run
to get the next point and connect the
dots
...
Example 167
...
(-2,-1)
Remember a negative slope is downhill!
Find the intersection point, ( − 2, − 1)
( − 2, − 1)
Our Solution
As we are graphing our lines, it is possible to have one of two unexpected results
...
Example 168
...
The two lines do not intersect! They
are parallel! If the lines do not intersect we know that there is no point
that works in both equations, there is
no solution
∅ No Solution
We also could have noticed that both lines had the same slope
...
Example 169
...
Both equations are the same line! As
one line is directly on top of the other
line, we can say that the lines “intersect” at all the points! Here we say we
have infinite solutions
Once we had both equations in slope-intercept form we could have noticed that
the equations were the same
...
World View Note: The Babylonians were the first to work with systems of
equations with two variables
...
1 Practice - Graphing
Solve each equation by graphing
...
2
Systems of Equations - Substitution
Objective: Solve systems of equations using substitution
...
First, it requires the
graph to be perfectly drawn, if the lines are not straight we may arrive at the
wrong answer
...
2134 for example
...
Instead, an algebraic approach will be used
...
We will build the concepts of
substitution through several example, then end with a five-step process to solve
problems using this method
...
x=5
y = 2x − 3
y = 2(5) − 3
y = 10 − 3
y=7
(5, 7)
We already know x = 5, substitute this into the other equation
Evaluate, multiply first
Subtract
We now also have y
Our Solution
When we know what one variable equals we can plug that value (or expression) in
for the variable in the other equation
...
The reason for this is shown in the
next example
...
2x − 3y = 7
y = 3x − 7
2x − 3(3x − 7) = 7
We know y = 3x − 7, substitute this into the other equation
Solve this equation, distributing − 3 first
139
2x − 9x + 21 = 7
− 7x + 21 = 7
− 21 − 21
− 7x = − 14
−7 −7
x=2
y = 3(2) − 7
y=6−7
y=−1
(2, − 1)
Combine like terms 2x − 9x
Subtract 21
Divide by − 7
We now have our x, plug into the y = equation to find y
Evaluate, multiply first
Subtract
We now also have y
Our Solution
By using the entire expression 3x − 7 to replace y in the other equation we were
able to reduce the system to a single linear equation which we can easily solve for
our first variable
...
If this happens we can isolate it by
solving for the lone variable
...
3x + 2y = 1
x − 5y = 6
+ 5y + 5y
x = 6 + 5y
3(6 + 5y) + 2y = 1
18 + 15y + 2y = 1
18 + 17y = 1
− 18
− 18
17y = − 17
17
17
y=−1
x = 6 + 5( − 1)
x=6−5
x=1
(1, − 1)
Lone variable is x, isolate by adding 5y to both sides
...
This process is described and illustrated in the following table which lists
the five steps to solving by substitution
...
Find the lone variable
2x + y = − 5
− 2x
− 2x
2
...
Substitute into the untouched equation
4x − 2( − 5 − 2x) = 2
4x + 10 + 4x = 2
8x + 10 = 2
− 10 − 10
4
...
Plug into lone variable equation and evaluate y = − 5 + 2
y=−3
Solution
( − 1, − 3)
Problem
Sometimes we have several lone variables in a problem
...
Example 173
...
We will chose x in the first
Solve for the lone variable, subtract y from both sides
Plug into the untouched equation, the second equation
Solve, parenthesis are not needed here, combine like terms
Subtract 5 from both sides
Divide both sides by − 2
We have our y!
Plug into lone variable equation, evaluate
Now we have our x
141
(2, 3)
Our Solution
Just as with graphing it is possible to have no solution ∅ (parallel lines) or
infinite solutions (same line) with the substitution method
...
Example 174
...
Our Solution
Because we had a true statement, and no variables, we know that anything that
works in the first equation, will also work in the second equation
...
Example 175
...
Our Solution
Because we had a false statement, and no variables, we know that nothing will
work in both equations
...
It was in this book that he suggested using
letters from the end of the alphabet for unknown values
...
One more question needs to be considered, what if there is no lone variable? If
there is no lone variable substitution can still work to solve, we will just have to
select one variable to solve for and use fractions as we solve
...
5x − 6y = − 14
− 2x + 4y = 12
5x − 6y = − 14
+ 6y + 6y
5x = − 14 + 6y
5
5
5
− 14 6y
x=
+
5
5
− 14 6y
+
+ 4y = 12
−2
5
5
28 12y
−
+ 4y = 12
5
5
28(5) 12y(5)
−
+ 4y(5) = 12(5)
5
5
28 − 12y + 20y = 60
28 + 8y = 60
− 28
− 28
8y = 32
8 8
y=4
− 14 6(4)
+
x=
5
5
− 14 24
x=
+
5
5
10
x=
5
x=2
(2, 4)
No lone variable,
we will solve for x in the first equation
Solve for our variable, add 6y to both sides
Divide each term by 5
Plug into untouched equation
Solve, distribute through parenthesis
Clear fractions by multiplying by 5
Reduce fractions and multiply
Combine like terms − 12y + 20y
Subtract 28 from both sides
Divide both sides by 8
We have our y
Plug into lone variable equation, multiply
Add fractions
Reduce fraction
Now we have our x
Our Solution
Using the fractions does make the problem a bit more tricky
...
143
4
...
1) y = − 3x
y = 6x − 9
2) y = x + 5
y = − 2x − 4
3) y = − 2x − 9
y = 2x − 1
4) y = − 6x + 3
y = 6x + 3
5) y = 6x + 4
y = − 3x − 5
6) y = 3x + 13
y = − 2x − 22
7) y = 3x + 2
y = − 3x + 8
8) y = − 2x − 9
y = − 5x − 21
9) y = 2x − 3
y = − 2x + 9
10) y = 7x − 24
y = − 3x + 16
11) y = 6x − 6
− 3x − 3y = − 24
12) − x + 3y = 12
y = 6x + 21
13) y = − 6
3x − 6y = 30
14) 6x − 4y = − 8
y = − 6x + 2
15) y = − 5
3x + 4y = − 17
16) 7x + 2y = − 7
y = 5x + 5
17) − 2x + 2y = 18
y = 7x + 15
18) y = x + 4
3x − 4y = − 19
19) y = − 8x + 19
− x + 6y = 16
20) y = − 2x + 8
− 7x − 6y = − 8
21) 7x − 2y = − 7
y=7
22) x − 2y = − 13
4x + 2y = 18
23) x − 5y = 7
2x + 7y = − 20
24) 3x − 4y = 15
7x + y = 4
25) − 2x − y = − 5
x − 8y = − 23
26) 6x + 4y = 16
− 2x + y = − 3
27) − 6x + y = 20
− 3x − 3y = − 18
28) 7x + 5y = − 13
x − 4y = − 16
29) 3x + y = 9
2x + 8y = − 16
30) − 5x − 5y = − 20
− 2x + y = 7
144
31) 2x + y = 2
3x + 7y = 14
32) 2x + y = − 7
5x + 3y = − 21
33) x + 5y = 15
− 3x + 2y = 6
34) 2x + 3y = − 10
7x + y = 3
35) − 2x + 4y = − 16
y=−2
36) − 2x + 2y = − 22
− 5x − 7y = − 19
37) − 6x + 6y = − 12
8x − 3y = 16
38) − 8x + 2y = − 6
− 2x + 3y = 11
39) 2x + 3y = 16
− 7x − y = 20
40) − x − 4y = − 14
− 6x + 8y = 12
145
4
...
When solving systems we have found that graphing is very limited when solving
equations
...
This is
probably the most used idea in solving systems in various areas of algebra
...
This leads us to
our second method for solving systems of equations
...
We will set up the process in the following examples, then define the five step process we can use to solve by elimination
...
3x − 4y = 8
5x + 4y = − 24
8x
= − 16
8
8
x=−2
5( − 2) + 4y = − 24
− 10 + 4y = − 24
+ 10
+ 10
4y = − 14
4
4
−7
y=
2
−7
− 2,
2
Notice opposites in front of y ′s
...
Solve for x, divide by 8
We have our x!
Plug into either original equation, simplify
Add 10 to both sides
Divide by 4
Now we have our y!
Our Solution
In the previous example one variable had opposites in front of it, − 4y and 4y
...
This allowed us to solve for
the x
...
However, generally we won’t
have opposites in front of one of the variables
...
This is shown in the next example
...
− 6x + 5y = 22
2x + 3y = 2
3(2x + 3y) = (2)3
We can get opposites in front of x, by multiplying the
second equation by 3, to get − 6x and + 6x
Distribute to get new second equation
...
What we are looking for with our opposites is the least common multiple (LCM) of the coefficients
...
The LCM of
3 and 5 is 15
...
This illustrates an important point,
some problems we will have to multiply both equations by a constant (on both
sides) to get the opposites we want
...
3x + 6y = − 9
2x + 9y = − 26
3(3x + 6y) = ( − 9)3
9x + 18y = − 27
− 2(2x + 9y) = ( − 26)( − 2)
− 4x − 18y = 52
9x + 18y = − 27
− 4x − 18y = 52
5x
= 25
5
5
x=5
3(5) + 6y = − 9
15 + 6y = − 9
− 15
− 15
We can get opposites in front of x, find LCM of 6 and 9,
The LCM is 18
...
This is illustrated
in the next example which includes the five steps we will go through to solve a
problem using elimination
...
Line up the variables and constants
2
...
Add
4
...
Plug into either original and solve
Solution
2x − 5y = − 13
− 3y + 4 = − 5x
Second Equation:
− 3y + 4 = − 5x
+ 5x − 4 + 5x − 4
5x − 3y = − 4
2x − 5y = − 13
5x − 3y = − 4
First Equation: multiply by − 5
− 5(2x − 5y) = ( − 13)( − 5)
− 10x + 25y = 65
Second Equation: multiply by 2
2(5x − 3y) = ( − 4)2
10x − 6y = − 8
− 10x + 25y = 65
10x − 6y = − 8
19y = 57
19y = 57
19 19
y=3
2x − 5(3) = − 13
2x − 15 = − 13
+ 15 + 15
2x
=2
2
2
x=1
(1, 3)
World View Note: The famous mathematical text, The Nine Chapters on the
Mathematical Art, which was printed around 179 AD in China describes a formula very similar to Gaussian elimination which is very similar to the addition
method
...
Just as with substitution, if the variables all disappear
from our problem, a true statment will indicate infinite solutions and a false statment will indicate no solution
...
2x − 5y = 3
− 6x + 15y = − 9
To get opposites in front of x, multiply first equation by 3
3(2x − 5y) = (3)3
6x − 15y = 9
Distribute
6x − 15y = 9
− 6x + 15y = − 9
0=0
Infinite solutions
Add equations together
True statement
Our Solution
Example 181
...
Multiply first equation by 3
Multiply second equation by − 2
Add both new equations together
False statement
Our Solution
We have covered three different methods that can be used to solve a system of
two equations with two variables
...
Substitution works great
when we have a lone variable, and addition works great when the other two
methods fail
...
149
4
...
1) 4x + 2y = 0
− 4x − 9y = − 28
3) − 9x + 5y = − 22
9x − 5y = 13
5) − 6x + 9y = 3
6x − 9y = − 9
7) 4x − 6y = − 10
4x − 6y = − 14
9) − x − 5y = 28
− x + 4y = − 17
11) 2x − y = 5
5x + 2y = − 28
13) 10x + 6y = 24
− 6x + y = 4
15) 2x + 4y = 24
4x − 12y = 8
17) − 7x + 4y = − 4
10x − 8y = − 8
19) 5x + 10y = 20
− 6x − 5y = − 3
21) − 7x − 3y = 12
− 6x − 5y = 20
23) 9x − 2y = − 18
5x − 7y = − 10
25) 9x + 6y = − 21
− 10x − 9y = 28
27) − 7x + 5y = − 8
− 3x − 3y = 12
29) − 8x − 8y = − 8
10x + 9y = 1
31) 9y = 7 − x
− 18y + 4x = − 26
33) 0 = 9x + 5y
2
y = 7x
2) − 7x + y = − 10
− 9x − y = − 22
4) − x − 2y = − 7
x + 2y = 7
6) 5x − 5y = − 15
5x − 5y = − 15
8) − 3x + 3y = − 12
− 3x + 9y = − 24
10) − 10x − 5y = 0
− 10x − 10y = − 30
12) − 5x + 6y = − 17
x − 2y = 5
14) x + 3y = − 1
10x + 6y = − 10
16) − 6x + 4y = 12
12x + 6y = 18
18) − 6x + 4y = 4
− 3x − y = 26
20) − 9x − 5y = − 19
3x − 7y = − 11
22) − 5x + 4y = 4
− 7x − 10y = − 10
24) 3x + 7y = − 8
4x + 6y = − 4
26) − 4x − 5y = 12
− 10x + 6y = 30
28) 8x + 7y = − 24
6x + 3y = − 18
30) − 7x + 10y = 13
4x + 9y = 22
32) 0 = − 9x − 21 + 12y
4
7
1 + 3 y + 3 x=0
34) − 6 − 42y = − 12x
1
7
x− 2 − 2y =0
150
4
...
Solving systems of equations with 3 variables is very similar to how we solve systems with two variables
...
With three variables
we will reduce the system down to one with two variables (usually by addition),
which we can then solve by either addition or substitution
...
We will use addition with two equations to eliminate one variable
...
Then we will use a different pair of equations
and use addition to eliminate the same variable
...
Once we have done this we will have two equations (A) and (B)
with the same two variables that we can solve using either method
...
Example 182
...
We will solve by addition
Multiply (A) by − 1
Add to the second equation, unchanged
Solve, divide by 2
We now have x! Plug this into either (A) or (B)
We plug it into (A), solve this equation, subtract 2
Divide by 2
We now have z! Plug this and x into any original equation
We use the first, multiply 3(2) = 6 and combine with − 1
Solve, subtract 5
Divide by 2
We now have y!
Our Solution
As we are solving for x, y, and z we will have an ordered triplet (x, y, z) instead of
152
just the ordered pair (x, y)
...
However, sometimes we may have to do a bit of work to get
a variable to eliminate
...
As we do this remmeber it is improtant to eliminate the
same variable both times using two different pairs of equations
...
4x − 3y + 2z = − 29
6x + 2y − z = − 16
− 8x − y + 3z = 23
No variable will easily eliminate
...
4x − 3y + 2z = − 29
6x + 2y − z = − 16
Start with first two equations
...
Make the first equation have 12x, the second − 12x
3(4x − 3y + 2z) = ( − 29)3
12x − 9y + 6z = − 87
− 2(6x + 2y − z) = ( − 16)( − 2)
− 12x − 4y + 2z = 32
12x − 9y + 6z = − 87
− 12x − 4y + 2z = 32
(A) − 13y + 8z = − 55
6x + 2y − z = − 16
− 8x − y + 3z = 23
Multiply the first equation by 3
Multiply the second equation by − 2
Add these two equations together
This is our (A) equation
Now use the second two equations (a different pair)
The LCM of 6 and − 8 is 24
...
Multiply and combine like terms
Add 13
Divide by 4
We have our x!
Our Solution!
World View Note: Around 250, The Nine Chapters on the Mathematical Art
were published in China
...
One problem had four equations with five variables!
Just as with two variables and two equations, we can have special cases come up
with three variables and three equations
...
Example 184
...
Multiply the first equation by 2
Add this to the second equation, unchanged
A true statment
Our Solution
Example 185
...
It is possible to solve each
system several different ways
...
155
4
...
1) a − 2b + c = 5
2a + b − c = − 1
3a + 3b − 2c = − 4
2) 2x + 3y = z − 1
3x = 8z − 1
5y + 7z = − 1
3) 3x + y − z = 11
x + 3y = z + 13
x + y − 3z = 11
4) x + y + z = 2
6x − 4y + 5z = 31
5x + 2y + 2z = 13
5) x + 6y + 3z = 4
2x + y + 2z = 3
3x − 2y + z = 0
6) x − y + 2z = − 3
x + 2y + 3z = 4
2x + y + z = − 3
7) x + y + z = 6
2x − y − z = − 3
x − 2y + 3z = 6
8) x + y − z = 0
x + 2y − 4z = 0
2x + y + z = 0
9) x + y − z = 0
x− y −z=0
x + y + 2z = 0
10) x + 2y − z = 4
4x − 3y + z = 8
5x − y = 12
11) − 2x + y − 3z = 1
x − 4y + z = 6
4x + 16y + 4z = 24
12) 4x + 12y + 16z = 4
3x + 4y + 5z = 3
x + 8y + 11z = 1
13) 2x + y − 3z = 0
x − 4y + z = 0
4x + 16y + 4z = 0
14) 4x + 12y + 16z = 0
3x + 4y + 5z = 0
x + 8y + 11z = 0
15) 3x + 2y + 2z = 3
x + 2y − z = 5
2x − 4y + z = 0
16) p + q + r = 1
p + 2q + 3r = 4
4p + 5q + 6r = 7
17) x − 2y + 3z = 4
2x − y + z = − 1
4x + y + z = 1
18) x + 2y − 3z = 9
2x − y + 2z = − 8
3x − y − 4z = 3
19) x − y + 2z = 0
x − 2y + 3z = − 1
2x − 2y + z = − 3
20) 4x − 7y + 3z = 1
3x + y − 2z = 4
4x − 7y + 3z = 6
21) 4x − 3y + 2z = 40
5x + 9y − 7z = 47
9x + 8y − 3z = 97
22) 3x + y − z = 10
8x − y − 6z = − 3
5x − 2y − 5z = 1
156
23) 3x + 3y − 2z = 13
6x + 2y − 5z = 13
5x − 2y − 5z = − 1
24) 2x − 3y + 5z = 1
3x + 2y − z = 4
4x + 7y − 7z = 7
25) 3x − 4y + 2z = 1
2x + 3y − 3z = − 1
x + 10y − 8z = 7
26) 2x + y = z
4x + z = 4y
y=x+1
27) m + 6n + 3p = 8
3m + 4n = − 3
5m + 7n = 1
28) 3x + 2y = z + 2
y = 1 − 2x
3z = − 2y
29) − 2w + 2x + 2y − 2z = − 10
w+x+ y+z =−5
3w + 2x + 2y + 4z = − 11
w + 3x − 2y + 2z = − 6
30) − w + 2x − 3y + z = − 8
−w+x+ y −z =−4
w + x + y + z = 22
− w + x − y − z = − 14
31) w + x + y + z = 2
w + 2x + 2y + 4z = 1
−w+x− y −z=−6
− w + 3x + y − z = − 2
32) w + x − y + z = 0
− w + 2x + 2y + z = 5
− w + 3x + y − z = − 4
− 2w + x + y − 3z = − 7
157
4
...
One application of system of equations are known as value problems
...
For example, if our
variable is the number of nickles in a person’s pocket, those nickles would have a
value of five cents each
...
The basic structure of the table is shown below
...
Quite
often, this will be our variables
...
The third column is used for the total value which we calculate by
multiplying the number by the value
...
The last row of the table is
for totals
...
This means sometimes this row may have some blanks in it
...
This is shown in the following example
...
In a child’s bank are 11 coins that have a value of S1
...
The coins are either
quarters or dimes
...
This is the number total
...
85 for the final total,
Write final total in cents (185)
Because 25 and 10 are cents
q + d = 11
25q + 10d = 185
− 10(q + d) = (11)( − 10)
− 10q − 10d = − 110
− 10q − 10d = − 110
25q + 10d = 185
15q
= 75
15
15
q=5
(5) + d = 11
−5
−5
d=6
First and last columns are our equations by adding
Solve by either addition or substitution
...
On the back of the dime it says “one dime” (not 10 cents)
...
On the penny it
says “one cent” (not 1 cent)
...
Ticket sales also have a value
...
These problems can be solve in much the same way
...
There were 41 tickets sold for an event
...
50 and tickets
for adults cost S2
...
Total receipts for the event were S73
...
How many of each
type of ticket were sold?
Child
Adult
Total
Child
Adult
Total
Number Value Total
c
1
...
5
1
...
5
1
...
5
c + a = 41
1
...
5
c + a = 41
−c
−c
a = 41 − c
1
...
5
1
...
5
− 0
...
5
− 82 − 82
− 0
...
5
Using our value table, c for child, a for adult
Child tickets have value 1
...
00
(we can drop the zeros after the decimal point)
Multiply number by value to get totals
We have 41 tickets sold
...
50
Write in dollars as 1
...
Solve for a by subtracting c
Substitute into untouched equation
Distribute
Combine like terms
Subtract 82 from both sides
Divide both sides by − 0
...
5
− 0
...
Instead they
will give a relationship between the items
...
While it is clear that we need to
multiply one variable by 2, it may not be clear which variable gets multiplied by
2
...
If there are
twice as many dimes, than we multiply the other variable (nickels) by two
...
This type of problem is in the next example
...
A man has a collection of stamps made up of 5 cent stamps and 8 cent stamps
...
The total value of
all the stamps is S3
...
How many of each stamp does he have?
Number Value Total
Five
f
5
5f
Eight
3f
8
24f
Total
348
Use value table, f for five cent stamp, and e for eight
Also list value of each stamp under value column
Number Value Total
Five
f
5
5f
Eight
e
8
8e
Total
Multiply number by value to get total
Five
Eight
Total
Number Value Total
f
5
5f
e
8
8e
348
e = 3f
5f + 8e = 348
5f + 8(3f ) = 348
5f + 24f = 348
29f = 348
29
29
f = 12
e = 3(12)
The final total was 338(written in cents)
We do not know the total number, this is left blank
...
There are 12 five cent stamps
Plug into first equation
161
e = 36
12 five cent, 36 eight cent stamps
We have e, There are 36 eight cent stamps
Our Solution
The same process for solving value problems can be applied to solving interest
problems
...
Invest Rate Interest
Account 1
Account 2
Total
Our first column is for the amount invested in each account
...
Just as before, we multiply the investment amount by the rate to find the final column, the interest
earned
...
Example 189
...
At the end of the year she had earned S270 in interest
...
06
Account 2
y
0
...
06 0
...
09 0
...
Account 1
Account 2
Total
Invest Rate Interest
x
0
...
06x
y
0
...
09y
4000
270
x + y = 4000
0
...
09y = 270
− 0
...
06)
− 0
...
06y = − 240
Total investment is 4000,
Total interest was 276
First and last column give our two equations
Solve by either substitution or addition
Use Addition, multiply first equation by − 0
...
06x − 0
...
06x + 0
...
03y = 30
0
...
03
y = 1000
x + 1000 = 4000
− 1000 − 1000
x = 3000
S1000 at 9% and S3000 at 6%
Add equations together
Divide both sides by 0
...
Example 190
...
He earned S1230 in interest after one year
...
04
Invest Rate
Interest
5000
x
5000x
8000 x + 0
...
04 8000x + 320
Total
1230
5000x + 8000x + 320 = 1230
13000x + 320 = 1230
− 320 − 320
13000x = 910
13000 13000
x = 0
...
07) + 0
...
11
S5000 at 7% and S8000 at 11%
Our investment table
...
04
Be sure to write this rate as a decimal!
Multiply to fill in interest column
...
04)
Total interest was 1230
...
5 Practice - Value Problems
Solve
...
25
...
How many of each is there?
2) A collection of half dollars and nickels is worth S13
...
There are 34 coins in
all
...
Admission was S2
...
50 for children
...
00
...
90 made up of dimes and quarters
...
25 in nickels and dimes
...
75 is made up of quarters and half dollars
...
25
...
25 in dimes and nickels, were distributed among 45 boys
...
Admission was S1 each for adults and 75
cents each for children
...
50
...
Tickets for
students were 50 cents each and for adults 75 cents each
...
50
...
For activity-card holders,
the price was S1
...
The
total amount of money collected was S310
...
50 each and the hamburgers sold
for S2
...
There were 131 total sandwiches sold for a total value of S342
...
The cost of
a student ticket was S1
...
50
...
How many students and how many nonstudents attented the game?
164
14) A bank contains 27 coins in dimes and quarters
...
95
...
15) A coin purse contains 18 coins in nickels and dimes
...
15
...
16) A business executive bought 40 stamps for S9
...
The purchase included 25c
stamps and 20c stamps
...
Altogether, 15
stamps were sold for a total cost of S3
...
How many of each type of stamps
were sold?
18) A drawer contains 15c stamps and 18c stamps
...
The total value of all
the stamps is S1
...
How many 15c stamps are in the drawer?
19) The total value of dimes and quarters in a bank is S6
...
There are six more
quarters than dimes
...
20) A child’s piggy bank contains 44 coins in quarters and dimes
...
60
...
21) A coin bank contains nickels and dimes
...
The total value of all the coins is S2
...
Find the
number of each type of coin in the bank
...
Some of the bills are one dollar bills, and
the rest are five dollar bills
...
Find
the number of each type of bill in the cash box
...
In all, twelve bills were handed to the customer
...
24) A collection of stamps consists of 22c stamps and 40c stamps
...
The
total value of the stamps is S8
...
Find the number of 22c stamps in the
collection
...
The
total interest after one year is S3385
...
5%
...
How much was invested at each rate?
27) A total of S9000 is invested, part of it at 10% and the rest at 12%
...
How much was invested at each rate?
28) A total of S18000 is invested, part of it at 6% and the rest at 9%
...
How much was invested at each rate?
29) An inheritance of S10000 is invested in 2 ways, part at 9
...
The combined annual interest was S1038
...
How much
was invested at each rate?
165
30) Kerry earned a total of S900 last year on his investments
...
31) Jason earned S256 interest last year on his investments
...
32) Millicent earned S435 last year in interest
...
33) A total of S8500 is invested, part of it at 6% and the rest at 3
...
The total
interest after one year is S385
...
5%
...
How much was invested at each rate?
35) A total of S15000 is invested, part of it at 8% and the rest at 11%
...
How much was invested at each rate?
36) A total of S17500 is invested, part of it at 7
...
5%
...
50
...
25% and the rest at 5
...
The
total interest after one year is S300
...
5% and the rest at 9%
...
How much was invested at each rate?
39) A total of S11000 is invested, part of it at 6
...
2%
...
How much was invested at each rate?
40) An investment portfolio earned S2010 in interest last year
...
41) Samantha earned S1480 in interest last year on her investments
...
42) A man has S5
...
There are twice as many
nickels as dimes and 3 more dimes than quarters
...
30 consists of nickels, dimes and quarters
...
75
...
6
Systems of Equations - Mixture Problems
Objective: Solve mixture problems by setting up a system of equations
...
Mixture problems
are ones where two different solutions are mixed together resulting in a new final
solution
...
The second column is
labeled “part”
...
If we mix prices we will put prices in this column
...
Then we can get an equation
by adding the amount and/or total columns that will help us solve the problem
and answer the questions
...
We will start with one variable problems
...
A chemist has 70 mL of a 50% methane solution
...
5
Add
x
0
...
We start with 70, but
don ′t know how much we add, that is x
...
5 for start, 0
...
167
Amount Part Total
Start
70
0
...
8
Final 70 + x 0
...
5
35
Add
x
0
...
8x
Final 70 + x 0
...
6x
35 + 0
...
6x
− 0
...
6x
35 + 0
...
2x = 7
0
...
2
x = 35
35 mL must be added
Add amount column to get final amount
...
6 because we want the
final solution to be 60% methane
...
be sure to distribute on the last row: (70 + x)0
...
6x
Subtract 35 from both sides
Divide both sides by 0
...
Example 192
...
50 by mixing two types of coffee
...
00
...
50 should the cafe mix with the first?
Amount Part Total
Start
40
3
Add
x
1
...
We know the starting
amount and its cost, S3
...
50
...
5
Final 40 + x 2
...
We want this final amount to sell for S2
...
168
Amount Part
Total
Start
40
3
120
Add
x
1
...
5x
Final 40 + x 2
...
5x
Multiply amount by part to get the total
...
5
120 + 1
...
5x
− 1
...
5x
120 = 100 + x
− 100 − 100
20 = x
20mL must be added
...
5x
Subtract 100 from both sides
We have our x
...
59
million metric tons of coffee a year! That is over three times as much coffee as
second place Vietnam!
The above problems illustrate how we can put the mixture table together and get
an equation to solve
...
The following example is one such problem
...
A farmer has two types of milk, one that is 24% butterfat and another which is
18% butterfat
...
24
Milk 2
y
0
...
2
We don ′t know either start value, but we do know
final is 42
...
24 0
...
18 0
...
2
8
...
x + y = 42
0
...
18y = 8
...
169
− 0
...
18)
− 0
...
18y = − 7
...
18x − 0
...
56
0
...
18y = 8
...
06x
= 0
...
06
0
...
Multiply first equation by − 0
...
06
We have our x, 14 gal of 24% butterfat
Plug into original equation to find y
Subtract 14 from both sides
We have our y, 28 gal of 18% butterfat
Our Solution
The same process can be used to solve mixtures of prices with two unknowns
...
In a candy shop, chocolate which sells for S4 a pound is mixed with nuts which
are sold for S2
...
50 a pound
...
5
Final
30
3
...
5 2
...
5 105
Multiply amount by part to get totals
c + n = 30
4c + 2
...
5n = 105
120 − 4n + 2
...
5n = 105
− 120
− 120
− 1
...
5 − 1
...
5
We have our n, 10 lbs of nuts
Plug into c = equation to find c
We have our c, 20 lbs of chocolate
Our Solution
With mixture problems we often are mixing with a pure solution or using water
which contains none of our chemical we are interested in
...
For water, the percentage is 0%
...
Example 195
...
How much of each should be used to make 70 L?
Amount Part Final
Antifreeze
a
1
Water
w
0
Final
70
0
...
65 45
...
5
(45
...
5
− 45
...
5
45
...
5L of water
We use a and w for our variables
...
Water has no antifreeze, its
percentage is 0
...
5 from both sides
We have our w
Our Solution
171
4
...
1) A tank contains 8000 liters of a solution that is 40% acid
...
How
many pounds of the second should be added to the first in order to get a
mixture of 5% salt?
4) How much alcohol must be added to 24 gallons of a 14% solution of alcohol in
order to produce a 20% solution?
5) How many pounds of a 4% solution of borax must be added to 24 pounds of a
12% solution of borax to obtain a 10% solution of borax?
6) How many grams of pure acid must be added to 40 grams of a 20% acid
solution to make a solution which is 36% acid?
7) A 100 LB bag of animal feed is 40% oats
...
How many ounces of the S400 alloy should be used
to make an alloy that costs S300 per ounce?
9) How many pounds of tea that cost S4
...
25 per pound to make a mixture that costs S3
...
50 per kilogram must be
mixed with 24 kg of jelly beans that cost S3
...
50 per kilogram?
12) How many kilograms of soil supplement that costs S7
...
50 per kilogram to
make a fertilizer that costs S4
...
60 per liter must be mixed with
18 L of anil that costs S2
...
90 per
liter?
15) Solution A is 50% acid and solution B is 80% acid
...
of a solution that is 68% acid?
172
16) A certain grade of milk contains 10% butter fat and a certain grade of cream
60% butter fat
...
How many gallons of each must be mixed to produce 60 gallons of
cream which is 19% butterfat?
18) A syrup manufacturer has some pure maple syrup and some which is 85%
maple syrup
...
How many milliliters of each solution
should the chemist use?
20) A hair dye is made by blending 7% hydrogen peroxide solution and a 4%
hydrogen peroxide solution
...
How many gallons of each must be used to make 60 gal of paint
that is 19% green dye?
22) A candy mix sells for S2
...
It contains chocolates worth S1
...
00 per kilogram
...
00/lb
...
00/lb
...
00/lb
...
of
the mixture?
24) A grocer is mixing 40 cent per lb
...
coffee to make a
mixture worth 54c per lb
...
of the mixture?
25) A grocer wishes to mix sugar at 9 cents per pound with sugar at 6 cents per
pound to make 60 pounds at 7 cents per pound
...
75 per pound is mixed with a
vitamin supplement that costs S3
...
How many pounds of each
should be used to make 5 lb of a mixture that costs S4
...
30 per ounce with an alloy that
costs S1
...
How many ounces of each were used to make a mixture
of 200 oz costing S2
...
How many
kilograms of each were used to make a 5 kg mixture that costs S4
...
How much of the 60% mixture is used?
30) How many ounces of water evaporated from 50 oz of a 12% salt solution to
produce a 15% salt solution?
31) A caterer made an ice cream punch by combining fruit juice that cost S2
...
25 per gallon
...
50 per pound?
32) A clothing manufacturer has some pure silk thread and some thread that is
85% silk
...
How many pounds of each fiber should be woven together to produce
600 lb of a fabric that is 28% wool?
34) How many pounds of coffee that is 40% java beans must be mixed with 80 lb
of coffee that is 30% java beans to make a coffee blend that is 32% java
beans?
35) The manager of a specialty food store combined almonds that cost S4
...
50 per pound
...
24 per pound?
36) A tea that is 20% jasmine is blended with a tea that is 15% jasmine
...
20 per
pound with meat that costs S4
...
How many pounds of each
were used to make a 50 lb mixture tha costs S3
...
10 per pound with
pine wood chips that cost S2
...
How many pounds of each were
used to make an 80 lb mixture costing S2
...
1 Exponent Properties
...
2 Negative Exponents
...
3 Scientific Notation
...
4 Introduction to Polynomials
...
5 Multiply Polynomials
...
6 Multiply Special Products
...
7 Divide Polynomials
...
1
Polynomials - Exponent Properties
Objective: Simplify expressions using the properties of exponents
...
Exponents represent repeated multiplication
...
World View Note: The word exponent comes from the Latin “expo” meaning
out of and “ponere” meaning place
...
a3a2
(aaa)(aa)
a5
Expand exponents to multiplication problem
Now we have 5a ′s being multiplied together
Our Solution
A quicker method to arrive at our answer would have been to just add the exponents: a3a2 = a3+2 = a5 This is known as the product rule of exponents
Product Rule of Exponents: aman = am+n
The product rule of exponents can be used to simplify many problems
...
This is shown in the following examples
Example 197
...
2x3 y 5z · 5xy 2z 3
10x4 y 7z 4
Multiply 2 · 5, add exponents on x, y and z
Our Solution
Rather than multiplying, we will now try to divide with exponents
Example 199
...
This is known as the quotient rule of exponents
...
This is shown in the following
examples
...
713
75
78
Same base, subtract the exponents
Our Solution
Example 201
...
This is investigated in the following example
...
3
a2
a2 · a2 · a2
a6
This means we have a2 three times
Add exponents
Our solution
A quicker method to arrive at the solution would have been to just multiply the
exponents, (a2)3 = a2·3 = a6
...
Power of a Power Rule of Exponents: (am)n = amn
This property is often combined with two other properties which we will investigate now
...
(ab)3
(ab)(ab)(ab)
a3b3
This means we have (ab) three times
Three a ′s and three b ′s can be written with exponents
Our Solution
178
A quicker method to arrive at the solution would have been to take the exponent
of three and put it on each factor in parenthesis, (ab)3 = a3b3
...
Power of a Product Rule of Exponents: (ab)m = ambm
It is important to be careful to only use the power of a product rule with multiplication inside parenthesis
...
Warning 204
...
Example 205
...
This is known
as the power of a quotient rule of exponents
...
This is shown in the following examples
...
(x3 yz 2)4
x12y 4z 8
Put the exponent of 4 on each factor, multiplying powers
Our solution
179
Example 207
...
An expressions such as 53 does not mean we multipy 5 by 3,
rather we multiply 5 three times, 5 × 5 × 5 = 125
...
Example 208
...
Never multipy a base by the exponent
...
In this lesson we have discussed 5 different exponent properties
...
Rules of Exponents
aman = am+n
am
Quotient Rule of Exponents
= am−n
an
Power of a Power Rule of Exponents
(am)n = amn
Power of a Product Rule of Exponents (ab)m = ambm
a m am
Power of a Quotient Rule of Exponents
= m
b
b
Product Rule of Exponents
These five properties are often mixed up in the same problem
...
However, order of operations still
applies to a problem
...
This is illustrated in the next few examples
...
(4x3 y · 5x4 y 2)3
(20x7 y 3)3
203x21y 9
8000x21y 9
In parenthesis simplify using product rule, adding exponents
With power rules, put three on each factor, multiplying exponents
Evaluate 203
Our Solution
180
Example 210
...
3a3b · 10a4b3
2a4b2
Simplify numerator with product rule, adding exponents
30a7b4
2a4b2
Now use the quotient rule to subtract exponents
15a3b2
Our Solution
Example 212
...
3ab2(2a4b2)3
6a5b7
2
3ab2(8a12b6)
6a5b7
2
24a13b8
6a5b7
2
Simplify inside parenthesis first, using power rule in numerator
Simplify numerator using product rule
4a8b)2
16a16b2
Simplify using the quotient rule
Now that the parenthesis are simplified, use the power rules
Our Solution
Clearly these problems can quickly become quite involved
...
181
5
...
2) 4 · 44 · 42
1) 4 · 44 · 44
3) 4 · 22
4) 3 · 33 · 32
5) 3m · 4mn
6) 3x · 4x2
7) 2m4n2 · 4nm2
8) x2 y 4 · xy 2
9) (33)4
10) (43)4
11) (44)2
12) (32)3
13) (2u3v 2)2
14) (xy)3
15) (2a4)4
17)
16) (2xy)4
45
18)
19)
21)
23)
4x3 y 4
3xy 3
25) (x3 y 4 · 2x2 y 3)2
27) 2x(x4 y 4)4
29)
31)
33)
35)
26) (u2v 2 · 2u4)3
3vu5 · 2v3
uv 2 · 2u3v
30)
2ba7 · 2b4
ba2 · 3a3b4
32)
2a2b2a7
(ba4)2
34)
yx2 · (y 4)2
2y 4
36)
n3(n4)2
2mn
38)
(2y 3x2)2
2x2 y 4 · x2
40)
2
2x4 y 5 · 2z 10 x2 y 7
(xy 2z 2)4
42)
2q 3 p3r 4 · 2p3
(qrp3)2
x3
2y 17
(2x2 y 4)4
3
3
2m n4 · 2m4n4
mn4
37)
2xy 5 · 2x2 y 3
2xy 4 · y 3
39)
q 3r 2 · (2p2 q 2r 3)2
2p3
41)
zy 3 · z 3 x4 y 4
x3 y 3z 3
43)
xy 3
4xy
28)
2x 7 y 5
3x3 y · 4x2 y 3
(2x)3
x2 y 4
4xy
24)
3nm2
3n
34
3
22)
32
3
37
33
20)
43
4
2x2 y 2z 6 · 2zx2 y 2
(x2z 3)2
182
4
5
...
There are a few special exponent properties that deal with exponents that are not
positive
...
a3
a3
Use the quotient rule to subtract exponents
a0
Our Solution, but now we consider the problem a the second way:
a3
a3
Rewrite exponents as repeated multiplication
aaa
aaa
Reduce out all the a ′s
1
=1
1
Our Solution, when we combine the two solutions we get:
a0 = 1
Our final result
...
This is illustrated in the following example
...
(3x2)0
1
Zero power rule
Our Solution
Another property we will consider here deals with negative exponents
...
183
Example 216
...
Rewrite exponents as repeated multiplication
Reduce three a ′s out of top and bottom
1
aa
1
a2
a−2 =
Simplify to exponents
Our Solution, putting these solutions together gives:
1
a2
Our Final Solution
This example illustrates an important property of exponents
...
Once we take the reciprical the exponent is now
positive
...
Following are
the rules of negative exponents
a−m =
Rules of Negative Exponets:
1
a−m
a
b
1
m
= am
−m
=
bm
am
Negative exponents can be combined in several different ways
...
When the base with exponent moves,
the exponent is now positive
...
Example 217
...
Also, it is important to remember that exponents only effect
what they are attached to
...
We now have the following nine properties of exponents
...
Properties of Exponents
aman = am+n
(ab)m = ambm
am
= am−n
an
(am)n = amn
a
b
m
=
am
bm
a0 = 1
a−m =
1
a−m
a
b
1
am
= am
−m
=
bm
am
World View Note: Nicolas Chuquet, the French mathematician of the 15th cen¯
tury wrote 121m to indicate 12x−1
...
Simplifying with negative exponents is much the same as simplifying with positive
exponents
...
As we do this it is important to be very careful of rules for
adding, subtracting, and multiplying with negatives
...
4x−5 y −3 · 3x3 y −2
6x−5 y 3
12x−2 y −5
6x−5 y 3
2x3 y −8
2x3
y8
Simplify numerator with product rule, adding exponents
Quotient rule to subtract exponets, be careful with negatives!
( − 2) − ( − 5) = ( − 2) + 5 = 3
( − 5) − 3 = ( − 5) + ( − 3) = − 8
Negative exponent needs to move down to denominator
Our Solution
185
Example 219
...
We
did not make the number negative! Negative exponents never make the bases negative, they simply mean we have to take the reciprocal of the base
...
Example 220
...
2 Practice - Negative Exponents
Simplify
...
1) 2x4 y −2 · (2xy 3)4
2) 2a−2b−3 · (2a0b4)4
3) (a4b−3)3 · 2a3b−2
4) 2x3 y 2 · (2x3)0
5) (2x2 y 2)4x−4
6) (m0n3 · 2m−3n−3)0
7) (x3 y 4)3 · x−4 y 4
8) 2m−1n−3 · (2m−1n−3)4
9)
2x−3 y 2
3x −3 y 3 · 3x0
10)
3y 3
3yx3 · 2x4 y −3
3x3 y 2
11)
4xy −3 · x−4 y 0
4y −1
12)
13)
u2v −1
2u0v 4 · 2uv
14)
2xy 2 · 4x3 y −4
4x−4 y −4 · 4x
15)
u2
4u0v 3 · 3v 2
16)
2x −2 y 2
4yx2
17)
2y
(x0 y 2)4
18)
(a4)4
2b
19) (
2a2b3 4
)
a−1
4y −2 · 3x−2 y −4
20) (
2y −4 −2
)
x2
21)
2nm4
(2m2n2)4
22)
23)
(2mn)4
m0n−2
24)
2x −3
(x4 y −3) −1
25)
y 3 · x −3 y 2
(x4 y 2)3
26)
2x −2 y 0 · 2xy 4
(xy 0)−1
27)
2u −2v 3 · (2uv 4)−1
2u−4v 0
28)
2yx2 · x −2
(2x0 y 4)−1
30)
u −3v −4
2v(2u −3v 4)0
29) (
2x0 · y 4 3
)
y4
2y 2
(x4 y 0)−4
31)
y(2x4 y 2)2
2x4 y 0
32)
33)
2yzx2
2x4 y 4z −2 · (zy 2)4
34)
35)
2kh0 · 2h −3k0
(2kj 3)2
36) (
37)
(cb3)2 · 2a −3b2
(a3b−2c3)3
38)
39)
(yx −4z 2)−1
z 3 · x2 y 3z −1
40)
187
b−1
(2a4b0)0 · 2a −3b2
2b4c−2 · (2b3c2)−4
a −2b4
(2x−3 y 0z −1)3 · x −3 y 2 −2
)
2x3
2q 4 · m2 p2 q 4
(2m−4 p2)3
2mpn −3
(m0n −4 p2)3 · 2n2 p0
5
...
One application of exponent properties comes from scientific notation
...
An example of
really large numbers would be the distance that light travels in a year in miles
...
Doing basic operations such as multiplication and division with these
numbers would normally be very combersome
...
First we will take a look at what scientific notation is
...
Scientific Notation: a × 10b where 1
a < 10
The exponent, b, is very important to how we convert between scientific notation
and normal numbers, or standard notation
...
Multiplying by 10 in affect moves the decimal point
one place
...
To decide which direction to
move the decimal (left or right) we simply need to remember that positive exponents mean in standard notation we have a big number (bigger than ten) and negative exponents mean in standard notation we have a small number (less than
one)
...
Example 221
...
42
× 104
1
...
Convert 0
...
2
× 10−3
4
...
Convert 3
...
Move decimal right 5 places
Our Solution
Example 224
...
4 × 10−3 to standard notation
0
...
Move decimal left 3 places
Our Solution
Converting between standard notation and scientific notation is important to
understand how scientific notation works and what it does
...
The way we do this is first do the operation with the front
number (multiply or divide) then use exponent properties to simplify the 10’s
...
The negative exponent simply informs us that we are
dealing with small numbers
...
Example 225
...
1 × 10−7)(3
...
1)(3
...
77
10−7105 = 10−2
7
...
4
...
1 × 10−3
Deal with numbers and 10 ′s separately
4
...
6
3
...
6 × 10
7
Be careful with negatives, 4 − ( − 3) = 4 + 3 = 7
Our Solution
189
Example 227
...
8 × 10−4)3
1
...
832
(10−4)3 = 10−12
5
...
83
Multiply exponents
Our Solution
Often when we multiply or divide in scientific notation the end result is not in scientific notation
...
This is shown in the following examples
...
(4
...
1 × 109)
(4
...
1) = 28
...
867 × 101
10110−3109 = 107
2
...
He used his system to calculate the number of grains of sand it
would take to fill the universe
...
Example 229
...
014 × 10−3
3
...
014
= 0
...
8
Divide numbers
0
...
3 × 10−1
10 −110−3
= 103
10−7
5
...
3 Practice - Scientific Notation
Write each number in scientific notiation
1) 885
2) 0
...
081
4) 1
...
039
6) 15000
Write each number in standard notation
7) 8
...
56 x 102
9) 9 x 10−4
10) 5 x 104
11) 2 x 100
12) 6 x 10−5
Simplify
...
13) (7 x 10−1)(2 x 10−3)
15) (5
...
16 x 10−2)
17) (2
...
8 × 10−5)
16) (5
...
84 × 10−1)
18)
7
...
7 × 10−4
19)
4
...
7 × 10−3
20)
21)
5
...
62 × 10−2
7
...
32 × 10−1
22)
3
...
02 × 100
23) (5
...
8 × 10 )
4 −4
27) (8
...
1 × 10−6
5
...
8 × 10 )
35)
37)
3
...
4 × 10−6
6
...
8 × 10−3
26) (5
...
88 × 10−4)(4
...
6 × 100)(6
...
83 × 10−2
24) (9
...
4 × 105
7 × 10−2
32) (3
...
58 × 10−2
1
...
3 × 101)5
38)
5 × 106
6
...
4
Polynomials - Introduction to Polynomials
Objective: Evaluate, add, and subtract polynomials
...
Polynomials are made up of terms
...
For example, 5x, 2y 2, − 5, ab3c, and x are all terms
...
Expressions are often named based on
the number of terms in them
...
A binomial has two terms, such as a2 − b2
...
The term polynomial means many terms
...
If we know what the variable in a polynomial represents we can replace the variable with the number and evaluate the polynomial as shown in the following
example
...
2x2 − 4x + 6 when x = − 4
2( − 4)2 − 4( − 4) + 6
2(16) − 4( − 4) + 6
32 + 16 + 6
54
Replace variable x with − 4
Exponents first
Multiplication (we can do all terms at once)
Add
Our Solution
It is important to be careful with negative variables and exponents
...
This means −
32 = − 9 because the exponent is only attached to the 3
...
When
we replace a variable with parenthesis like in the previous example, the substituted value is in parenthesis
...
However, consider the next example
...
− x2 + 2x + 6 when x = 3
− (3)2 + 2(3) + 6
− 9 + 2(3) + 6
−9+6+6
3
Replace variable x with 3
Exponent only on the 3, not negative
Multiply
Add
Our Solution
192
World View Note: Ada Lovelace in 1842 described a Difference Engine that
would be used to caluclate values of polynomials
...
Generally when working with polynomials we do not know the value of the variable, so we will try and simplify instead
...
When adding polynomials we are mearly combining like terms
...
(4x3 − 2x + 8) + (3x3 − 9x2 − 11)
7x3 − 9x2 − 2x − 3
Combine like terms 4x3 + 3x3 and 8 − 11
Our Solution
Generally final answers for polynomials are written so the exponent on the variable counts down
...
Subtracting polynomials is almost as fast
...
When we have a negative in
front of parenthesis we distribute it through, changing the signs of everything
inside
...
Example 233
...
Example 234
...
4 Practice - Introduction to Polynomials
Simplify each expression
...
5
Polynomials - Multiplying Polynomials
Objective: Multiply polynomials
...
We will first look at multiplying monomials, then monomials by
polynomials and finish with polynomials by polynomials
...
This is shown in the next example
...
(4x3 y 4z)(2x2 y 6z 3)
8x5 y 10z 4
Multiply numbers and add exponents for x, y, and z
Our Solution
In the previous example it is important to remember that the z has an exponent
of 1 when no exponent is written
...
Be very careful with exponents in polynomials
...
Next we consider multiplying a monomial by a polynomial
...
Here we will see the exact
same process
...
4x3(5x2 − 2x + 5)
20x5 − 8x4 + 20x3
Distribute the 4x3, multiplying numbers, adding exponents
Our Solution
Following is another example with more variables
...
Example 237
...
All of which
work, often students prefer the method they are first taught
...
All three methods will be used to solve the same two multiplication problems
...
As we do this we take each term of the second polynomial and put it
in front of the first polynomial
...
(4x + 7y)(3x − 2y)
3x(4x + 7y) − 2y(4x + 7y)
12x2 + 21xy − 8xy − 14y 2
12x2 + 13xy − 14y 2
Distribute (4x + 7y) through parenthesis
Distribute the 3x and − 2y
Combine like terms 21xy − 8xy
Our Solution
This example illustrates an important point, the negative/subtraction sign stays
with the 2y
...
Multiplying by distributing can easily be extended to problems with more terms
...
(2x − 5)(4x2 − 7x + 3)
4x2(2x − 5) − 7x(2x − 5) + 3(2x − 5)
8x3 − 20x2 − 14x2 + 35x + 6x − 15
8x3 − 34x2 + 41x − 15
Distribute (2x − 5) through parenthesis
Distribute again through each parenthesis
Combine like terms
Our Solution
This process of multiplying by distributing can easily be reversed to do an important procedure known as factoring
...
Multiply by FOIL
Another form of multiplying is known as FOIL
...
The
letters of FOIL help us remember every combination
...
O stand for Outside, we multiply the outside
two terms
...
L stands for
Last, we multiply the last term of each binomial
...
(4x + 7y)(3x − 2y)
(4x)(3x) = 12x2
(4x)( − 2y) = − 8xy
(7y)(3x) = 21xy
(7y)( − 2y) = − 14y 2
12x2 − 8xy + 21xy − 14y 2
12x2 + 13xy − 14y 2
Use FOIL to multiply
F − First terms (4x)(3x)
O − Outside terms (4x)( − 2y)
I − Inside terms (7y)(3x)
L − Last terms (7y)( − 2y)
Combine like terms − 8xy + 21xy
Our Solution
197
Some students like to think of the FOIL method as distributing the first term 4x
through the (3x − 2y) and distributing the second term 7y through the (3x − 2y)
...
Example 241
...
Multiplying in rows
A third method for multiplying polynomials looks very similar to multiplying
numbers
...
The same process can be done with polynomials
...
Example 242
...
Line up like terms
Add like terms to get Our Solution
This same process is easily expanded to a problem with more terms
...
(2x − 5)(4x2 − 7x + 3)
4x3 − 7x + 3
× 2x − 5
2
− 20x + 35x − 15
3
8x − 14x2 + 6x
8x3 − 34x2 + 41x − 15
Rewrite as vertical problem
Put polynomial with most terms on top
Multiply − 5 by each term
Multiply 2x by each term
...
It is suggested
that you are very comfortable with at least one of these methods as you work
through the practice problems
...
Example 244
...
This is shown in the last example
...
3(2x − 4)(x + 5)
3(2x + 10x − 4x − 20)
3(2x2 + 6x − 20)
6x2 + 18x − 60
2
Multiply the binomials, we will use FOIL
Combine like terms
Distribute the 3
Our Solution
A common error students do is distribute the three at the start into both parenthesis
...
Be careful of this error
...
199
5
...
1) 6(p − 7)
2) 4k(8k + 4)
3) 2(6x + 3)
4) 3n2(6n + 7)
5) 5m4(4m + 4)
6) 3(4r − 7)
7) (4n + 6)(8n + 8)
8) (2x + 1)(x − 4)
9) (8b + 3)(7b − 5)
10) (r + 8)(4r + 8)
11) (4x + 5)(2x + 3)
12) (7n − 6)(n + 7)
13) (3v − 4)(5v − 2)
14) (6a + 4)(a − 8)
15) (6x − 7)(4x + 1)
16) (5x − 6)(4x − 1)
17) (5x + y)(6x − 4y)
18) (2u + 3v)(8u − 7v)
19) (x + 3y)(3x + 4y)
20) (8u + 6v)(5u − 8v)
21) (7x + 5y)(8x + 3y)
22) (5a + 8b)(a − 3b)
23) (r − 7)(6r 2 − r + 5)
24) (4x + 8)(4x2 + 3x + 5)
25) (6n − 4)(2n2 − 2n + 5)
26) (2b − 3)(4b2 + 4b + 4)
27) (6x + 3y)(6x2 − 7xy + 4y 2)
28) (3m − 2n)(7m2 + 6mn + 4n2)
29) (8n2 + 4n + 6)(6n2 − 5n + 6)
30) (2a2 + 6a + 3)(7a2 − 6a + 1)
31) (5k 2 + 3k + 3)(3k 2 + 3k + 6)
32) (7u2 + 8uv − 6v 2)(6u2 + 4uv + 3v 2)
33) 3(3x − 4)(2x + 1)
34) 5(x − 4)(2x − 3)
35) 3(2x + 1)(4x − 5)
36) 2(4x + 1)(2x − 6)
37) 7(x − 5)(x − 2)
38) 5(2x − 1)(4x + 1)
39) 6(4x − 1)(4x + 1)
40) 3(2x + 3)(6x + 9)
200
5
...
There are a few shortcuts that we can take when multiplying polynomials
...
These shortcuts will also be useful to us as our study of algebra continues
...
A sum and a difference is easily recognized as the numbers and variables are exactly the same, but
the sign in the middle is different (one sum, one difference)
...
Example 246
...
Rather than going through all this work, when we have a sum and a difference we
will jump right to our solution by squaring the first term and squaring the last
term, putting a subtraction between them
...
(x − 5)(x + 5)
x2 − 25
Recognize sum and difference
Square both, put subtraction between
...
Often students ask if they can just multiply out using another method
and not learn the shortcut
...
For this
reason it is very important to be able to recognize these shortcuts
...
201
Example 248
...
Our Solution
Example 249
...
Our Solution
It is interesting to note that while we can multiply and get an answer like a2 − b2
(with subtraction), it is impossible to multiply real numbers and end up with a
product such as a2 + b2 (with addition)
...
These are easy
to recognize as we will have a binomial with a 2 in the exponent
...
(a + b)2
(a + b)(a + b)
a(a + b) + b(a + b)
a2 + ab + ab + b2
a2 + 2ab + b2
Squared is same as multiplying by itself
Distribute (a + b)
Distribute again through final parenthesis
Combine like terms ab + ab
Our Solution
This problem also helps us find our shortcut for multiplying
...
The middle term is 2 times
the first term times the second term
...
This can be shortened to square the first, twice the product, square the last
...
This is illustrated in the
following example
Example 251
...
A common error is to do the following: (x − 5)2 = x2 − 25
(or x2 + 25)
...
This is
why it is important to use the shortcut to help us find the correct solution
...
This is illustrated in the next
examples
...
(2x + 5)2
(2x)2 = 4x2
2(2x)(5) = 20x
52 = 25
4x2 + 20x + 25
Recognize perfect square
Square the first
Twice the product
Square the last
Our Solution
Example 253
...
Our Solution
Example 254
...
Our Solution
These two formulas will be important to commit to memory
...
The final
example covers both types of problems (two perfect squares, one positive, one
negative), be sure to notice the difference between the examples and how each formula is used
Example 255
...
While French mathematician Blaise
Pascal often gets credit for working with these expansions of binomials in the 17th
century, Chinese mathematicians had been working with them almost 400 years
earlier!
203
5
...
1) (x + 8)(x − 8)
2) (a − 4)(a + 4)
3) (1 + 3p)(1 − 3p)
4) (x − 3)(x + 3)
5) (1 − 7n)(1 + 7n)
6) (8m + 5)(8m − 5)
7) (5n − 8)(5n + 8)
8) (2r + 3)(2r − 3)
9) (4x + 8)(4x − 8)
10) (b − 7)(b + 7)
11) (4y − x)(4y + x)
13) (4m − 8n)(4m + 8n)
15) (6x − 2y)(6x + 2y)
2
17) (a + 5)
12) (7a + 7b)(7a − 7b)
14) (3y − 3x)(3y + 3x)
16) (1 + 5n)2
18) (v + 4)2
2
20) (1 − 6n)2
2
22) (7k − 7)2
19) (x − 8)
21) (p + 7)
23) (7 − 5n)2
25) (5m − 8)2
27) (5x + 7y)2
29) (2x + 2y)2
31) (5 + 2r)2
24) (4x − 5)2
26) (3a + 3b)2
28) (4m − n)2
30) (8x + 5y)2
32) (m − 7)2
34) (8n + 7)(8n − 7)
33) (2 + 5x)2
36) (b + 4)(b − 4)
35) (4v − 7) (4v + 7)
38) (7x + 7)2
37) (n − 5)(n + 5)
40) (3a − 8)(3a + 8)
39) (4k + 2)2
204
5
...
Dividing polynomials is a process very similar to long division of whole numbers
...
The way we do this is very similar to distributing,
but the operation we distribute is the division, dividing each term by the monomial and reducing the resulting expression
...
9x5 + 6x4 − 18x3 − 24x2
3x2
9x5 6x4 18x3 24x2
+
−
−
3x2
3x2
3x2 3x2
3x3 + 2x2 − 6x − 8
Divide each term in the numerator by 3x2
Reduce each fraction, subtracting exponents
Our Solution
Example 257
...
Also
4x2
interesting in this problem is the second term 4x2 divided out completely
...
Long division is required when we divide by more than just a monomial
...
205
An example is given to review the (general) steps that are used with whole numbers that we will also use with polynomials
Example 258
...
The only difference is we
will replace the word “number” with the word “term”
Dividing Polynomials
1
...
Multiply this term by the divisor
206
3
...
Bring down the next term
5
...
Be sure not to miss this step! This process is illustrated in
the following two examples
...
3x3 − 5x2 − 32x + 7
x−4
Rewrite problem as long division
3x3
= 3x2
x
x − 4|3x3 − 5x2 − 32x + 7
Divide front terms:
3x2
x − 4|3x3 − 5x2 − 32x + 7
− 3x3 + 12x2
7x2 − 32x
Multiply this term by divisor: 3x2(x − 4) = 3x3 − 12x2
Change the signs and combine
Bring down the next term
3x2 + 7x
x − 4|3x3 − 5x2 − 32x + 7
− 3x3 + 12x2
7x2 − 32x
− 7x2 + 28x
− 4x + 7
3x2 + 7x − 4
x − 4|3x3 − 5x2 − 32x + 7
− 3x3 + 12x2
7x2 − 32x
− 7x2 + 28x
− 4x + 7
+ 4x − 16
−9
Repeat, divide front terms:
7x2
= 7x
x
Multiply this term by divisor: 7x(x − 4) = 7x2 − 28x
Change the signs and combine
Bring down the next term
Repeat, divide front terms:
− 4x
=−4
x
Multiply this term by divisor: − 4(x − 4) = − 4x + 16
Change the signs and combine
Remainder put over divisor and subtracted (due to negative)
207
3x2 + 7x − 4 −
9
x−4
Our Solution
Example 260
...
This is very
important in long division, the variables must count down and no exponent can
be skipped
...
If an exponent
is skipped we will have to add a term to the problem, with zero for its coefficient
...
208
Example 261
...
If so we will have to adjust the
problem
...
World View Note: Paolo Ruffini was an Italian Mathematician of the early
19th century
...
209
5
...
1)
20x4 + x3 + 2x2
4x3
2)
5x4 + 45x3 + 4x2
9x
3)
20n4 + n3 + 40n2
10n
4)
3k 3 + 4k2 + 2k
8k
5)
12x4 + 24x3 + 3x2
6x
6)
5p4 + 16p3 + 16p2
4p
7)
10n4 + 50n3 + 2n2
10n2
8)
3m4 + 18m3 + 27m2
9m2
9)
x2 − 2x − 71
x+8
10)
r 2 − 3r − 53
r−9
11)
n2 + 13n + 32
n+5
12)
13)
v 2 − 2v − 89
v − 10
b2 − 10b + 16
b−7
14)
x2 + 4x − 26
x+7
15)
a2 − 4a − 38
a−8
16)
17)
45p2 + 56p + 19
9p + 4
x2 − 10x + 22
x−4
18)
19)
10x2 − 32x + 9
10x − 2
48k2 − 70k + 16
6k − 2
20)
n2 + 7n + 15
n+4
21)
4r 2 − r − 1
4r + 3
22)
23)
n2 − 4
n−2
3m2 + 9m − 9
3m − 3
24)
2x2 − 5x − 8
2x + 3
25)
27b2 + 87b + 35
3b + 8
26)
27)
4x2 − 33x + 28
4x − 5
3v 2 − 32
3v − 9
28)
4n2 − 23n − 38
4n + 5
29)
a3 + 15a2 + 49a − 55
a+7
30)
31)
x3 − 26x − 41
x+4
8k 3 − 66k 2 + 12k + 37
k−8
32)
33)
3n3 + 9n2 − 64n − 68
n+6
x3 − 16x2 + 71x − 56
x−8
34)
35)
x3 − 46x + 22
x+7
k3 − 4k 2 − 6k + 4
k−1
36)
37)
9p3 + 45p2 + 27p − 5
9p + 9
2n3 + 21n2 + 25n
2n + 3
38)
39)
r 3 − r 2 − 16r + 8
r −4
8m3 − 57m2 + 42
8m + 7
40)
41)
12n3 + 12n2 − 15n − 4
2n + 3
2x3 + 12x2 + 4x − 37
2x + 6
42)
43)
4v 3 − 21v 2 + 6v + 19
4v + 3
24b3 − 38b2 + 29b − 60
4b − 7
210
Chapter 6 : Factoring
6
...
212
6
...
216
6
...
221
6
...
226
6
...
229
6
...
234
6
...
237
211
6
...
The opposite of multiplying polynomials together is factoring polynomials
...
We use factored polynomials to
help us solve equations, learn behaviors of graphs, work with fractions and more
...
In this lesson we will focus on factoring using the greatest common factor or GCF
of a polynomial
...
In this lesson we will work the same problem backwards
...
To do this we have to be able to first identify what is the GCF of a polynomial
...
To
find a GCF of sevearal numbers we are looking for the largest number that can be
divided by each of the numbers
...
Find the GCF of 15, 24, and 27
15
24
27
= 5,
= 6,
=9
3
3
3
GCF = 3
Each of the numbers can be divided by 3
Our Solution
When there are variables in our problem we can first find the GCF of the num-
212
bers using mental math, then we take any variables that are in common with each
term, using the lowest exponent
...
GCF of 24x4 y 2z, 18x2 y 4, and 12x3 yz 5
24
18
12
= 4,
= 3,
=2
6
6
6
x2 y
GCF = 6x2 y
Each number can be divided by 6
x and y are in all 3, using lowest exponets
Our Solution
To factor out a GCF from a polynomial we first need to identify the GCF of all
the terms, this is the part that goes in front of the parenthesis, then we divide
each term by the GCF, the answer is what is left inside the parenthesis
...
4x2 − 20x + 16
4x
− 20x
16
= x2,
= − 5x,
=4
4
4
4
4(x2 − 5x + 4)
GCF is 4, divide each term by 4
2
This is what is left inside the parenthesis
Our Solution
With factoring we can always check our solutions by multiplying (distributing in
this case) out the answer and the solution should be the original equation
...
25x4 − 15x3 + 20x2
25x
− 15x
20x2
= 5x2,
= − 3x,
=4
5x2
5x2
5x2
5x2(5x2 − 3x + 4)
4
GCF is 5x2, divide each term by this
3
This is what is left inside the parenthesis
Our Solution
Example 266
...
21x3 + 14x2 + 7x
14x2
7x
21x
= 3x2,
= 2x,
=1
7x
7x
7x
7x(3x2 + 2x + 1)
GCF is 7x, divide each term by this
3
This is what is left inside the parenthesis
Our Solution
It is important to note in the previous example, that when the GCF was 7x and
7x was one of the terms, dividing gave an answer of 1
...
Anything divided by itself is 1, be sure to not forget to put the 1 into
the solution
...
We can
simply identify the GCF and put it in front of the parenthesis as shown in the following two examples
...
12x5 y 2 − 6x4 y 4 + 8x3 y 5
2x3 y 2(6x2 − 3xy 2 + 4y 3)
GCF is 2x3 y 2, put this in front of parenthesis and divide
Our Solution
Example 269
...
Be very careful that each term is accounted for in your final solution
...
1 Practice - Greatest Common Factor
Factor the common factor out of each expression
...
2
Factoring - Grouping
Objective: Factor polynomials with four terms using grouping
...
This
GCF is often a monomial like in the problem 5x y + 10xz the GCF is the monomial 5x, so we would have 5x(y + 2z)
...
To see this, consider the following two example
...
3ax − 7bx
x(3a − 7b)
Both have x in common, factor it out
Our Solution
Now the same problem, but instead of x we have (2a + 5b)
...
3a(2a + 5b) − 7b(2a + 5b)
(2a + 5b)(3a − 7b)
Both have (2a + 5b) in common, factor it out
Our Solution
In the same way we factored out a GCF of x we can factor out a GCF which is a
binomial, (2a + 5b)
...
Here we will have to use another strategy to factor
...
Grouping is how we will factor if there are four
terms in the problem
...
Example 272
...
We arrived at the solution by looking at the
two parts, 5b(2a + 3) and 2(2a + 3)
...
Then we can factor the GCF out of both the left and right sides
...
If they match we can pull this matching GCF out front, putting the rest in
parenthesis and we will be factored
...
216
Example 273
...
If there is any difference between
the two we either have to do some adjusting or it can’t be factored using the
grouping method
...
Example 274
...
Instead of 7 we will use − 7
...
3x(2x + 3y) − 7(2x + 3y)
(2x + 3y)(3x − 7)
(2x + 3y) is the same! Factor out this GCF
Our Solution
Often we can recognize early that we need to use the opposite of the GCF when
factoring
...
If it is negative
then we will use the opposite of the GCF to be sure they match
...
5xy − 8x − 10y + 16
5xy − 8x − 10y + 16
x(5y − 8) − 2(5y − 8)
(5y − 8)(x − 2)
Split the problem into two groups
GCF on left is x, on right we need a negative,
so we use − 2
(5y − 8) is the same! Factor out this GCF
Our Solution
217
Sometimes when factoring the GCF out of the left or right side there is no GCF
to factor out
...
Often this is all
we need to be sure the two binomials match
...
12ab − 14a − 6b + 7
12ab − 14a − 6b + 7
2a(6b − 7) − 1(6b − 7)
(6b − 7)(2a − 1)
Split the problem into two groups
GCF on left is 2a, on right, no GCF, use − 1
(6b − 7) is the same! Factor out this GCF
Our Solution
Example 277
...
In this case we may have to adjust the problem slightly
...
To do this we will move
the second term to the end of the problem and see if that helps us use grouping
...
4a2 − 21b3 + 6ab − 14ab2
2
4a − 21b
3
+ 6ab − 14ab
2
1(4a2 − 21b3) + 2ab(3 − 7b)
4a2 + 6ab − 14ab2 − 21b3
4a2 + 6ab − 14ab2 − 21b3
2a(2a + 3b) − 7b2(2a + 3b)
(2a + 3b)(2a − 7b2)
Split the problem into two groups
GCF on left is 1, on right is 2ab
Binomials don ′t match! Move second term to end
Start over, split the problem into two groups
GCF on left is 2a, on right is − 7b2
(2a + 3b) is the same! Factor out this GCF
Our Solution
When rearranging terms the problem can still be out of order
...
There are two ways that this can
happen, one with addition, one with subtraction
...
This is because addition is the same in either order (5 + 3 = 3 + 5 = 8)
...
7 + y − 3xy − 21x
7 + y − 3xy − 21x
1(7 + y) − 3x(y + 7)
(y + 7)(1 − 3x)
Split the problem into two groups
GCF on left is 1, on the right is − 3x
y + 7 and 7 + y are the same, use either one
Our Solution
However, if the binomial has subtraction, then we need to be a bit more careful
...
Notice what happens when we
factor out − 1
...
(b − a)
− 1( − b + a)
− 1(a − b)
Factor out − 1
Addition can be in either order, switch order
The order of the subtraction has been switched!
Generally we won’t show all the above steps, we will simply factor out the opposite of the GCF and switch the order of the subtraction to make it match the
other binomial
...
8xy − 12y + 15 − 10x
8xy − 12y 15 − 10x
4y(2x − 3) + 5(3 − 2x)
4y(2y − 3) − 5(2x − 3)
(2x − 3)(4y − 5)
Split the problem into two groups
GCF on left is 4y, on right, 5
Need to switch subtraction order, use − 5 in middle
Now 2x − 3 match on both! Factor out this GCF
Our Solution
World View Note: Sofia Kovalevskaya of Russia was the first woman on the
editorial staff of a mathematical journal in the late 19th century
...
219
6
...
1) 40r 3 − 8r 2 − 25r + 5
2) 35x3 − 10x2 − 56x + 16
3) 3n3 − 2n2 − 9n + 6
4) 14v 3 + 10v 2 − 7v − 5
5) 15b3 + 21b2 − 35b − 49
6) 6x3 − 48x2 + 5x − 40
7) 3x3 + 15x2 + 2x + 10
8) 28p3 + 21p2 + 20p + 15
9) 35x3 − 28x2 − 20x + 16
10) 7n3 + 21n2 − 5n − 15
11) 7xy − 49x + 5y − 35
12) 42r3 − 49r 2 + 18r − 21
13) 32xy + 40x2 + 12y + 15x
14) 15ab − 6a + 5b3 − 2b2
15) 16xy − 56x + 2y − 7
16) 3mn − 8m + 15n − 40
17) 2xy − 8x2 + 7y 3 − 28y 2x
18) 5mn + 2m − 25n − 10
19) 40xy + 35x − 8y 2 − 7y
20) 8xy + 56x − y − 7
21) 32uv − 20u + 24v − 15
22) 4uv + 14u2 + 12v + 42u
23) 10xy + 30 + 25x + 12y
24) 24xy + 25y 2 − 20x − 30y 3
25) 3uv + 14u − 6u2 − 7v
26) 56ab + 14 − 49a − 16b
27) 16xy − 3x − 6x2 + 8y
220
6
...
Factoring with three terms, or trinomials, is the most important type of factoring
to be able to master
...
Example 282
...
This is because it is grouping! The GCF of the left two terms is x and
the GCF of the second two terms is − 4
...
This
is shown in the following example, the same problem worked backwards
Example 283
...
Why did
we pick + 6x − 4x and not + 5x − 3x? The reason is because 6x − 4x is the only
combination that works! So how do we know what is the one combination that
works? To find the correct way to split the middle term we will use what is called
the ac method
...
The way the ac method works is we find a pair of numers that multiply to a certain number and add to another number
...
In the previous
221
example that would mean we wanted to multiply to − 24 and add to + 2
...
This process is shown in the next few examples
Example 284
...
x2 − 4x + 3
x2 − 3x − x + 3
x(x − 3) − 1(x − 3)
(x − 3)(x − 1)
Want to multiply to 3, add to − 4
− 3 and − 1, split the middle term
Factor by grouping
Our Solution
Example 286
...
These problems solve just like problems with one variable, using the coefficients to decide how to split the middle
term
Example 287
...
Consier the
following example, done incorrectly, ignoring negative signs
Warning 288
...
Now
the problem will be done correctly
...
x2 + 5x − 6
x2 + 6x − x − 6
x(x + 6) − 1(x + 6)
(x + 6)(x − 1)
Want to multiply to − 6, add to 5
6 and − 1, split the middle term
Factor by grouping
Our Solution
You may have noticed a shortcut for factoring these problems
...
This pattern
does not always work, so be careful getting in the habit of using it
...
In all
the problems we have factored in this lesson there is no number in front of x2
...
This is shown in the next few
examples
...
x2 − 7x − 18
(x − 9)(x + 2)
Want to multiply to − 18, add to − 7
− 9 and 2, write the factors
Our Solution
223
Example 291
...
If there is no combination of
numbers that multiplies and adds to the correct numbers, then we say we cannot
factor the polynomial, or we say the polynomial is prime
...
Example 292
...
If all the terms in a
problem have a common factor we will want to first factor out the GCF before we
factor using any other method
...
3x2 − 24x + 45
3(x2 − 8x + 15)
3(x − 5)(x − 3)
GCF of all terms is 3, factor this out
Want to multiply to 15, add to − 8
− 5 and − 3, write the factors
Our Solution
Again it is important to comment on the shortcut of jumping right to the factors,
this only works if there is no coefficient on x2
...
Be
careful not to use this shortcut on all factoring problems!
World View Note: The first person to use letters for unknown values was Francois Vieta in 1591 in France
...
224
6
...
1) p2 + 17p + 72
2) x2 + x − 72
3) n2 − 9n + 8
4) x2 + x − 30
5) x2 − 9x − 10
6) x2 + 13x + 40
7) b2 + 12b + 32
8) b2 − 17b + 70
9) x2 + 3x − 70
10) x2 + 3x − 18
12) a2 − 6a − 27
11) n2 − 8n + 15
14) p2 + 7p − 30
2
13) p + 15p + 54
16) m2 − 15mn + 50n2
15) n2 − 15n + 56
2
17) u − 8uv + 15v
2
19) m2 + 2mn − 8n2
21) x2 − 11xy + 18y 2
23) x2 + xy − 12y 2
25) x2 + 4xy − 12y 2
27) 5a2 + 60a + 100
29) 6a2 + 24a − 192
31) 6x2 + 18xy + 12y 2
18) m2 − 3mn − 40n2
20) x2 + 10xy + 16y 2
22) u2 − 9uv + 14v 2
24) x2 + 14xy + 45y 2
26) 4x2 + 52x + 168
28) 5n2 − 45n + 40
30) 5v 2 + 20v − 25
32) 5m2 + 30mn − 90n2
34) 6m2 − 36mn − 162n2
33) 6x2 + 96xy + 378y 2
225
6
...
When factoring trinomials we used the ac method to split the middle term and
then factor by grouping
...
World View Note: It was French philosopher Rene Descartes who first used letters from the beginning of the alphabet to represent values we know (a, b, c) and
letters from the end to represent letters we don’t know and are solving for (x, y,
z)
...
In the previous lesson we always multiplied to just c because there
was no number in front of x2
...
Now we will have a number in front of x2 so we will be
looking for numbers that multiply to ac and add to b
...
Example 294
...
The previous example illustrates an important point, the shortcut does not work when a 1
...
Example 295
...
10x2 − 27x + 5
10x2 − 25x − 2x + 5
5x(2x − 5) − 1(2x − 5)
(2x − 5)(5x − 1)
Multiply to ac or (10)(5) = 50, add to − 27
The numbers are − 25 and − 2, split the middle term
Factor by grouping
Our Solution
The same process works with two variables in the problem
Example 297
...
Factoring out the GCF first also has the added
bonus of making the numbers smaller so the ac method becomes easier
...
18x3 + 33x2 − 30x
3x[6x2 + 11x − 10]
3x[6x2 + 15x − 4x − 10]
3x[3x(2x + 5) − 2(2x + 5)]
3x(2x + 5)(3x − 2)
GCF = 3x, factor this out first
Multiply to ac or (6)( − 10) = − 60, add to 11
The numbers are 15 and − 4, split the middle term
Factor by grouping
Our Solution
As was the case with trinomials when a = 1, not all trinomials can be factored
...
Example 299
...
4 Practice - Trinomials where a
Factor each completely
...
5
Factoring - Factoring Special Products
Objective: Identify and factor special products including a difference of
squares, perfect squares, and sum and difference of cubes
...
The first is one we have seen before
...
Here we will use this special product to help us factor
Difference of Squares: a2 − b2 = (a + b)(a − b)
If we are subtracting two perfect squares then it will always factor to the sum and
difference of the square roots
...
x2 − 16
(x + 4)(x − 4)
Subtracting two perfect squares, the square roots are x and 4
Our Solution
Example 301
...
It is always
prime
...
Example 302
...
Multiply to 36, add to 0
No combinations that multiply to 36 add to 0
Our Solution
229
It turns out that a sum of squares is always prime
...
Because the square root of a fourth power is a square ( a4 = a2),
we can factor a difference of fourth powers just like we factor a difference of
squares, to a sum and difference of the square roots
...
This is shown in the following examples
...
a4 − b4
(a2 + b2)(a2 − b2)
(a2 + b2)(a + b)(a − b)
Difference of squares with roots a2 and b2
The first factor is prime, the second is a difference of squares!
Our Solution
Example 304
...
We had a shortcut for multiplying a perfect square which can be reversed to help us factor a perfect square
Perfect Square: a2 + 2ab + b2 = (a + b)2
A perfect square can be difficult to recognize at first glance, but if we use the ac
method and get two of the same numbers we know we have a perfect square
...
This is shown in the following examples
...
x2 − 6x + 9
(x − 3)2
Multiply to 9, add to − 6
The numbers are − 3 and − 3, the same! Perfect square
Use square roots from first and last terms and sign from the middle
230
Example 306
...
Problems would be written as “three sheafs of a
good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop sold for 29
dou
...
Another factoring shortcut has cubes
...
Both sum and difference of cubes have very similar factoring
formulas
Sum of Cubes: a3 + b3 = (a + b)(a2 − ab + b2)
Difference of Cubes: a3 − b3 = (a − b)(a2 + ab + b2)
Comparing the formulas you may notice that the only difference is the signs in
between the terms
...
S stands for Same sign as the problem
...
O stands for Opposite sign
...
Finally, AP stands for Always Positive
...
The following examples show factoring with cubes
...
m3 − 27
(m 3)(m2 3m 9)
(m − 3)(m2 + 3m + 9)
We have cube roots m and 3
Use formula, use SOAP to fill in signs
Our Solution
Example 308
...
When we fill in the trinomial’s first and last terms we square the cube roots 5p and 2r
...
Notice that when done
correctly, both get cubed
...
As a general rule, this factor will always be
prime (unless there is a GCF which should have been factored out before using
cubes rule)
...
Only after checking for a GCF should we be using the special products
...
72x2 − 2
2(36x2 − 1)
2(6x + 1)(6x − 1)
GCF is 2
Difference of Squares, square roots are 6x and 1
Our Solution
Example 310
...
128a4b2 + 54ab5
2ab2(64a3 + 27b3)
2ab2(4a + 3b)(16a2 − 12ab + 9b2)
GCF is 2ab2
Sum of cubes! Cube roots are 4a and 3b
Our Solution
232
6
...
1) r 2 − 16
2) x2 − 9
3) v 2 − 25
4) x2 − 1
5) p2 − 4
6) 4v 2 − 1
8) 9a2 − 1
7) 9k 2 − 4
10) 5n2 − 20
9) 3x2 − 27
12) 125x2 + 45y 2
11) 16x2 − 36
14) 4m2 + 64n2
13) 18a2 − 50b2
16) k 2 + 4k + 4
15) a2 − 2a + 1
18) n2 − 8n + 16
17) x2 + 6x + 9
20) k 2 − 4k + 4
19) x2 − 6x + 9
22) x2 + 2x + 1
2
21) 25p − 10p + 1
2
23) 25a + 30ab + 9b
2
2
2
25) 4a − 20ab + 25b
2
27) 8x − 24xy + 18y
29) 8 − m
2
24) x2 + 8xy + 16y 2
26) 18m2 − 24mn + 8n2
28) 20x2 + 20xy + 5y 2
30) x3 + 64
3
32) x3 + 8
3
31) x − 64
33) 216 − u3
35) 125a3 − 64
34) 125x3 − 216
36) 64x3 − 27
38) 32m3 − 108n3
37) 64x3 + 27y 3
40) 375m3 + 648n3
39) 54x3 + 250y 3
42) x4 − 256
41) a4 − 81
43) 16 − z 4
45) x4 − y 4
44) n4 − 1
46) 16a4 − b4
48) 81c4 − 16d4
47) m4 − 81b4
233
6
...
With so many different tools used to factor, it is easy to get lost as to which tool
to use when
...
A large part of deciding how to solve a problem is based on how many
terms are in the problem
...
Factoring Strategy (GCF First!!!!!)
•
2 terms: sum or difference of squares or cubes:
a2 − b2 = (a + b)(a − b)
a2 + b2 = Prime
a3 + b3 = (a + b)(a2 − ab + b2)
a3 − b3 = (a − b)(a2 + ab + b2)
•
3 terms: ac method, watch for perfect square!
a2 + 2ab + b2 = (a + b)2
Multiply to ac and add to b
•
4 terms: grouping
We will use the above strategy to factor each of the following examples
...
Example 312
...
5x 2 y + 15xy − 35x2 − 105x
5x(xy + 3y − 7x − 21)
5x[y(x + 3) − 7(x + 3)]
5x(x + 3)(y − 7)
GCF first, 5x
Four terms, try grouping
(x + 3) match!
Our Solution
Example 314
...
108x3 y 2 − 39x2 y 2 + 3xy 2
3xy 2(36x2 − 13x + 1)
3xy 2(36x2 − 9x − 4x + 1)
3xy 2[9x(4x − 1) − 1(4x − 1)]
3xy 2(4x − 1)(9x − 1)
GCF first, 3xy 2
Thee terms, ac method, multiply to 36, add to − 13
− 9 and − 4, split middle term
Factor by grouping
Our Solution
World View Note: Variables originated in ancient Greece where Aristotle would
use a single capital letter to represent a number
...
5 + 625y 3
5(1 + 125y 3)
5(1 + 5y)(1 − 5y + 25y 2)
GCF first, 5
Two terms, sum of cubes
Our Solution
It is important to be comfortable and confident not just with using all the factoring methods, but decided on which method to use
...
6 Practice - Factoring Strategy
Factor each completely
...
7
Factoring - Solve by Factoring
Objective: Solve quadratic equation by factoring and using the zero
product rule
...
However, when we have x2 (or a
higher power of x) we cannot just isolate the variable as we did with the linear
equations
...
We can use this to help us
solve factored polynomials as in the following example
...
(2x − 3)(5x + 1) = 0
2x − 3 = 0 or 5x + 1 = 0
+3+3
−1−1
2x = 3 or 5x = − 1
2 2
5
5
3
−1
x = or
2
5
One factor must be zero
Set each factor equal to zero
Solve each equation
Our Solution
For the zero product rule to work we must have factors to set equal to zero
...
Example 318
...
If it does not, we must move terms around so it does
equal zero
...
Example 319
...
(x − 7)(x + 3) = − 9
x2 − 7x + 3x − 21 = − 9
x2 − 4x − 21 = − 9
+9 +9
x2 − 4x − 12 = 0
(x − 6)(x + 2) = 0
x − 6 = 0 or x + 2 = 0
+6+6
−2−2
x = 6 or − 2
Not equal to zero, multiply first, use FOIL
Combine like terms
Move − 9 to other side so equation equals zero
Factor using the ac method, mutiply to − 12, add to − 4
The numbers are 6 and − 2
Set each factor equal to zero
Solve each equation
Our Solution
Example 321
...
However, it is possible to
have only one solution as the next example illustrates
...
4x2 = 12x − 9
− 12x + 9 − 12x + 9
4x2 − 12x + 9 = 0
(2x − 3)2 = 0
2x − 3 = 0
+3+3
2x = 3
2 2
3
x=
2
Set equal to zero by moving terms to left
Factor using the ac method, multiply to 36, add to − 12
− 6 and − 6, a perfect square!
Set this factor equal to zero
Solve the equation
Our Solution
As always it will be important to factor out the GCF first if we have one
...
This may give us more than just two solution
...
Example 323
...
2x3 − 14x2 + 24x = 0
2x(x2 − 7x + 12) = 0
2x(x − 3)(x − 4) = 0
2x = 0 or x − 3 = 0 or x − 4 = 0
2 2
+3+3
+4+4
x = 0 or 3 or 4
Factor out the GCF of 2x
Factor with ac method, multiply to 12, add to − 7
The numbers are − 3 and − 4
Set each factor equal to zero
Solve each equation
Our Solutions
Example 325
...
When we set this
factor equal to zero we got a false statement
...
Often a student will skip setting the GCF factor equal to zero if there is no variables in the GCF
...
If an equation does not factor we will have to solve it using another
method
...
World View Note: While factoring works great to solve problems with x2,
Tartaglia, in 16th century Italy, developed a method to solve problems with x3
...
To this day the formula is known as
Cardan’s Formula
...
While it is possible, there are a few properties of square roots that we have not covered yet and thus it is common to
break a rule of roots that we are not aware of at this point
...
When we talk about roots we will come back to problems like
these and see how we can solve using square roots in a method called completing
the square
...
7 Practice - Solve by Factoring
Solve each equation by factoring
...
1 Reduce Rational Expressions
...
2 Multiply and Divide
...
3 Least Common Denominator
...
4 Add and Subtract
...
5 Complex Fractions
...
6 Proportions
...
7 Solving Rational Equations
...
8 Application: Dimensional Analysis
...
1
Rational Expressions - Reduce Rational Expressions
Objective: Reduce rational expressions by dividing out common factors
...
Examples of rational expressions include:
x2 − x − 12
x2 − 9x + 20
and
3
x−2
and
a−b
b−a
and
3
2
As rational expressions are a special type of fraction, it is important to remember
with fractions we cannot have zero in the denominator of a fraction
...
Example 326
...
3
We
can however, evaluate any other value in the expression
...
It was the Mayans of Central America
who first used zero to aid in the use of their base 20 system as a place holder!
Rational expressions are easily evaluated by simply substituting the value for the
variable and using order of operations
...
x2 − 4
when x = − 6
x2 + 6x + 8
( − 6)2 − 4
( − 6)2 + 6( − 6) + 8
36 − 4
36 + 6( − 6) + 8
36 − 4
36 − 36 + 8
32
8
4
Substitute − 5 in for each variable
Exponents first
Multiply
Add and subtract
Reduce
Our Solution
Just as we reduced the previous example, often a rational expression can be
reduced, even without knowing the value of the variable
...
We have already seen this with monomials when we
discussed properties of exponents
...
Example 328
...
Negative exponents move to denominator
Our Solution
244
However, if there is more than just one term in either the numerator or denominator, we can’t divide out common factors unless we first factor the numerator
and denominator
...
28
8x2 − 16
28
8(x2 − 2)
7
2(x2 − 2)
Denominator has a common factor of 8
Reduce by dividing 24 and 8 by 4
Our Solution
Example 330
...
x2 − 25
x2 + 8x + 15
(x + 5)(x − 5)
(x + 3)(x + 5)
x−5
x+3
Numerator is difference of squares, denominator is factored using ac
Divide out common factor (x + 5)
Our Solution
It is important to remember we cannot reduce terms, only factors
...
In the prex−5
vious example we had the solution x + 3 , we cannot divide out the x’s because
they are terms (separated by + or − ) not factors (separated by multiplication)
...
1 Practice - Reduce Rational Expressions
Evaluate
1)
4v + 2
6
3)
x−3
x2 − 4x + 3
5)
2)
b−3
3b − 9
when x = − 4
4)
a+2
a2 + 3a + 2
when b = 0
6)
n2 − n − 6
n−3
27p
18p2 − 36p
when v = 4
b+2
b2 + 4b + 4
when b = − 2
when a = − 1
when n = 4
State the excluded values for each
...
17)
21x2
18x
18)
12n
4n2
19)
24a
40a2
20)
21k
24k2
21)
32x3
8x4
22)
90x2
20x
23)
18m − 24
60
24)
10
81n3 + 36n2
25)
20
4p + 2
26)
n−9
9n − 81
27)
x+1
x2 + 8x + 7
28)
28m + 12
36
29)
32x2
28x2 + 28x
30)
49r + 56
56r
32)
b2 + 14b + 48
b2 + 15b + 56
34)
9v + 54
v 2 − 4v − 60
30x − 90
50x + 40
36)
35)
12x2 − 42x
30x2 − 42x
k2 − 12k + 32
k2 − 64
38)
37)
6a − 10
10a + 4
9p + 18
p2 + 4p + 4
40)
39)
2n2 + 19n − 10
9n + 90
3x2 − 29x + 40
5x2 − 30x − 80
31)
33)
n2 + 4n − 12
n2 − 7n + 10
246
41)
8m + 16
20m − 12
42)
56x − 48
24x2 + 56x + 32
43)
2x2 − 10x + 8
3x2 − 7x + 4
44)
50b − 80
50b + 20
45)
7n2 − 32n + 16
4n − 16
46)
35v + 35
21v + 7
47)
n2 − 2n + 1
6n + 6
48)
56x − 48
24x2 + 56x + 32
49)
7a2 − 26a − 45
6a2 − 34a + 20
50)
4k3 − 2k 2 − 2k
9k 3 − 18k 2 + 9k
247
7
...
Multiplying and dividing rational expressions is very similar to the process we use
to multiply and divide fractions
...
15 14
·
49 45
1 2
·
7 3
2
21
First reduce common factors from numerator and denominator (15 and 7)
Multiply numerators across and denominators across
Our Solution
The process is identical for division with the extra first step of multiplying by the
reciprocal
...
248
Example 333
...
Example 334
...
Example 335
...
Example 336
...
To solve we still need to factor, and we use the reciprocal of the divided
fraction
...
a2 + 7a + 10
a+1
a−1
·
÷
a2 + 6a + 5 a2 + 4a + 4 a + 2
Factor each expression
(a + 1)
(a − 1)
(a + 5)(a + 2)
·
÷
(a + 5)(a + 1) (a + 2)(a + 2) (a + 2)
Reciprocal of last fraction
(a + 5)(a + 2)
(a + 1)
(a + 2)
·
·
(a + 5)(a + 1) (a + 2)(a + 2) (a − 1)
Divide out common factors
1
a−1
(a + 2), (a + 2), (a + 1), (a + 5)
Our Solution
World View Note: Indian mathematician Aryabhata, in the 6th century, pubn(n + 1)(n + 2)
lished a work which included the rational expression
for the sum of
6
1
2
2
2
the first n squares (1 + 2 + 3 +
...
2 Practice - Multiply and Divide
Simplify each expression
...
3
Rational Expressions - Least Common Denominators
Objective: Idenfity the least common denominator and build up
denominators to match this common denominator
...
The process we use to find the LCD is based
on the process used to find the LCD of intergers
...
Find the LCD of 8 and 6
8, 16, 24
...
Example 339
...
However, we must first factor each polynomial so we can identify all the factors to
be used (attaching highest exponent if necessary)
...
Find the LCD of x2 + 2x − 3 and x2 − x − 12
(x − 1)(x + 3) and (x − 4)(x + 3)
(x − 1)(x + 3)(x − 4)
Factor each polynomial
LCD uses all unique factors
Our Solution
Notice we only used (x + 3) once in our LCD
...
The only time we need to repeat a factor or use
an exponent on a factor is if there are exponents when one of the polynomials is
factored
253
Example 341
...
Then we would have used (x − 5) twice in the LCD
because it showed up twice in one of the polynomials
...
Once we know the LCD, our goal will be to build up fractions so they have
matching denominators
...
We can build up a fraction’s denominator by multipliplying the numerator and denoinator by any factors
that are not already in the denominator
...
5a
?
=
3a2b 6a5b3
2a3b2
5a 2a3b2
3a2b 2a3b2
10a4b2
6a5b3
Idenfity what factors we need to match denominators
3 · 2 = 6 and we need three more a ′s and two more b ′s
Multiply numerator and denominator by this
Our Solution
Example 343
...
The reason for this is to add and subtract fractions
we will want to be able to combine like terms in the numerator, then when we
reduce at the end we will want our denominators factored
...
Example 344
...
Build up each fraction so they have a common denominator
5x
x2 − 5x − 6
and
(x − 6)(x + 1)
x−2
x2 + 4x + 3
(x + 1)(x + 3)
LCD: (x − 6)(x + 1)(x + 3)
First: (x + 3) Second: (x − 6)
5x
x+3
(x − 6)(x + 1) x + 3
and
x−2
x−6
(x + 1)(x + 3) x − 6
x2 − 8x + 12
5x2 + 15x
and
(x − 6)(x + 1)(x + 3)
(x − 6)(x + 1)(x + 3)
Factor to find LCD
Use factors to find LCD
Identify which factors are missing
Multiply fractions by missing factors
Multiply numerators
Our Solution
World View Note: When the Egyptians began working with fractions, they
4
expressed all fractions as a sum of unit fraction
...
An interesting problem with this system is
this is not a unique solution,
4
5
is also equal to the sum
255
1
3
1
1
1
+ 5 + 6 + 10
...
3 Practice - Least Common Denominator
Build up denominators
...
4
Rational Expressions - Add & Subtract
Objective: Add and subtract rational expressions with and without
common denominators
...
Recall that when adding with a common denominator we add the
numerators and keep the denominator
...
Remember to reduce, if possible, your final answer
...
x−4
x2 − 2x − 8
+
x+8
x2 − 2x − 8
2x + 4
x2 − 2x − 8
2(x + 2)
(x + 2)(x − 4)
2
x−4
Same denominator, add numerators, combine like terms
Factor numerator and denominator
Divide out (x + 2)
Our Solution
257
Subtraction with common denominator follows the same pattern, though the subtraction can cause problems if we are not careful with it
...
Then we can treat it
like an addition problem
...
Example 347
...
When we don’t have a common denominator we will have to find the least
common denominator (LCD) and build up each fraction so the denominators
match
...
Example 348
...
Build up, multiply 6 by 2 and 4 by 3
Multiply
Add numerators
258
13
12
Our Solution
The same process is used with variables
...
4b
7a
+
2b
6ab4
3a
2b3 7a
4b a
+
2b3 3a2b 6ab4 a
4ab
14ab3
+ 2 4
2b4
6a
6a b
The LCD is 6a2b4
...
Example 350
...
Build up denominators
Multiply first fraction by 4a, second by 5
Our Solution
If our denominators have more than one term in them we will need to factor first
to find the LCD
...
Example 351
...
Example 352
...
4 Practice - Add and Subtract
Add or subtract the rational expressions
...
1)
2
a+3
3)
t2 + 4t
t−1
5)
4
+ a+3
2t − 7
t−1
+
2x2 + 3
5
6r
9)
8
9t3
4)
x2 − 5x + 9
x2
x−2
− x2 − 6x + 5
5
− 8r
5
a2 + 3a
6)
3
x
4
7
xy 2
− a2 + 5a − 6
4
+ x2
3
+ x2 y
10)
+ 6t2
6x − 8
x−2
−
a2 + 5a − 6
8)
x2 − 6x + 5
7)
2)
x+5
8
+
x−3
12
11)
a+2
2
−
a−4
4
12)
2a − 1
3a2
+
5a + 1
9a
13)
x−1
4x
−
2x + 3
x
14)
2c − d
c2d
−
c+d
cd2
15)
5x + 3y
2x2 y
16)
2
x−1
+ x+1
17)
2z
z −1
3z
18)
2
x−5
+ 4x
19)
8
x2 − 4
3
20)
4x
x2 − 25
21)
t
t−3
5
22)
2
x+3
23)
2
5x2 + 5x
24)
3a
4a − 20
25)
t
y −t
26)
x
x−5
27)
x
x2 + 5x + 6
28)
2x
x2 − 1
− x2 + 5x + 4
29)
x
x2 + 15x + 56
30)
2x
x2 − 9
+ x2 + x − 6
31)
5x
x2 − x − 6
32)
4x
x2 − 2x − 3
− x2 − 5x + 6
33)
2x
x2 − 1
34)
x−1
x2 + 3x + 2
+ x2 + 4x + 3
35)
x+1
x2 − 2x − 35
36)
3x + 2
3x + 6
+ 4 − x2
37)
4 − a2
a2 − 9
− 3−a
38)
4y
y2 − 1
− y − y+1
39)
2z
1 − 2z
+ 2z + 1 − 4z 2 − 1
40)
2r
r 2 − s2
41)
2x − 3
x2 + 3x + 2
+ x2 + 5x + 6
3x − 1
42)
x+2
x2 − 4x + 3
+ x2 + 4x − 5
43)
2x + 7
x2 − 2x − 3
− x2 + 6x + 5
3x − 2
44)
3x − 8
x2 + 6x + 8
+ x2 + 3x + 2
−
3x + 4y
xy 2
− z+1
− x+2
− 4t − 12
4
− 3x + 3
y
− y+t
2
− x2 + 3x + 2
7
− x2 + 13x + 42
18
− x2 − 9
4
− x2 + 2x − 3
x+6
+ x2 + 7x + 10
a −2
3z
3
261
2
3
x
+ x+5
4
+ (x + 3)2
9a
+ 6a − 30
+
x−5
x
3
5
3
x+5
x
2
2
1
1
+ r+s − r−s
4x + 5
2x − 3
7
...
Complex fractions have fractions in either the numerator, or denominator, or usually both
...
This will be illustrated first with integers, then we will consider how the process can be expanded
to include expressions with variables
...
Example 353
...
Generally we prefer a different method, to multiply the numerator and denominator of the large fraction (in effect each term in the complex fraction) by the
least common denominator (LCD)
...
We will simplify the same problem using this second method
...
2
3
5
6
1
−4
1
+2
LCD is 12, multiply each term
262
2(12)
3
5(12)
6
−
+
1(12)
4
1(12)
2
2(4) − 1(3)
5(2) + 1(6)
8−3
10 + 6
5
16
Reduce each fraction
Multiply
Add and subtract
Our Solution
Clearly the second method is a much cleaner and faster method to arrive at our
solution
...
We
will first find the LCD of the small fractions, and multiply each term by this LCD
so we can clear the small fractions and simplify
...
1
1 − x2
Identify LCD (use highest exponent)
1
1− x
LCD = x2
1(x2) −
1(x2) −
Multiply each term by LCD
1(x2)
x2
1(x2)
x
Reduce fractions (subtract exponents)
1(x2) − 1
1(x2) − x
Multiply
x2 − 1
x2 − x
Factor
(x + 1)(x − 1)
x(x − 1)
Divide out (x − 1) factor
x+1
x
Our Solution
The process is the same if the LCD is a binomial, we will need to distribute
3
x+4
−2
2
5 + x+4
Multiply each term by LCD, (x + 4)
263
3(x + 4)
x+4
− 2(x + 4)
5(x + 4) +
Reduce fractions
2(x + 4)
x+4
3 − 2(x + 4)
5(x + 4) + 2
Distribute
3 − 2x − 8
5x + 20 + 2
Combine like terms
− 2x − 5
5x + 22
Our Solution
The more fractions we have in our problem, the more we repeat the same process
...
2
3
1
− ab3 + ab
ab2
4
1
+ ab − ab
a2b
Idenfity LCD (highest exponents)
LCD = a2b3
3(a2b3)
1(a2b3)
2(a2b3)
− ab3 + ab
2
ab
1(a2b3)
4(a2b3)
+ ab(a2b3) − ab
2b
a
2ab − 3a + ab2
4b2 + a3b4 − ab2
Multiply each term by LCD
Reduce each fraction (subtract exponents)
Our Solution
World View Note: Sophie Germain is one of the most famous women in mathematics, many primes, which are important to finding an LCD, carry her name
...
The
largest known Germain prime (at the time of printing) is 183027 · 2265440 − 1 which
has 79911 digits!
Some problems may require us to FOIL as we simplify
...
Example 357
...
Remember, the exponent is only on the factor
it is attached to, not the whole term
...
m−2 + 2m−1
m + 4m−2
1
m2
1(m2)
m2
2
+m
m+
Make each negative exponent into a fraction
4
m2
2(m2)
m
4(m2)
m(m2) + m2
+
1 + 2m
m3 + 4
Multiply each term by LCD, m2
Reduce the fractions
Our Solution
Once we convert each negative exponent into a fraction, the problem solves
exactly like the other complex fraction problems
...
5 Practice - Complex Fractions
Solve
...
31)
x−2 − y −2
x−1 + y −1
32)
x−2 y + xy −2
x−2 − y −2
33)
x−3 y − xy −3
x−2 − y −2
34)
4 − 4x−1 + x−2
4 − x−2
36)
x−3 + y −3
x−2 − x−1 y −1 + y −2
x−2 − 6x−1 + 9
35)
x2 − 9
267
7
...
This definition can be
generalized to two equal rational expressions
...
Cross Product: If
a
c
= then ad = bc
b
d
The cross product tells us we can multiply diagonally to get an equation with no
fractions that we can solve
...
20 x
=
6
9
Calculate cross product
268
(20)(9) = 6x
180 = 6x
6
6
30 = x
Multiply
Divide both sides by 6
Our Solution
World View Note: The first clear definition of a proportion and the notation
for a proportion came from the German Leibniz who wrote, “I write dy: x = dt: a;
for dy is to x as dt is to a, is indeed the same as, dy divided by x is equal to dt
divided by a
...
”
If the proportion has more than one term in either numerator or denominator, we
will have to distribute while calculating the cross product
...
x+3 2
=
5
4
5(x + 3) = (4)(2)
5x + 15 = 8
− 15 − 15
5x = − 7
5
5
7
x=−
5
Calculate cross product
Multiply and distribute
Solve
Subtract 15 from both sides
Divide both sides by 5
Our Solution
This same idea can be seen when the variable appears in several parts of the proportion
...
4
6
=
x 3x + 2
4(3x + 2) = 6x
12x + 8 = 6x
− 12x − 12x
8 = − 6x
−6 −6
4
− =x
3
Calculate cross product
Distribute
Move variables to one side
Subtract 12x from both sides
Divide both sides by − 6
Our Solution
269
Example 362
...
We can solve this quadratic in the same way we solved quadratics in the
past, either factoring, completing the square or the quadratic formula
...
Example 363
...
We can use them to compare different quantities and make conclusions about how quantities are related
...
This consistency of the numerator
and denominator is essential in setting up our proportions
...
A six foot tall man casts a shadow that is 3
...
If the shadow of a flag
pole is 8 feet long, how tall is the flag pole?
shadow
height
We will put shadows in numerator, heights in denomintor
270
3
...
5 8
=
x
6
3
...
5x = 48
3
...
5
x = 13
...
5 feet and a height of 6 feet
The flagpole has a shadow of 8 feet, but we don ′t know the height
This gives us our proportion, calculate cross product
Multiply
Divide both sides by 3
...
In a basketball game, the home team was down by 9 points at the end of the
game
...
What was the final score of the game?
home
visiter
We will put home in numerator, visitor in denominator
x−9
x
Don ′t know visitor score, but home is 9 points less
6
7
Home team scored 6 for every 7 the visitor scored
x−9 6
=
x
7
7(x − 9) = 6x
7x − 63 = 6x
− 7x
− 7x
− 63 = − x
−1 −1
63 = x
63 − 9 = 54
63 to 54
This gives our proportion, calculate the cross product
Distribute
Move variables to one side
Subtract 7x from both sides
Divide both sides by − 1
We used x for the visitor score
...
6 Practice - Proportions
Solve each proportion
...
Round your answer to the nearest tenth
...
31) The currency in Western Samoa is the Tala
...
70 to 1 Tala
...
3 Tala?
32) If you can buy one plantain for S0
...
84?
272
33) Kali reduced the size of a painting to a height of 1
...
What is the new
width if it was originally 5
...
tall and 10 in
...
2 in : 2
...
If the model train is 5 in tall then
how tall is the real train?
35) A bird bath that is 5
...
4 ft long
...
2 ft adult elephant casts
...
2 mi from each other
...
9 in : 10
...
If the Vikings
scored 3 points for every 2 points the Timberwolves scored, what was the
half time score?
38) Sarah worked 10 more hours than Josh
...
charges S8 more for a repair than Low Cost
Computer Repair
...
If Christina drives 12
minutes for every 17 minutes Kelsey drives, how long is each commute?
273
7
...
When solving equations that are made up of rational expressions we will solve
them using the same strategy we used to solve linear equations with fractions
...
2
5 3
x− =
3
6 4
2(12)
5(12) 3(12)
x−
=
3
6
4
2(4)x − 5(2) = 3(3)
8x − 10 = 9
+ 10 + 10
8x = 19
8 8
19
x=
8
Multiply each term by LCD, 12
Reduce fractions
Multiply
Solve
Add 10 to both sides
Divide both sides by 8
Our Solution
We will use the same process to solve rational equations, the only difference is our
274
LCD will be more involved
...
If our
LCD equals zero, the solution is undefined
...
Example 367
...
As the LCD gets more complex, it is important to remember the process we are using to solve is still the
same
...
x
1
5
+
=
x + 2 x + 1 (x + 1)(x + 2)
Multiply terms by LCD, (x + 1)(x + 2)
x(x + 1)(x + 2) 1(x + 1)(x + 2) 5(x + 1)(x + 2)
+
=
x+2
x+1
(x + 1)(x + 2)
Reduce fractions
275
Distribute
Combine like terms
Make equatino equal zero
Subtract 6 from both sides
Factor
Set each factor equal to zero
Solve each equation
x(x + 1) + 1(x + 2) = 5
x2 + x + x + 2 = 5
x2 + 2x + 2 = 5
−5−5
2
x + 2x − 3 = 0
(x + 3)(x − 1) = 0
x + 3 = 0 or x − 1 = 0
−3−3
+1+1
x = − 3 or x = 1
( − 3 + 1)( − 3 + 2) = ( − 2)( − 1) = 2
(1 + 1)(1 + 2) = (2)(3) = 6
x = − 3 or 1
Check solutions, LCD can ′t be zero
Check − 3 in (x + 1)(x + 2), it works
Check 1 in (x + 1)(x + 2), it works
Our Solution
In the previous example the denominators were factored for us
...
1
11
x
−
=
2 − 3x + 2
x
x−1 x−2
(x − 1)(x − 2)
LCD = (x − 1)(x − 2)
x(x − 1)(x − 2) 1(x − 1)(x − 2)
11(x − 1)(x − 2)
−
=
x−1
x−2
(x − 1)(x − 2)
x(x − 2) − 1(x − 1) = 11
x2 − 2x − x + 1 = 11
x2 − 3x + 1 = 11
− 11 − 11
x2 − 3x − 10 = 0
(x − 5)(x + 2) = 0
x − 5 = 0 or x + 2 = 0
+5+5
−2−2
x = 5 or x = − 2
(5 − 1)(5 − 2) = (4)(3) = 12
( − 2 − 1)( − 2 − 2) = ( − 3)( − 4) = 12
276
Factor denominator
Identify LCD
Multiply each term by LCD, reduce
Distribute
Combine like terms
Make equation equal zero
Subtract 11 from both sides
Factor
Set each factor equal to zero
Solve each equation
Check answers, LCD can ′t be 0
Check 5 in (x − 1)(x − 2), it works
Check − 2 in (x − 1)(x − 2), it works
x = 5 or − 2
Our Solution
World View Note: Maria Agnesi was the first women to publish a math textbook in 1748, it took her over 10 years to write! This textbook covered everything
from arithmetic thorugh differential equations and was over 1,000 pages!
If we are subtracting a fraction in the problem, it may be easier to avoid a future
sign error by first distributing the negative through the numerator
...
x−2 x+2 5
−
=
x−3 x+2 8
x−2 −x−2 5
+
=
x+2
8
x−3
Distribute negative through numerator
Identify LCD, 8(x − 3)(x + 2), multiply each term
(x − 2)8(x − 3)(x + 2) ( − x − 2)8(x − 3)(x + 2) 5 · 8(x − 3)(x + 2)
+
=
Reduce
x−3
x+2
8
8(x − 2)(x + 2) + 8( − x − 2)(x − 3) = 5(x − 3)(x + 2)
8(x2 − 4) + 8( − x2 + x + 6) = 5(x2 − x − 6)
8x2 − 32 − 8x2 + 8x + 48 = 5x2 − 5x − 30
8x + 16 = 5x2 − 5x − 30
− 8x − 16
− 8x − 16
2
0 = 5x − 13x − 46
0 = (5x − 23)(x + 2)
5x − 23 = 0 or x + 2 = 0
+ 23 + 23
−2−2
5x = 23 or x = − 2
5
5
23
x=
or − 2
5
8
23
−3
5
23
8
+2 =8
5
5
33
5
=
2112
25
8( − 2 − 3)( − 2 + 2) = 8( − 5)(0) = 0
23
x=
5
FOIL
Distribute
Combine like terms
Make equation equal zero
Subtract 8x and 16
Factor
Set each factor equal to zero
Solve each equation
Check solutions, LCD can ′t be 0
23
in 8(x − 3)(x + 2), it works
5
Check − 2 in 8(x − 3)(x + 2), can ′t be 0!
Check
Our Solution
In the previous example, one of the solutions we found made the LCD zero
...
277
7
...
8
Rational Expressions - Dimensional Analysis
Objective: Use dimensional analysis to preform single unit, dual unit,
square unit, and cubed unit conversions
...
When we convert units of measure we can do so by multiplying several fractions together in a
process known as dimensional analysis
...
When multiplying, if we multiply by 1, the value of the expression
does not change
...
Notice the numerator and denominator are not identical in appearance, but rather identical in
value
...
1 4
= =
1 4
1
2
2
4
=
100cm
1lb
1hr
60 min
=
=
=
1m
16oz 60 min
1hr
The last few fractions that include units are called conversion factors
...
For example, 1 mile = 5280 feet
...
5280ft
Similarly we could make a conversion factor 1mi
...
The idea behind dimensional analysis is we will multiply by a fraction in such a
way that the units we don’t want will divide out of the problem
...
It is the same with units
...
Example 371
...
37 miles to feet
17
...
37mi
??mi
1
17
...
37
1
1
91, 713
...
37 miles as a fraction, put it over 1
To divide out the miles we need miles in the denominator
We are converting to feet, so this will go in the numerator
Fill in the relationship described above, 1 mile = 5280 feet
Divide out the miles and multiply across
Our Solution
5280ft
In the previous example, we had to use the conversion factor 1mi so the miles
1 mi
would divide out
...
This is why when doing dimensional analysis it is very important to
use units in the set-up of the problem, so we know how to correctly set up the
conversion factor
...
If 1 pound = 16 ounces, how many pounds 435 ounces?
435oz
1
435oz
1
Write 435 as a fraction, put it over 1
?? lbs
??oz
To divide out oz,
put it in the denominator and lbs in numerator
435oz
1
1 lbs
16oz
Fill in the given relationship, 1 pound = 16 ounces
280
435
1
1 lbs
435 lbs
=
16
16
Divide out oz, multiply across
...
1875 lbs
Our Solution
The same process can be used to convert problems with several units in them
...
Example 373
...
What was the student’s speed in
feet per second?
45mi
hr
45mi
hr
′′
per ′′ is the fraction bar, put hr in denominator
5280ft
1mi
To clear mi they must go in denominator and become ft
45mi
hr
5280ft
1mi
1hr
3600 sec
To clear hr they must go in numerator and become sec
45
1
5280ft
1
1
3600 sec
Divide out mi and hr
...
So if we are converting square inches (in2) to square ft (ft2), the conversion
factor would be squared,
the convesion factor
...
Similarly if the units are cubed, we will cube
Example 374
...
296296yd3
Divide out ft3
Multiply across and divide
Our Solution
When calculating area or volume, be sure to use the units and multiply them as
well
...
A room is 10 ft by 12 ft
...
33yd2
Our solution
282
To focus on the process of conversions, a conversion sheet has been included at
the end of this lesson which includes several conversion factors for length, volume,
mass and time in both English and Metric units
...
If we can identify relationships that represent the same value we can make
them into a conversion factor
...
A child is perscribed a dosage of 12 mg of a certain drug and is allowed to refill
his prescription twice
...
283
Conversion Factors
Length
Area
English
Metric
12 in = 1 ft 1000 mm = 1 m
1 yd = 3 ft
10 mm = 1 cm
1 yd = 36 in
100 cm = 1 m
1 mi = 5280 ft 10 dm = 1 m
1 dam = 10 m
1 hm = 100 m
1 km = 1000 m
English
Metric
1 ft2 = 144 in2
1 a = 100 m2
1 yd2 = 9 ft2
1 ha = 100 a
2
1 acre = 43,560 ft
640 acres = 1 mi2
English/Metric
1 ha = 2
...
54 cm = 1 in
1 m = 3
...
61 km = 1 mi
Weight (Mass)
English
1 lb = 16 oz
1 T = 2000 lb
Volume
Metric
1 g = 1000 mg
1 g = 100 cg
1000 g = 1 kg
1000 kg = 1 t
English
Metric
1 c = 8 oz 1 mL = 1 cc = 1 cm3
1 pt = 2 c
1 L = 1000 mL
1 qt = 2 pt
1 L = 100 cL
1 gal = 4 qt
1 L = 10 dL
1000 L = 1 kL
English/Metric
28
...
2 lb = 1 kg
English/Metric
16
...
06 qt = 1 L
3
...
8 Practice - Dimensional Analysis
Use dimensional analysis to convert the following:
1) 7 mi
...
to tons
3) 11
...
35 km to centimeters
5) 9,800,000 mm (milimeters) to miles
6) 4
...
0065 km3 to cubic meters
10) 14
...
5 mph (miles per hour) to feet per second
13) 185 yd
...
to miles per hour
14) 153 ft/s (feet per second) to miles per hour
15) 248 mph to meters per second
16) 186,000 mph to kilometers per year
17) 7
...
How many
miles per 1-gallon did she travel? How many yards per 1-ounce?
20) A chair lift at the Divide ski resort in Cold Springs, WY is 4806 feet long and
takes 9 minutes
...
Determine this printer’s
output in pages per day, and reams per month
...
If an average person lives
to the age of 75, how many times does the average heart beat in a lifetime?
23) Blood sugar levels are measured in miligrams of gluclose per deciliter of blood
volume
...
The
carpet cost S18 per square yard
...
How fast is it going in miles per hour?
in meters per second?
26) A cargo container is 50 ft long, 10 ft wide, and 8 ft tall
...
27) A local zoning ordinance says that a house’s “footprint” (area of its ground
1
floor) cannot occupy more than 4 of the lot it is built on
...
How
many typical pages of text can be stored on a 700-megabyte compact disc?
Assume that one typical page of text contains 2000 characters
...
Its purpose was to restore the river
and the habitants along its bank
...
About how much water was
released during the 1-week flood?
30) The largest single rough diamond ever found, the Cullinan diamond, weighed
3106 carats; how much does the diamond weigh in miligrams? in pounds?
(1 carat - 0
...
1 Square Roots
...
2 Higher Roots
...
3 Adding Radicals
...
4 Multiply and Divide Radicals
...
5 Rationalize Denominators
...
6 Rational Exponents
...
7 Radicals of Mixed Index
...
8 Complex Numbers
...
1
Radicals - Square Roots
Objective: Simplify expressions with square roots
...
A square root “unsquares” a number
...
√
The square root of 25 is written as 25
...
The R came from the
latin, “radix”, which can be translated as “source” or “foundation”
...
√
√
1=1
121 = 11
√
√
4=2
625 = 25
√
√
9=3
− 81 = Undefined
√
The final example, − 81 is currently undefined as negatives have no square root
...
Thus we can only take square roots of positive numbers
...
√
Not all numbers have a nice even square root
...
828427124746190097603377448419
...
To be as accurate as possible, we will never use the calculator to find decimal approximations of
square roots
...
We will do
this using a property known as the product rule of radicals
Product Rule of Square Roots:
√
a·b =
√
a·
√
b
√
We can use the product rule to simplify an expression such as 36 · 5 by spliting
√
√
√
it into two roots, 36 · 5 , and simplifying the first root, 6 5
...
There are several ways this can be done
...
This is shown in the next example
...
√
75
25 · 3
√
√
25 · 3
√
5 3
√
75 is divisible by 25, a perfect square
Split into factors
Product rule, take the square root of 25
Our Solution
If there is a coefficient in front of the radical to begin with, the problem merely
becomes a big multiplication problem
...
√
5 63
√
5 9·7
√ √
5 9· 7
√
5·3 7
√
15 7
63 is divisible by 9, a perfect square
Split into factors
Product rule, take the square root of 9
Multiply coefficients
Our Solution
As we simplify radicals using this method it is important to be sure our final
answer can be simplified no more
...
√
72
√
9·8
√ √
9· 8
√
3 8
√
3 4·2
√ √
3 4· 2
√
3·2 2
72 is divisible by 9, a perfect square
Split into factors
Product rule, take the square root of 9
But 8 is also divisible by a perfect square, 4
Split into factors
Product rule, take the square root of 4
Multiply
289
√
6 2
Our Solution
...
Variables often are part of the radicand as well
...
For example, x8 = x4, because we
divide the exponent of 8 by 2
...
When squaring, we multiply the exponent by two, so when
taking a square root we divide the exponent by 2
...
Example 381
...
Sometimes we
have a remainder
...
This
is shown in the following example
...
20x5 y 9z 6
5 9 6
√
√
4· 5·
√
4 · 5x y z
√
x5 · y 9 · z 6
√
2x2 y 4z 3 5xy
20 is divisible by 4, a perfect square
Split into factors
Simplify, divide exponents by 2, remainder is left inside
Our Solution
290
8
...
√
1) 245
√
3) 36
√
5) 12
√
7) 3 12
√
9) 6 128
√
11) − 8 392
√
13) 192n
√
15) 196v 2
√
17) 252x2
√
19) − 100k 4
√
21) − 7 64x4
√
23) − 5 36m
25)
45x2 y 2
27)
16x3 y 3
29)
320x4 y 4
31) 6
80xy 2
245x2 y 3
√
35) − 2 180u3v
33) 5
37) − 8
39) 2
180x4 y 2z 4
80hj 4k
41) − 4
54mnp2
2)
√
125
4)
√
196
6)
√
72
√
8) 5 32
√
10) 7 128
√
12) − 7 63
√
14) 343b
√
16) 100n3
√
18) 200a3
20) − 4
175p4
√
22) − 2 128n
24) 8 112p2
√
26) 72a3b4
√
28) 512a4b2
√
30) 512m4n3
√
32) 8 98mn
34) 2
72x2 y 2
36) − 5 72x3 y 4
√
38) 6 50a4bc2
40) −
42) − 8
291
32xy 2z 3
32m2 p4 q
8
...
While square roots are the most common type of radical we work with, we can
take higher roots of numbers as well: cube roots, fourth roots, fifth roots, etc
...
√
m
a = b if bm = a
The small letter m inside the radical is called the index
...
For square roots the index is 2
...
World View Note: The word for root comes from the French mathematician
Franciscus Vieta in the late 16th century
...
Example 383
...
First its impor√
√
tant not to forget to check the index on the root
...
This is
because 92 = 81 and 34 = 81
...
We can take an odd root of a negative number, because a negative number
raised to an odd power is still negative
...
In a later section we will discuss
how to work with roots of negative, but for now we will simply say they are undefined
...
√
√ √
Product Property of Radicals: m ab = m a · m b
Often we are not as familiar with higher powers as we are with squares
...
292
Example 384
...
Test 3, 33 = 27, 54 is divisible by 27!
Write as factors
Product rule, take cubed root of 27
Our Solution
Just as with square roots, if we have a coefficient, we multiply the new coefficients
together
...
√
3 4 48
24 = 16
√
3 4 16 · 3
√
√
3 4 16 · 4 3
√
3·2 4 3
√
64 3
We are working with a fourth root, want fourth powers
Test 2, 24 = 16, 48 is divisible by 16!
Write as factors
Product rule, take fourth root of 16
Multiply coefficients
Our Solution
We can also take higher roots of variables
...
Any whole answer is how many of that varible will
come out of the square root
...
This is shown in the following examples
...
5
x25y 17z 3
5 3 5
x y
2 3
yz
Divide each exponent by 5, whole number outside, remainder inside
Our Solution
25
In the previous example, for the x, we divided 5 = 5 R 0, so x5 came out, no x’s
17
remain inside
...
For the z, when we divided 5 = 0R 3, all three or z 3 remained inside
...
Example 387
...
The number 8 works!
Take cube root of 8, dividing exponents on variables by 3
Remainders are left in radical
...
2 Practice - Higher Roots
Simplify
...
3
Radicals - Adding Radicals
Objective: Add like radicals by first simplifying each radical
...
Consider the following example
...
5x + 3x − 2x
6x
Combine like terms
Our Solution
√
√
√
5 11 + 3 11 − 2 11
√
6 11
Combine like terms
Our Solution
√
Notice that when we combined the terms with 11 it was just like combining
terms with x
...
We add and subtract the coefficients in front of the
295
radical, and the radical stays the same
...
Example 389
...
Often
problems we solve have no like radicals, however, if we simplify the radicals first
we may find we do in fact have like radicals
...
√
√
√
√
5 45 + 6 18 − 2 98 + 20
√
√
√
√
5 9 · 5 + 6 9 · 2 − 2 49 · 2 + 4 · 5
√
√
√
√
5·3 5 +6·3 2 −2·7 2 +2 5
√
√
√
√
15 5 + 18 2 − 14 2 + 2 5
√
√
17 5 + 4 2
Simplify radicals, find perfect square factors
Take roots where possible
Multiply coefficients
Combine like terms
Our Solution
World View Note: The Arab writers of the 16th century used the symbol similar to the greater than symbol with a dot underneath for radicals
...
√
√
√
4 3 54 − 9 3 16 + 5 3 9
√
√
√
4 3 27 · 2 − 9 3 8 · 2 + 5 3 9
√
√
√
4·3 3 2 −9·2 3 2 +5 3 9
√
√
√
12 3 2 − 18 3 2 + 5 3 9
√
√
−6 3 2 +5 3 9
Simplify each radical, finding perfect cube factors
Take roots where possible
Multiply coefficients
√
√
Combine like terms 12 3 2 − 18 3 2
Our Solution
296
8
...
4
Radicals - Multiply and Divide Radicals
Objective: Multiply and divide radicals using the product and quotient
rules of radicals
...
The
prodcut rule of radicals which we have already been using can be generalized as
follows:
√
√
√
Product Rule of Radicals: a m b · c m d = ac m bd
Another way of stating this rule is we are allowed to multiply the factors outside
the radical and we are allowed to multiply the factors inside the radicals, as long
as the index matches
...
Example 392
...
√
√
2 3 18 · 6 3 15
√
12 3 270
√
12 3 27 · 10
√
12 · 3 3 10
√
36 3 10
Multiply outside and inside the radical
Simplify the radical, divisible by 27
Take cube root where possible
Multiply coefficients
Our Solution
When multiplying with radicals we can still use the distributive property or FOIL
just as we could with variables
...
√ √
√
7 6 (3 10 − 5 15 )
√
√
21 60 − 35 90
√
√
21 4 · 15 − 35 9 · 10
√
√
21 · 2 15 − 35 · 3 10
√
√
42 15 − 105 10
Distribute, following rules for multiplying radicals
Simplify each radical, finding perfect square factors
Take square root where possible
Multiply coefficients
Our Solution
Example 395
...
In one of the tables
√
there is an approximation of 2 accurate to five decimal places (1
...
√
√
√
√
(2 5 − 3 6 )(7 2 − 8 7 )
√
√
√
√
14 10 − 16 35 − 21 12 − 24 42
√
√
√
√
14 10 − 16 35 − 21 4 · 3 − 24 42
√
√
√
√
14 10 − 16 35 − 21 · 2 3 − 24 42
√
√
√
√
14 10 − 16 35 − 42 3 − 24 42
FOIL, following rules for multiplying radicals
Simplify radicals, find perfect square factors
Take square root where possible
Multiply coefficient
Our Solution
As we are multiplying we always look at our final solution to check if all the radicals are simplified and all like radicals or like terms have been combined
...
√
m
a b
a
b
Quotient Rule of Radicals: m = m
√
c
d
c d
Example 397
...
If there is a radical in the
denominator we will rationalize it, or clear out any radicals in the denominator
...
The problems we will consider here will all have a monomial in the denominator
...
The index tells us how many of each factor we will need to clear the radical
...
This
is shown in the following examples
...
√
6
√
5
√
6
√
5
√
5
√
5
√
Multiply numerator and denominator by 5
√
30
5
Example 399
...
√
43 2
The 25 can be written as 52
...
However, this would have made the numbers very large
and we would have needed to reduce our soultion at the end
...
We will also always want to reduce our fractions (inside and out of the radical)
before we rationalize
...
√
6 14
√
12 22
Reduce coefficients and inside radical
√
7
√
2 11
√
11
√
11
√
7
√
2 11
Index is 2, need two elevens, need 1 more
√
Multiply numerator and denominator by 11
√
77
2 · 11
√
77
22
Multiply denominator
Our Solution
The same process can be used to rationalize fractions with variables
...
18 4 6x3 y 4z
8 4 10xy 6z 3
9
44
9
4
4
√
4
3x2
5y 2z 3
√
4
3x2
5y 2z 3
4
53 y 2z
4
53 y 2z
Reduce coefficients and inside radical
Index is 4
...
Multiply numerator and denominator by 4 53 y 2z
9 4 375x2 y 2z
4 · 5yz
Multiply denominator
9 4 375x2 y 2z
20yz
Our Solution
301
8
...
√
√
1) 3 5 · − 4 16
√
√
3) 12m · 15m
√
√
3
3
5) 4x3 · 2x4
√ √
7) 6 ( 2 + 2)
√
√
9) − 5 15 (3 3 + 2)
√
√
11) 5 10 (5n + 2 )
√
√
13) (2 + 2 2 )( − 3 + 2 )
√
√
15) ( 5 − 5)(2 5 − 1)
√
√
√
√
17) ( 2a + 2 3a )(3 2a + 5a )
√
√
19) ( − 5 − 4 3 )( − 3 − 4 3 )
√
√
√
2) − 5 10 · 15
√
√
4) 5r 3 · − 5 10r 2
√
√
3
3
6) 3 4a4 · 10a3
√ √
√
8) 10 ( 5 + 2 )
√
√
10) 5 15 (3 3 + 2)
√ √
√
12) 15 ( 5 − 3 3v )
√
√
14) ( − 2 + 3 )( − 5 + 2 3 )
√
√
√
√
16) (2 3 + 5 )(5 3 + 2 4 )
√ √
√
√
18) ( − 2 2p + 5 5 )( 5p + 5p )
√
√
20) (5 2 − 1)( − 2m + 5)
21)
12
√
5 100
22)
23)
√
5
√
4 125
24)
25)
27)
29)
√
√
15
√
2 4
√
12
√
3
√
10
√
6
26)
2
√
3 5
√
2 4
√
3 3
28)
√
4 3
√
15
5x2
4
3x3 y 3
30)
31)
2p2
√
3p
32)
33)
√
33 10
√
53 27
34)
√
3
4
3xy 4
5
√
8n2
√
10n
√
3
15
√
3
64
√
4
35)
5
√
3
4 4
36)
2
√
4
2 64
37)
√
4
5 5r 4
√
4
8r 2
38)
√
4
302
4
64m4n2
8
...
It is considered bad practice to have a radical in the denominator of a fraction
...
In the lesson on dividing radicals we talked
about how this was done with monomials
...
If the binomial is in the numerator the process to rationalize the denominator is
essentially the same as with monomials
...
Example 403
...
When we have addition or subtraction in the numerator or denominator we must divide all terms by the same
number
...
√
√
2 20x5 − 12x2
√
18x
√
√
2 4 · 5x3 − 4 · 3x2
√
9 · 2x
√
√
2 · 2x2 5x − 2x 3
√
3 2x
√
√
4x2 5x − 2x 3
√
3 2x
√
√
√
(4x2 5x − 2x 3 )
2x
√
√
2x
3 2x
√
√
4x2 10x2 − 2x 6x
3 · 2x
√
√
4x3 10 − 2x 6x
6x
Simplify radicals by finding perfect squares
Simplify roots, divide exponents by 2
...
We ask ourselves, can the fraction be
reduced? Can the radicals be simplified? These steps may happen several times
on our way to the solution
...
Consider √
, if we were to multiply the denominator by
3−5
√
√
3 we would have to distribute it and we would end up with 3 − 5 3
...
So our
current method will not work
...
A
conjugate is made up of the same terms, with the opposite sign in the middle
...
The advantage of a conjugate is when we multiply them together we have
√
√
( 3 − 5)( 3 + 5), which is a sum and a difference
...
Squaring 3 and 5, with subtraction in the
middle gives the product 3 − 25 = − 22
...
This is exactly what we want
...
√
2
3−5
√
3 +5
2
√
√
3−5
3 +5
√
2 3 + 10
3 − 25
√
2 3 + 10
− 22
√
− 3−5
11
Multiply numerator and denominator by conjugate
Distribute numerator, difference of squares in denominator
Simplify denoinator
Reduce by dividing all terms by − 2
Our Solution
In the previous example, we could have reduced by dividng by 2, giving the solu√
tion
3+5
,
− 11
both answers are correct
...
√
√
15
√
5+ 3
Multiply by conjugate,
305
√
5−
√
3
√
15
√
√
5+ 3
√
√
5− 3
√
√
5− 3
√
√
√
75 − 45
5−3
√
25 · 3 − 9 · 5
2
√
√
5 3−3 5
2
Distribute numerator, denominator is difference of squares
Simplify radicals in numerator, subtract in denominator
Take square roots where possible
Our Solution
Example 406
...
We just need to remember to FOIL out the numerator
...
√
3− 5
√
2− 3
√
3− 5
√
2− 3
√
2+ 3
√
2+ 3
√
√
√
6 + 3 3 − 2 5 − 15
4−3
Multiply by conjugate, 2 +
√
3
FOIL in numerator, denominator is difference of squares
Simplify denominator
√
√
√
6 + 3 3 − 2 5 − 15
1
Divide each term by 1
√
√
√
6 + 3 3 − 2 5 − 15
Our Solution
306
Example 408
...
√
√
3x 2x + 4x3
√
5x − 3x
√
√
3x 2x + 4x3
√
5x − 3x
√
Multiply by the conjugate, 5x + 3x
√
5x + 3x
√
5x + 3x
FOIL in numerator,
denominator is difference of squares
√
√
√
√
15x2 2x + 3x 6x2 + 5x 4x3 + 12x4
25x2 − 3x
Simplify radicals
√
√
√
√
15x2 2x + 3x2 6 + 10x2 x + 2x2 3
25x2 − 3x
Divide each term by x
√
√
√
√
15x 2x + 3x 6 + 10x x + 2x 3
25x − 3
Our Solution
World View Note: During the 5th century BC in India, Aryabhata published a
treatise on astronomy
...
307
8
...
1)
√
4+2 3
√
9
2)
√
−4+ 3
√
4 9
3)
√
4+2 3
√
5 4
4)
√
2 3−2
√
2 16
5)
√
2−5 5
√
4 13
6)
7)
9)
√
√
2−3 3
√
3
8)
5
√
√
3 5+ 2
11)
2
√
5+
√
5+4
√
4 17
√
√
5− 2
√
3 6
5
√
3+4 5
10)
12)
2
√
5
√
√
2 3− 2
4
13)
3
√
4−3 3
14)
√
15)
4
√
3+ 5
16)
2
√
√
2 5 +2 3
18)
4
√
√
4 3− 5
20)
√
3+ 3
√
3−1
22)
√
2 + 10
√
√
2+ 5
17) −
4
√
4−4 2
19)
1
√
1+ 2
21)
√
√
14 − 2
√
7− 2
√
√
2−2
√
23)
√
ab − a
√
b− a
24)
25)
√
a + ab
√
√
a+ b
26)
√
a + ab
√
√
a+ b
28)
√
√
2 5+ 3
√
1− 3
27)
2+
2+
√
√
6
3
√
29)
a−
31)
6
√
√
3 2−2 3
33)
a−b
√
√
a b −b a
35)
a+
√
2−
b
5
√
−3+ 5
14 −
14 +
√
a−b
√
7
7
30)
√
32)
ab
√
√
a b −b a
34)
√
4 2 +3
√
√
3 2+ 3
36)
√
−1+ 5
√
√
2 5 +5 2
b
√
√
308
a+
b
37)
√
√
5 2+ 3
√
5+5 2
38)
309
√
√
3+ 2
√
√
2 3− 2
8
...
When we simplify radicals with exponents, we divide the exponent by the index
...
This idea is how we will
define rational exponents
...
We can
use this property to change any radical expression into an exponential expression
...
3
5
√
√
(6 3x )5 = (3x) 6
(5 x )3 = x 5
3
2
1
1
−
−
= (xy) 3
√ 3 =a 7 √
2
(3 xy )
(7 a )
Index is denominator
Negative exponents from reciprocals
We can also change any rational exponent into a radical expression by using the
denominator as the index
...
5
2
√
√
a 3 = (3 a )5 (2mn) 7 = (7 2mn )2
4
2
1
1
−
−
x 5 = √ 4 (xy) 9 = √
9 xy )2
5
(
( x)
Index is denominator
Negative exponent means reciprocals
World View Note: Nicole Oresme, a Mathematician born in Normandy was the
1
1
first to use rational exponents
...
However his notation went largely unnoticed
...
Example 412
...
The following table reviews all of our exponent properties
...
When multiplying we only need to multiply the numerators together and denominators together
...
Example 413
...
1
3
x y
2
5
1
3
4
Multiply
3
by each exponent
4
3
x 4 y 10
Our Solution
Example 415
...
− 1
2
5
1
3
2
25x y 3
4
−
9x 5 y 2
5
15
25x y
12
9x 15 y
− 1
4
10
2
15
− 10
Using order of operations, simplify inside parenthesis first
Need common denominators before we can subtract exponents
Subtract exponents, be careful of the negative:
15
15 19
4
4
− −
+
=
=
10
10 10 10
10
1
7
25x
− 15
y
−2
19
10
The negative exponent will flip the fraction
9
1
2
9
7
25x
− 15
y
The exponent
19
10
1
goes on each factor
2
1
92
1
25 2 x
1
7
− 30
19
1
Evaluate 9 2 and 25 2 and move negative exponent
y 20
7
3x 30
19
Our Solution
5y 20
It is important to remember that as we simplify with rational exponents we are
using the exact same properties we used when simplifying integer exponents
...
It may be worth
reviewing your notes on exponent properties to be sure your comfortable with
using the properties
...
6 Practice - Rational Exponents
Write each expression in radical form
...
√
1
6) v
5) √ 3
( 6x )
7)
1
√
(4 n )7
8)
√
5a
Evaluate
...
Your answer should contain only positive exponents
...
7
Radicals - Radicals of Mixed Index
Objective: Reduce the index on a radical and multiply or divide radicals of different index
...
One thing we are allowed to do is
reduce, not just the radicand, but the index as well
...
Example 417
...
If we notice a common
factor in the index and all the exponnets on every factor we can reduce by
dividing by that common factor
...
√
a6b9c15
√
8
a2b3c5
24
Index and all exponents are divisible by 3
Our Solution
We can use the same process when there are coefficients in the problem
...
Example 419
...
First we will consider an example using rational exponents, then identify
the pattern we can use
...
√
√
3
4
ab2 a2b
1
3
(ab2) (a2b)
1
3
2
3
2
4
a b a b
4
12
8
12
6
12
1
4
1
4
3
12
a b a b
√
12 4 8 6 3
abab
√
12 10 11
a b
Rewrite as rational exponents
Multiply exponents
To have one radical need a common denominator, 12
Write under a single radical with common index, 12
Add exponents
Our Solution
To combine the radicals we need a common index (just like the common denominator)
...
This process is
shown in the next example
...
√
√
6
4
a2b3 a2b
√
12
a6b9a4b2
√
12 10 11
a b
Common index is 12
...
Example 422
...
Multiply first index and exponents by 3, second by 5
Add exponents
Simplify by dividing exponents by index, remainder is left inside
Our Solution
Just as with reducing the index, we will rewrite coefficients as exponential expressions
...
Example 423
...
Multiply first index and exponents by 4, second by 3
Add exponents (even on the 2)
217x11y 13
Simplify by dividing exponents by index, remainder is left inside
12
Simplify 25
5 11
2x y
2y 12 32x11y
Our Solution
315
If there is a binomial in the radical then we need to keep that binomial together
through the entire problem
...
3x(y + z) 3 9x(y + z)2
3x(y + z) 3 32x(y + z)2
6
3 3
3 4 2
3 x (y + z) 3 x (y + z)
6
3(y + z)
4
37x5(y + z)7
6
5
3x (y + z)
Rewrite 9 as 32
Common index: 6
...
In 1637 Rene Descartes was the first to put
a line over the entire radical expression
...
The only difference is our final answer cannot have a radical over the
denominator
...
6
x4 y 3z 2
8
x7 y 2z
24
x16y 12z 8
x21y 6z 3
Subtract exponents
x−5 y 6z 5
Negative exponent moves to denominator
24
24
24
Common index is 24
...
7 Practice - Radicals of Mixed Index
Reduce the following radicals
...
√ √
13) 3 5 6
√ √
15) x 3 7y
√ √
17) x 3 x − 2
√
19) 5 x2 y xy
xy 2 3 x2 y
√
√
4
5
23) a2bc2 a2b3c
√ √
4
25) a a3
√ √
5
27) b2 b3
21)
4
xy 3
3
x2 y
√
√
4
31) 9ab3 3a4b
29)
33)
3
16)
18)
20)
22)
41)
3xy 2z 4 9x3 yz 2
34)
2x3 y 3
√
36)
√
3
a2
√
4 a
x2 y 3
√
3 xy
√
ab3c
√
5 2 3 −1
a b c
4
6
32)
38)
4
39)
√ √
3
74 5
√ √
3 y 5 3z
√ √
4
3x y + 4
√ √
5
ab 2a2b2
√
√
5 2 3 4 2
ab
ab
x2 yz 3 5 x2 yz 2
√ √
3
6
26) x2 x5
√ √
4
3
28) a3 a2
√
√
5
30) a3b ab
24)
27a5(b + 1) 3 81a(b + 1)4
35)
37)
14)
(3x − 1)3
43)
5
(3x − 1)3
3
45)
(2x + 1)2
5
(2x + 1)2
40)
√
3
x2
√
5 x
√
5 4 2
a b
√
3
ab2
5
4
4
317
x3 y 4z 9
xy −2z
3
46)
√
3
4xy 2
ab2c
8x (y + z)5
42)
44)
a4b3c4
3
3
(2 + 5x)2
(2 + 5x)
(5 − 3x)3
(5 − 3x)2
3
4x2(y + z)2
8
...
World View Note: When mathematics was first used, the primary purpose was
for counting
...
However, the ancient Egyptians quickly developed the need for “a
part” and so they made up a new type of number, the ratio or fraction
...
The Mayans of Central America later made up the number
zero when they found use for it as a placeholder
...
In mathematics, when the current number system does not provide the tools to
solve the problems the culture is working with, we tend to make up new ways for
dealing with the problem that can solve the problem
...
This is also the case for the square roots of negative numbers
...
Definition of Imaginary Numbers: i2 = − 1 (thus i =
√
− 1)
√
3
Examples of imaginary numbers include 3i, − 6i, 5 i and 3i 5
...
With this definition, the square root of a negative number is no longer undefined
...
First we will consider exponents on imaginary numbers
...
If we multiply both sides of the definition
by i, the equation becomes i3 = − i
...
Multiplying again by i gives i5 = i
...
And if this
pattern continues we see a cycle forming, the exponents on i change we cycle
through simplified answers of i, − 1, − i, 1
...
i35
8R3
i3
−i
Divide exponent by 4
Use remainder as exponent on i
Simplify
Our Solution
i124
31R0
i0
1
Divide exponent by 4
Use remainder as exponent on i
Simplify
Our Solution
Example 427
...
This means when adding and subtracting complex numbers we simply add or combine like terms
...
(2 + 5i) + (4 − 7i)
6 − 2i
Combine like terms 2 + 4 and 5i − 7i
Our Solution
It is important to notice what operation we are doing
...
We only use FOIL to multiply
...
For subtraction problems the idea is the same, we need to remember to first distribute the negative onto all the terms in the parentheses
...
(4 − 8i) − (3 − 5i)
4 − 8i − 3 + 5i
1 − 3i
Distribute the negative
Combine like terms 4 − 3 and − 8i + 5i
Our Solution
Addition and subtraction can be combined into one problem
...
(5i) − (3 + 8i) + ( − 4 + 7i)
5i − 3 − 8i − 4 + 7i
− 7 + 4i
Distribute the negative
Combine like terms 5i − 8i + 7i and − 3 − 4
Our Solution
Multiplying with complex numbers is the same as multiplying with variables with
one exception, we will want to simplify our final answer so there are no exponents
on i
...
(3i)(7i)
21i2
21( − 1)
− 21
Multilpy coefficients and i ′s
Simplify i2 = − 1
Multiply
Our Solution
Example 432
...
(2 − 4i)(3 + 5i)
6 + 10i − 12i − 20i2
6 + 10i − 12i − 20( − 1)
6 + 10i − 12i + 20
26 − 2i
FOIL
Simplify i2 = − 1
Multiply
Combine like terms 6 + 20 and 10i − 12i
Our Solution
Example 434
...
The next example uses
the shortcut
Example 435
...
If i is
√
− 1 , and it is in the denominator of a fraction, then we have a radical in the
denominator! This means we will want to rationalize our denominator so there
are no i’s
...
Example 436
...
Example 437
...
We will use the
product rule and simplify the negative as a factor of negative one
...
Example 438
...
√
√
− 24
−1·4·6
√
2i 6
Find perfect square factors, including − 1
Square root of − 1 is i, square root of 4 is 2
Our Solution
When simplifying complex radicals it is important that we take the − 1 out of the
radical (as an i) before we combine radicals
...
√ √
−6 −3
√
√
(i 6 )(i 3 )
√
− 18
√
− 9·2
√
−3 2
Simplify the negatives, bringing i out of radicals
Multiply i by i is i2 = − 1, also multiply radicals
Simplify the radical
Take square root of 9
Our Solution
If there are fractions, we need to make sure to reduce each term by the same
number
...
Example 441
...
This will be explored in more detail in a later section
...
8 Practice - Complex Numbers
Simplify
...
1 Solving with Radicals
...
2 Solving with Exponents
...
3 Complete the Square
...
4 Quadratic Formula
...
5 Build Quadratics From Roots
...
6 Quadratic in Form
...
7 Application: Rectangles
...
8 Application: Teamwork
...
9 Simultaneous Products
...
10 Application: Revenue and Distance
...
11 Graphs of Quadratics
...
1
Quadratics - Solving with Radicals
Objective: Solve equations with radicals and check for extraneous solutions
...
As you might expect,
to clear a root we can raise both sides to an exponent
...
To clear a cubed root we can raise
both sides to a third power
...
This will only happen if the index on the root is even, and it will not happen all
the time
...
If a value does not work it is called an extraneous solution
and not included in the final solution
...
√
7x + 2 = 4
( 7x + 2 )2 = 42
7x + 2 = 16
−2 −2
7x = 14
7 7
x=2
7(2) + 2 = 4
√
14 + 2 = 4
√
16 = 4
4=4
x=2
√
Even index! We will have to check answers
Square both sides, simplify exponents
Solve
Subtract 2 from both sides
Divide both sides by 7
Need to check answer in original problem
Multiply
Add
Square root
True! It works!
Our Solution
Example 443
...
√
4
3x + 6 = − 3
( 3x + 6 ) = ( − 3)4
3x + 6 = 81
−6 −6
3x = 75
3 3
x = 25
4
3(25) + 6 = − 3
√
4
75 + 6 = − 3
√
4
81 = − 3
3=−3
No Solution
√
4
Even index! We will have to check answers
Rise both sides to fourth power
Solve
Subtract 6 from both sides
Divide both sides by 3
Need to check answer in original problem
Multiply
Add
Take root
False, extraneous solution
Our Solution
If the radical is not alone on one side of the equation we will have to solve for the
radical before we raise it to an exponent
Example 445
...
In this case we remember to set the equation to zero and solve by
factoring
...
Sometimes both values work, sometimes only one, and sometimes neither
works
...
This means we must first isolate one of them before we square both sides
...
√
√
3x − 8 − x = 0
√
√
+ x+ x
√
√
3x − 8 = x
√
√
( 3x − 8 )2 = ( x )2
3x − 8 = x
− 3x − 3x
− 8 = − 2x
−2 −2
4=x
√
3(4) − 8 − 4 = 0
√
√
12 − 8 − 4 = 0
√
√
4 − 4 =0
Even index! We will have to check answers
√
Isolate first root by adding x to both sides
Square both sides
Evaluate exponents
Solve
Subtract 3x from both sides
Divide both sides by − 2
Need to check answer in original
Multiply
Subtract
Take roots
328
2−2=0
0=0
x=4
Subtract
True! It works
Our Solution
When there is more than one square root in the problem, after isolating one root
and squaring both sides we may still have a root remaining in the problem
...
When isolating, we will isolate the term with the square root
...
Example 447
...
√
√
3x + 9 − x + 4 = − 1
√
√
+ x+4 + x+4
√
√
3x + 9 = x + 4 − 1
√
√
( 3x + 9 )2 = ( x + 4 − 1)2
√
3x + 9 = x + 4 − 2 x + 4 + 1
√
3x + 9 = x + 5 − 2 x + 4
−x−5−x−5
√
2x + 4 = − 2 x + 4
√
(2x + 4)2 = ( − 2 x + 4 )2
4x2 + 16x + 16 = 4(x + 4)
4x2 + 16x + 16 = 4x + 16
− 4x − 16 − 4x − 16
4x2 + 12x = 0
4x(x + 3) = 0
4x = 0 or x + 3 = 0
4 4
−3−3
x = 0 or x = − 3
3(0) + 9 −
3( − 3) + 9 −
√
−9+9 −
√
(0) + 4 = − 1
√
9− 4=−1
3−2=−1
1=−1
( − 3) + 4 = − 1
( − 3) + 4 = − 1
√
√
0− 1=−1
0−1=−1
−1=−1
x=−3
Even index! We will have to check answers
√
Isolate the first root by adding x + 4
Square both sides
Evaluate exponents
Combine like terms
Isolate the term with radical
Subtract x and 5 from both sides
Square both sides
Evaluate exponents
Distribute
Make equation equal zero
Subtract 4x and 16 from both sides
Factor
Set each factor equal to zero
Solve
Check solutions in original
Check x = 0 first
Take roots
Subtract
False, extraneous solution
Check x = − 3
Add
Take roots
Subtract
True! It works
Our Solution
330
9
...
1)
√
3)
√
2x + 3 − 3 = 0
2)
√
6x − 5 − x = 0
4)
√
5) 3 + x =
7)
√
9)
√
√
6x + 13
3 − 3x − 1 = 2x
√
4x + 5 − x + 4 = 2
11)
√
13)
√
15)
√
5x + 1 − 4 = 0
√
x+2− x =2
6) x − 1 =
8)
√
√
12)
√
√
2x + 6 − x + 4 = 1
14)
√
6 − 2x −
16)
√
√
2x + 3 = 3
331
7−x
√
2x + 2 = 3 + 2x − 1
10)
√
2x + 4 − x + 3 = 1
√
√
3x + 4 − x + 2 = 2
√
7x + 2 − 3x + 6 = 6
4x − 3 −
2 − 3x −
√
√
3x + 1 = 1
3x + 7 = 3
9
...
Another type of equation we can solve is one with exponents
...
This is done with very few unexpected results when the exponent is odd
...
√
5
x5 = 32
√
x5 = 5 32
x=2
Use odd root property
Simplify roots
Our Solution
However, when the exponent is even we will have two results from taking an even
root of both sides
...
This is because
both 32 = 9 and ( − 3)2 = 9
...
x4 = 16
Use even root property ( ± )
332
√
4
√
x4 = ± 4 16
x=±2
Simplify roots
Our Solution
World View Note: In 1545, French Mathematicain Gerolamo Cardano published his book The Great Art, or the Rules of Algebra which included the solution of an equation with a fourth power, but it was considered absurd by many to
take a quantity to the fourth power because there are only three dimensions!
Example 451
...
If the roots did
not simplify to rational numbers we can keep the ± in the equation
...
(6x − 9)2 = 45
√
(6x − 9)2 = ± 45
√
6x − 9 = ± 3 5
+9 +9
√
6x = 9 ± 3 5
6
6
√
9±3 5
x=
6
√
3± 5
x=
2
Use even root property ( ± )
Simplify roots
Use one equation because root did not simplify to rational
Add 9 to both sides
Divide both sides by 6
Simplify, divide each term by 3
Our Solution
333
When solving with exponents, it is important to first isolate the part with the
exponent before taking any roots
...
(x + 4)3 − 6 = 119
+6 +6
(x + 4)3 = 125
√
3
(x + 4)3 = 125
x+4=5
−4−4
x=1
Isolate part with exponent
Use odd root property
Simplify roots
Solve
Subtract 4 from both sides
Our Solution
Example 454
...
Recall that a n = (n a )
...
Then we can clear the radical by raising both sides to an exponent
(remember to check answers if the index is even)
...
2
(4x + 1) 5 = 9
√
5
( 4x + 1 )2 = 9
√
√
(5 4x + 1 )2 = ± 9
Rewrite as a radical expression
Clear exponent first with even root property ( ± )
Simplify roots
334
√
5
4x + 1 = ± 3
( 4x + 1 )5 = ( ± 3)5
4x + 1 = ± 243
4x + 1 = 243 or 4x + 1 = − 243
−1 −1
−1
−1
4x = 242 or 4x = − 244
4
4
4
4
121
x=
, − 61
2
√
5
Clear radical by raising both sides to 5th power
Simplify exponents
Solve, need 2 equations!
Subtract 1 from both sides
Divide both sides by 4
Our Solution
Example 456
...
Raise both sides to 4th power
Solve
Add 2 to both sides
Divide both sides by 3
Need to check answer in radical form of problem
Multiply
Subtract
Evaluate root
Evaluate exponent
True! It works
Our Solution
With rational exponents it is very helpful to convert to radical form to be able to
see if we need a ± because we used the even root property, or to see if we need
to check our answer because there was an even root in the problem
...
335
9
...
1) x2 = 75
2) x3 = − 8
3) x2 + 5 = 13
4) 4x3 − 2 = 106
5) 3x2 + 1 = 73
6) (x − 4)2 = 49
7) (x + 2)5 = − 243
8) (5x + 1)4 = 16
9) (2x + 5)3 − 6 = 21
10) (2x + 1)2 + 3 = 21
2
11) (x − 1) 3 = 16
3
12) (x − 1) 2 = 8
3
13) (2 − x) 2 = 27
4
14) (2x + 3) 3 = 16
2
15) (2x − 3) 3 = 4
1
16) (x + 3)
−3
2
1 −3
17) (x + 2 )
=4
5
−3
=4
18) (x − 1)
= 32
20) (x + 3) 2 = − 8
21) (3x − 2) 5 = 16
22) (2x + 3) 2 = 27
3
5
19) (x − 1)
−2
= 32
4
3
3
23) (4x + 2) 5 = − 8
4
24) (3 − 2x) 3 = − 81
336
9
...
When solving quadratic equations in the past we have used factoring to solve for
our variable
...
Example 457
...
Consider
the following equation: x2 − 2x − 7 = 0
...
To find these two
solutions we will use a method known as completing the square
...
The next example reviews the square root
property
...
(x + 5)2 = 18
√
(x + 5)2 = ± 18
√
x+5=±3 2
−5 −5
√
x=−5±3 2
Square root of both sides
Simplify each radical
Subtract 5 from both sides
Our Solution
337
To complete the square, or make our problem into the form of the previous
example, we will be searching for the third term in a trinomial
...
This is shown in the following examples, where we
find the number that completes the square and then factor the perfect square
...
x2 + 8x + c
c=
1
·b
2
2
and our b = 8
2
1
· 8 = 42 = 16
2
x2 + 8x + 16
(x + 4)2
The third term to complete the square is 16
Our equation as a perfect square, factor
Our Solution
Example 460
...
5
x2 + x + c
3
1 5
·
2 3
2
=
5
6
2
=
25
36
c=
1
·b
2
2
and our b = 8
The third term to complete the square is
338
25
36
5
25
x2 + x +
3
36
x+
5
6
Our equation as a perfect square, factor
2
Our Solution
The process in the previous examples, combined with the even root property, is
used to solve quadratic equations by completing the square
...
3x2 + 18x − 6 = 0
+6+6
2
3x + 18x
=6
Problem
1
...
Divide each term by a
x + 6x
3
...
Add to both sides of equation
5
...
The advantage of this method is it can be used to solve any quadratic equation
...
Example 462
...
x2 − 3x − 2 = 0
+2+2
2
x − 3x
1
·3
2
2
=2
2
3
2
=
=
9
4
9 8 9 17
2 4
+ = + =
4 4 4
1 4
4
x2 − 3x +
9 8 9 17
= + =
4
4 4 4
3
x−
2
3
x−
2
2
2
=±
=
Separate constant from variables
Add 2 to both sides
No a, find number to complete the square
Add
9
to both sides,
4
Need common denominator (4) on right
Factor
17
4
Solve using the even root property
17
4
Simplify roots
√
3 ± 17
x− =
2
2
Add
3
to both sides,
2
340
1
·b
2
2
+
3 3
+
2 2
3±
x=
we already have a common denominator
√
17
Our Solution
2
Example 464
...
Once we get comfortable solving by completing the
square and using the five steps, any quadratic equation can be easily solved
...
3 Practice - Complete the Square
Find the value that completes the square and then rewrite as a perfect
square
...
11) v 2 − 8v + 45 = 0
15) 5k 2 − 10k + 48 = 0
17) x2 + 10x − 57 = 4
19) n2 − 16n + 67 = 4
21) 2x2 + 4x + 38 = − 6
23) 8b2 + 16b − 37 = 5
25) x2 = − 10x − 29
12) b2 + 2b + 43 = 0
16) 8a2 + 16a − 1 = 0
18) p2 − 16p − 52 = 0
20) m2 − 8m − 3 = 6
22) 6r 2 + 12r − 24 = − 6
24) 6n2 − 12n − 14 = 4
26) v 2 = 14v + 36
27) n2 = − 21 + 10n
28) a2 − 56 = − 10a
31) 2x2 + 63 = 8x
32) 5n2 = − 10n + 15
29) 3k 2 + 9 = 6k
33) p2 − 8p = − 55
30) 5x2 = − 26 + 10x
34) x2 + 8x + 15 = 8
35) 7n2 − n + 7 = 7n + 6n2
36) n2 + 4n = 12
39) 5x2 + 5x = − 31 − 5x
40) 8n2 + 16n = 64
37) 13b2 + 15b + 44 = − 5 + 7b2 + 3b
41) v 2 + 5v + 28 = 0
43) 7x2 − 6x + 40 = 0
38) − 3r 2 + 12r + 49 = − 6r 2
42) b2 + 7b − 33 = 0
44) 4x2 + 4x + 25 = 0
45) k 2 − 7k + 50 = 3
46) a2 − 5a + 25 = 3
49) m2 = − 15 + 9m
50) n2 − n = − 41
47) 5x2 + 8x − 40 = 8
51) 8r 2 + 10r = − 55
53) 5n2 − 8n + 60 = − 3n + 6 + 4n2
55) − 2x2 + 3x − 5 = − 4x2
48) 2p2 − p + 56 = − 8
52) 3x2 − 11x = − 18
54) 4b2 − 15b + 56 = 3b2
56) 10v 2 − 15v = 27 + 4v 2 − 6v
342
9
...
The general from of a quadratic is ax2 + bx + c = 0
...
ax2 + bc + c = 0
−c−c
ax2 + bx
=−c
a
a
a
b
−c
x2 + x =
a
a
2
2
b
1 b
b2
=
= 2
·
2a
2 a
4a
c 4a
b2
−
2
a 4a
4a
=
4ac b2 − 4ac
b2
− 2=
4a2
4a2 4a
b
b2
b2
4ac b2 − 4ac
x2 + x + 2 = 2 − 2 =
4a
4a
a
4a
4a2
b
x+
2a
b
x+
2a
2
2
=±
=
Separate constant from variables
Subtract c from both sides
Divide each term by a
Find the number that completes the square
Add to both sides,
Get common denominator on right
Factor
b2 − 4ac
4a2
Solve using the even root property
b2 − 4ac
4a2
Simplify roots
√
b
± b2 − 4ac
x+
=
2a
2a
√
− b ± b2 − 4ac
x=
2a
Subtract
b
from both sides
2a
Our Solution
This solution is a very important one to us
...
Once we identify what a, b, and c are in the quadratic, we can substitute those
343
−b±
b2 − 4ac
values into x =
and we will get our two solutions
...
However, at that time mathematics was not
done with variables and symbols, so the formula he gave was, “To the absolute
number multiplied by four times the square, add the square of the middle term;
the square root of the same, less the middle term, being divided by twice the
square
is
the
value
...
We can use the quadratic formula to solve any quadratic, this is shown in the following examples
...
x2 + 3x + 2 = 0
− 3 ± 32 − 4(1)(2)
x=
2(1)
√
−3± 9−8
x=
2 √
−3± 1
x=
2
−3±1
x=
2
−2
−4
x=
or
2
2
x = − 1 or − 2
a = 1, b = 3, c = 2, use quadratic formula
Evaluate exponent and multiplication
Evaluate subtraction under root
Evaluate root
Evaluate ± to get two answers
Simplify fractions
Our Solution
As we are solving using the quadratic formula, it is important to remember the
equation must fist be equal to zero
...
25x2 = 30x + 11
− 30x − 11 − 30x − 11
25x2 − 30x − 11 = 0
30 ± ( − 30)2 − 4(25)( − 11)
x=
2(25)
First set equal to zero
Subtract 30x and 11 from both sides
a = 25, b = − 30, c = − 11, use quadratic formula
Evaluate exponent and multiplication
344
x=
30 ±
√
900 + 1100
50 √
30 ± 2000
x=
50 √
30 ± 20 5
x=
50 √
3±2 5
x=
5
Evaluate addition inside root
Simplify root
Reduce fraction by dividing each term by 10
Our Solution
Example 468
...
We can end up with only one solution if the square root simplifies to zero
...
4x2 − 12x + 9 = 0
12 ± ( − 12)2 − 4(4)(9)
x=
2(4)
√
12 ± 144 − 144
x=
8
√
12 ± 0
x=
8
12 ± 0
x=
8
12
x=
8
3
x=
2
a = 4, b = − 12, c = 9, use quadratic formula
Evaluate exponents and multiplication
Evaluate subtraction inside root
Evaluate root
Evaluate ±
Reduce fraction
Our Solution
345
If a term is missing from the quadratic, we can still solve with the quadratic formula, we simply use zero for that term
...
Example 470
...
It is important to be familiar
with all three as each has its advantage to solving quadratics
...
1
...
If a = 1 and b is even, complete the square
3
...
Remember completing the square and quadratic formula will always work to solve
any quadratic
...
346
9
...
1) 4a2 + 6 = 0
2) 3k 2 + 2 = 0
3) 2x2 − 8x − 2 = 0
4) 6n2 − 1 = 0
5) 2m2 − 3 = 0
6) 5p2 + 2p + 6 = 0
7) 3r 2 − 2r − 1 = 0
8) 2x2 − 2x − 15 = 0
9) 4n2 − 36 = 0
10) 3b2 + 6 = 0
11) v 2 − 4v − 5 = − 8
12) 2x2 + 4x + 12 = 8
13) 2a2 + 3a + 14 = 6
14) 6n2 − 3n + 3 = − 4
15) 3k 2 + 3k − 4 = 7
16) 4x2 − 14 = − 2
17) 7x2 + 3x − 16 = − 2
18) 4n2 + 5n = 7
19) 2p2 + 6p − 16 = 4
20) m2 + 4m − 48 = − 3
21) 3n2 + 3n = − 3
22) 3b2 − 3 = 8b
23) 2x2 = − 7x + 49
24) 3r 2 + 4 = − 6r
25) 5x2 = 7x + 7
26) 6a2 = − 5a + 13
27) 8n2 = − 3n − 8
28) 6v 2 = 4 + 6v
29) 2x2 + 5x = − 3
30) x2 = 8
31) 4a2 − 64 = 0
32) 2k 2 + 6k − 16 = 2k
33) 4p2 + 5p − 36 = 3p2
34) 12x2 + x + 7 = 5x2 + 5x
35) − 5n2 − 3n − 52 = 2 − 7n2
36) 7m2 − 6m + 6 = − m
37) 7r 2 − 12 = − 3r
38) 3x2 − 3 = x2
39) 2n2 − 9 = 4
40) 6b2 = b2 + 7 − b
347
9
...
Up to this point we have found the solutions to quadratics by a method such as
factoring or completing the square
...
We will start with rational solutions
...
Once we have done this our expressions
will become the factors of the quadratic
...
The solutions are 4 and − 2
x = 4 or x = − 2
−4−4 +2 +2
x − 4 = 0 or x + 2 = 0
(x − 4)(x + 2) = 0
x2 + 2x − 4x − 8
x2 − 2x − 8 = 0
Set each solution equal to x
Make each equation equal zero
Subtract 4 from first, add 2 to second
These expressions are the factors
FOIL
Combine like terms
Our Solution
If one or both of the solutions are fractions we will clear the fractions by multiplying by the denominators
...
2
3
and
3
4
2
3
x = or x =
3
4
3x = 2 or 4x = 3
−2−2
−3−3
3x − 2 = 0 or 4x − 3 = 0
(3x − 2)(4x − 3) = 0
12x2 − 9x − 8x + 6 = 0
The solution are
Set each solution equal to x
Clear fractions by multiplying by denominators
Make each equation equal zero
Subtract 2 from the first, subtract 3 from the second
These expressions are the factors
FOIL
Combine like terms
348
12x2 − 17x + 6 = 0
Our Solution
If the solutions have radicals (or complex numbers) then we cannot use reverse
factoring
...
When there
are radicals the solutions will always come in pairs, one with a plus, one with a
minus, that can be combined into “one” solution using ±
...
This will clear the radical from our
problem
...
The solutions are
√
√
3 and − 3
√
x=± 3
x2 = 3
−3−3
x2 − 3 = 0
Write as ′′one ′′ expression equal to x
Square both sides
Make equal to zero
Subtract 3 from both sides
Our Solution
We may have to isolate the term with the square root (with plus or minus) by
adding or subtracting
...
√
√
The solutions are 2 − 5 2 and 2 + 5 2
√
x=2±5 2
−2−2
√
x−2=±5 2
x2 − 4x + 4 = 25 · 2
x2 − 4x + 4 = 50
− 50 − 50
2
x − 4x − 46 = 0
Write as ′′one ′′ expression equal to x
Isolate the square root term
Subtract 2 from both sides
Square both sides
Make equal to zero
Subtract 50
Our Solution
World View Note: Before the quadratic formula, before completing the square,
before factoring, quadratics were solved geometrically by the Greeks as early as
300 BC! In 1079 Omar Khayyam, a Persian mathematician solved cubic equations
geometrically!
If the solution is a fraction we will clear it just as before by multiplying by the
denominator
...
√
√
2+ 3
2− 3
The solutions are
and
4
4√
2± 3
x=
4√
4x = 2 ± 3
−2−2
√
4x − 2 = ± 3
16x2 − 16x + 4 = 3
−3−3
2
16x − 16x + 1 = 0
Write as ′′one ′′ expresion equal to x
Clear fraction by multiplying by 4
Isolate the square root term
Subtract 2 from both sides
Square both sides
Make equal to zero
Subtract 3
Our Solution
The process used for complex solutions is identical to the process used for radicals
...
The solutions are 4 − 5i and 4 + 5i
x = 4 ± 5i
−4−4
x − 4 = ± 5i
x2 − 8x + 16 = 25i2
x2 − 8x + 16 = − 25
+ 25 + 25
x2 − 8x + 41 = 0
Write as ′′one ′′ expression equal to x
Isolate the i term
Subtract 4 from both sides
Square both sides
i2 = − 1
Make equal to zero
Add 25 to both sides
Our Solution
Example 477
...
5 Practice - Build Quadratics from Roots
From each problem, find a quadratic equation with those numbers as
its solutions
...
6
Quadratics - Quadratic in Form
Objective: Solve equations that are quadratic in form by substitution
to create a quadratic equation
...
A quadratic is any equation of the form 0 =
a x2 + bx + c, however, we can use the skills learned to solve quadratics to solve
problems with higher (or sometimes lower) powers if the equation is in what is
called quadratic form
...
If this is the
case we can create a new variable, set it equal to the variable with smallest exponent
...
World View Note: Arab mathematicians around the year 1000 were the first to
use this method!
Example 478
...
We can solve this equation by factoring
Set each factor equal to zero
Solve each equation
Solutions for y, need x
...
The previous equation had four unique solutions
...
a−2 − a−1 − 6 = 0
b = a−1
b2 = a−2
b2 − b − 6 = 0
(b − 3)(b + 2) = 0
b − 3 = 0 or b + 2 = 0
+3+3
−2−2
b = 3 or b = − 2
−1
3=a
or − 2 = a−1
3−1 = a or ( − 2)−1 = a
1
1
a= ,−
3
2
Quadratic form, one exponent, − 2, is double the other, − 1
Make a new variable equal to the variable with lowest exponent
Square both sides
Substitute b2 for a−2 and b for a−1
Solve
...
We also can have irrational or complex solutions to our equations
...
±
2x4 + x2 = 6
−6−6
4
2
2x + x − 6 = 0
y = x2
y 2 = x4
2y 2 + y − 6 = 0
(2y − 3)(y + 2) = 0
2y − 3 = 0 or y + 2 = 0
−2−2
+3+3
2y = 3 or y = − 2
2 2
3
y = or y = − 2
2
3
2
= x or − 2 = x2
2
√
√
3 √ 2
= x or ± − 2 = x2
2
√
√
± 6
x=
,±i 2
2
Make equation equal to zero
Subtract 6 from both sides
Quadratic form, one exponent, 4, double the other, 2
New variable equal variable with smallest exponent
Square both sides
Solve
...
Substitute into y = x2
Square root of each side
Simplify each root, rationalize denominator
Our Solution
353
When we create a new variable for our substitution, it won’t always be equal to
just another variable
...
Example 481
...
Substitute into y = x − 7
Add 7
...
(x2 − 6x)2 = 7(x2 − 6x) − 12
− 7(x2 − 6x) + 12 − 7(x2 − 6x) + 12
(x2 − 6x)2 − 7(x2 − 6x) + 12 = 0
y = x2 − 6x
y 2 = (x2 − 6x)2
y 2 − 7y + 12 = 0
(y − 3)(y − 4) = 0
y − 3 = 0 or y − 4 = 0
+3+3
+4+4
y = 3 or y = 4
2
3 = x − 6x or 4 = x3 − 6x
2
1
· 6 = 32 = 9
2
2
12 = x − 6x + 9 or 13 = x2 − 6x + 9
Make equation equal zero
Move all terms to left
Quadratic form
Make new variable
Square both sides
Substitute into original equation
Solve by factoring
Set each factor equal to zero
Solve each equation
We have y, still need x
...
This is illustrated in
the following example, one with six solutions
...
x6 − 9x3 + 8 = 0
y = x3
y 2 = x6
y 2 − 9y + 8 = 0
(y − 1)(y − 8) = 0
y − 1 = 0 or y − 8 = 0
+1+1
+8+8
y = 1 or y = 8
x3 = 1 or x3 = 8
−1−1 −8−8
x3 − 1 = 0 or x3 − 8 = 0
(x − 1)(x2 + x + 1) = 0
x − 1 = 0 or x2 + x + 1 = 0
+1+1
x=1
√
− 1 ± 12 − 4(1)(1) 1 ± i 3
=
2(1)
2
(x − 2)(x2 + 2x + 4) = 0
x − 2 = 0 or x2 + 2x + 4 = 0
+2+2
x=2
√
− 2 ± 22 − 4(1)(4)
=−1±i 3
2(1)
√
√
1±i 3
,−1±i 3
x = 1, 2,
2
Quadratic form, one exponent, 6, double the other, 3
New variable equal to variable with lowest exponent
Square both sides
Substitute y 2 for x6 and y for x3
Solve
...
Set each factor equal to zero
Solve each equation
Solutions for y, we need x
...
Set each factor equal to zero
First equation is easy to solve
First solution
Quadratic formula on second factor
Factor the second difference of cubes
Set each factor equal to zero
...
6 Practice - Quadratic in Form
Solve each of the following equations
...
1) x4 − 5x2 + 4 = 0
2) y 4 − 9y 2 + 20 = 0
3) m4 − 7m2 − 8 = 0
4) y 4 − 29y 2 + 100 = 0
5) a4 − 50a2 + 49 = 0
6) b4 − 10b2 + 9 = 0
7) x4 − 25x2 + 144 = 0
8) y 4 − 40y 2 + 144 = 0
9) m4 − 20m2 + 64 = 0
10) x6 − 35x3 + 216 = 0
11) z 6 − 216 = 19z 3
12) y 4 − 2y 2 = 24
13) 6z 4 − z 2 = 12
14) x−2 − x−1 − 12 = 0
1
2
15) x 3 − 35 = 2x 3
16) 5y −2 − 20 = 21y −1
17) y −6 + 7y −3 = 8
18) x4 − 7x2 + 12 = 0
19) x4 − 2x2 − 3 = 0
20) x4 + 7x2 + 10 = 0
21) 2x4 − 5x2 + 2 = 0
22) 2x4 − x2 − 3 = 0
23) x4 − 9x2 + 8 = 0
24) x6 − 10x3 + 16 = 0
25) 8x6 − 9x3 + 1 = 0
26) 8x6 + 7x3 − 1 = 0
27) x8 − 17x4 + 16 = 0
28) (x − 1)2 − 4(x − 1) = 5
29) (y + b)2 − 4(y + b) = 21
30) (x + 1)2 + 6(x + 1) + 9 = 0
31) (y + 2)2 − 6(y + 2) = 16
32) (m − 1)2 − 5(m − 1) = 14
33) (x − 3)2 − 2(x − 3) = 35
34) (a + 1)2 + 2(a − 1) = 15
35) (r − 1)2 − 8(r − 1) = 20
36) 2(x − 1)2 − (x − 1) = 3
37) 3(y + 1)2 − 14(y + 1) = 5
39) (3x2 − 2x)2 + 5 = 6(3x2 − 2x)
2
1
38) (x2 − 3)2 − 2(x2 − 3) = 3
40) (x2 + x + 3)2 + 15 = 8(x2 + x + 3)
41) 2(3x + 1) 3 − 5(3x + 1) 3 = 88
42) (x2 + x)2 − 8(x2 + x) + 12 = 0
43) (x2 + 2x)2 − 2(x2 + 2x) = 3
44) (2x2 + 3x)2 = 8(2x2 + 3x) + 9
45) (2x2 − x)2 − 4(2x2 − x) + 3 = 0
46) (3x2 − 4x)2 = 3(3x2 − 4x) + 4
356
9
...
An application of solving quadratic equations comes from the formula for the area
of a rectangle
...
To solve problems with rectangles we will first draw a picture to
represent the problem and use the picture to set up our equation
...
The length of a rectangle is 3 more than the width
...
Length is 4 more, or x + 4, and area is 40
...
Length is x + 3, substitute 5 for x to find length
Our Solution
The above rectangle problem is very simple as there is only one rectangle
involved
...
Example 485
...
Find the
side of the original square
...
The length of a rectangle is 4 ft greater than the width
...
Find the dimensions of
the original rectangle
...
width becomes x + 3, length x + 4 + 3 = x + 7
x+7
(x + 3)(x + 7) = x(x + 4) + 33
x2 + 10x + 21 = x2 + 4x + 33
− x2
− x2
10x + 21 = 4x + 33
− 4x
− 4x
6x + 21 = 33
− 21 − 21
6x = 12
6 6
x=2
(2) + 4 = 6
2 ft by 6ft
Area is 33 more than original, x(x + 4) + 33
Set up equation, length times width is area
Subtract x2 from both sides
Move variables to one side
Subtract 4x from each side
Subtract 21 from both sides
Divide both sides by 6
x is the width of the original
x + 4 is the length
...
Perimeter is found by adding all
the sides of a polygon together
...
So we can use the equation P = 2l + 2w (twice the length plus
twice the width)
...
The area of a rectangle is 168 cm2
...
What are the dimensions of the rectangle?
x
y
xy = 168
2x + 2y = 52
− 2x
− 2x
2y = − 2x + 52
We don ′t know anything about length or width
Use two variables, x and y
Length times width gives the area
...
Solve by substitution, isolate y
Divide each term by 2
359
2
2
2
y = − x + 26
x( − x + 26) = 168
− x2 + 26x = 168
x2 − 26x = − 168
2
1
· 26 = 132 = 169
2
x2 − 26x + 324 = 1
(x − 13)2 = 1
x − 13 = ± 1
+ 13 + 13
x = 13 ± 1
x = 14 or 12
y = − (14) + 26 = 12
y = − (12) + 26 = 14
12 cm by 14cm
Substitute into area equation
Distribute
Divide each term by − 1, changing all the signs
Solve by completing the square
...
Substitute 14 into y = − x + 26
Substitute 12 into y = − x + 26
Both are the same rectangle, variables switched!
Our Solution
World View Note: Indian mathematical records from the 9th century demonstrate that their civilization had worked extensivly in geometry creating religious
alters of various shapes including rectangles
...
The
idea behind a frame problem is that a rectangle, such as a photograph, is centered
inside another rectangle, such as a frame
...
This is shown in
the following example
...
An 8 in by 12 in picture has a frame of uniform width around it
...
What is the width of the frame?
8
12 12 + 2x
8 + 2x
8 · 12 = 96
2 · 96 = 192
(12 + 2x)(8 + 2x) = 192
96 + 24x + 16x + 4x2 = 192
4x2 + 40x + 96 = 192
− 192 − 192
2
4x + 40x − 96 = 0
Draw picture, picture if 8 by 10
If frame has width x, on both sides, we add 2x
Area of the picture, length times width
Frame is the same as the picture
...
Area of everything, length times width
FOIL
Combine like terms
Make equation equal to zero by subtracting 192
Factor out GCF of 4
360
4(x2 + 10x − 24) = 0
4(x − 2)(x + 12) = 0
x − 2 = 0 or x + 12 = 0
− 12 − 12
+2+2
x = 2 or − 12
2 inches
Factor trinomial
Set each factor equal to zero
Solve each equation
Can ′t have negative frame width
...
A farmer has a field that is 400 rods by 200 rods
...
After
an hour of work, 72% of the field is left uncut
...
72) = 57600
(400 − 2x)(200 − 2x) = 57600
80000 − 800x − 400x + 4x2 = 57600
4x2 − 1200x + 80000 = 57600
− 57600 − 57600
2
4x − 1200x + 22400 = 0
4(x2 − 300x + 5600) = 0
4(x − 280)(x − 20) = 0
x − 280 = 0 or x − 20 = 0
+ 280 + 280
+ 20 + 20
x = 280 or 20
20 rods
Draw picture, outside is 200 by 400
If frame has width x on both sides,
subtract 2x from each side to get center
Area of entire field, length times width
Area of center, multiply by 28% as decimal
Area of center, length times width
FOIL
Combine like terms
Make equation equal zero
Factor out GCF of 4
Factor trinomial
Set each factor equal to zero
Solve each equation
The field is only 200 rods wide,
Can ′t cut 280 off two sides!
Our Solution
For each of the frame problems above we could have also completed the square or
use the quadratic formula to solve the trinomials
...
361
9
...
If 60 square meters are needed for the plants in the bed, what
should the dimensions of the rectangular bed be?
2) If the side of a square is increased by 5 the area is multiplied by 4
...
3) A rectangular lot is 20 yards longer than it is wide and its area is 2400 square
yards
...
4) The length of a room is 8 ft greater than it is width
...
ft
...
5) The length of a rectangular lot is 4 rods greater than its width, and its area is
60 square rods
...
6) The length of a rectangle is 15 ft greater than its width
...
Find the dimensions
...
The area of the rectangular piece is 108 in2
...
8) A room is one yard longer than it is wide
...
yd
...
50
...
9) The area of a rectangle is 48 ft2 and its perimeter is 32 ft
...
10) The dimensions of a picture inside a frame of uniform width are 12 by 16
inches
...
If the area of
the frame equals that of the mirror, what is the width of the frame
...
How wide a strip must be cut around it when
mowing the grass to have cut half of it
...
If the area of the path is equal to the area of the plot, determine
the width of the path
...
He needs an inner rectangular space
in the center for plants that must be 1 meter from the border of the bed and
362
that require 24 square meters for planting
...
How large must the page be if the length is to exceed the width by 2 inches?
16) A picture 10 inches long by 8 inches wide has a frame whose area is one half
the area of the picture
...
A strip of uniform
width is cut around the field, so that half the grain is left standing in the form
of a rectangular plot
...
If the
area of the frame equals the area of the picture find the width of the frame
...
The ring leaves 65% of the field uncut in the center
...
He starts cutting around the outside boundary spiraling around towards the
center
...
What is the width of the ring
that he has cut?
21) A frame is 15 in by 25 in and is of uniform width
...
What is the width of the
frame?
22) A farmer has a field 180 ft by 240 ft
...
How
wide a band should he cultivate?
23) The farmer in the previous problem has a neighber who has a field 325 ft by
420 ft
...
How wide a
band should his neighbor cultivate?
24) A third farmer has a field that is 500 ft by 550 ft
...
How wide a ring should he cultivate around the outside of his
field?
25) Donna has a garden that is 30 ft by 36 ft
...
How wide a ring around the outside should she cultivate?
26) A picture is 12 in by 25 in and is surrounded by a frame of uniform width
...
How wide is the
frame?
363
9
...
If it takes one person 4 hours to paint a room and another person 12 hours to
paint the same room, working together they could paint the room even quicker, it
turns out they would paint the room in 3 hours together
...
If the second person takes 12 hours to paint the room, he
1
1
1
paints 12 of the room each hour
...
Using a common denominator of 12 gives:
3
12
+
1
12
1
=
4
12
1
3
1
= 3
...
If is completed each
hour, it follows that it will take 3 hours to complete the entire room
...
If the first person does a job in
A, a second person does a job in B, and together they can do a job in T (total)
...
Teamwork Equation:
1
1
1
+ =
A B T
Often these problems will involve fractions
...
World View Note: When the Egyptians, who were the first to work with fractions, wrote fractions, they were all unit fractions (numerator of one)
...
Example 490
...
If his sister Maria helps, they can clean it in
2
2 5 hours
...
Example 491
...
Together they can complete the project in 10 hours
...
If Rachel is x, Mike is 2x
Using reciprocals, add individal times equaling total
Multiply each term by LCD, 10x
Combine like terms
We have our x, we said x was Rachel ′s time
Mike is double Rachel, this gives Mike ′s time
...
Example 492
...
If they built it
together it would take them 12 days
...
Find Britney, can ′t have negative time
Our Solution
In the previous example, when solving, one of the possible times ended up negative
...
Also, as we were solving, we had to factor x2 − 34x +
120
...
We could have also chosen to complete the square or use the quadratic formula to find our solutions
...
This means we may have
to convert minutes into hours to match the other units given in the problem
...
An electrician can complete a job in one hour less than his apprentice
...
How long would it take each of them
working alone?
1 hr 12 min = 1
1
12
hr
60
1 6
12
=1 =
5 5
60
Electrician: x − 1, Apprentice: x, Total:
6
5
1
1 5
+ =
x−1 x 6
1(6x(x − 1)) 1(6x(x − 1)) 5(6x(x − 1)
+
=
x
6
x−1
Change 1 hour 12 minutes to mixed number
Reduce and convert to fraction
Clearly define variables
Using reciprocals, make equation
Multiply each term by LCD 6x(x − 1)
366
6x + 6(x − 1) = 5x(x − 1)
6x + 6x − 6 = 5x2 − 5x
12x − 6 = 5x2 − 5x
− 12x + 6 − 12x + 6
0 = 5x2 − 17x + 6
0 = (5x − 2)(x − 3)
5x − 2 = 0 or x − 3 = 0
+2+2
+3+3
5x = 2 or x = 3
5 5
2
x = or x = 3
5
2
−3
−1=
or 3 − 1 = 2
5
5
Electrician: 2 hr, Apprentice: 3 hours
Reduce each fraction
Distribute
Combine like terms
Move all terms to one side of equation
Factor
Set each factor equal to zero
Solve each equation
Subtract 1 from each to find electrician
Ignore negative
...
A common example of this is a sink that is filled by a pipe
and emptied by a drain
...
This is shown in the
next example
...
A sink can be filled by a pipe in 5 minutes but it takes 7 minutes to drain a full
sink
...
5
17
...
8 Practice - Teamwork
1) Bills father can paint a room in two hours less than Bill can paint it
...
How much
time would each require working alone?
2) Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe
to fill a pool
...
If only the larger pipe is open, how many hours are required
to fill the pool?
3) Jack can wash and wax the family car in one hour less than Bob can
...
How much time would
each require if they worked alone?
4) If A can do a piece of work alone in 6 days and B can do it alone in 4 days,
how long will it take the two working together to complete the job?
5) Working alone it takes John 8 hours longer than Carlos to do a job
...
How long will it take each to do the
job working alone?
6) A can do a piece of work in 3 days, B in 4 days, and C in 5 days each working
alone
...
How long
will it take them to do the work together?
8) A cistern can be filled by one pipe in 20 minutes and by another in 30 minutes
...
If the
carpenter himself could do the work alone in 5 days, how long would the
assistant take to do the work alone?
11) If Sam can do a certain job in 3 days, while it takes Fred 6 days to do the
same job, how long will it take them, working together, to complete the job?
12) Tim can finish a certain job in 10 hours
...
If they work together, how long will it take them to
complete the job?
13) Two people working together can complete a job in 6 hours
...
It takes twice that long
for the outlet pipe to empty the tank
...
It takes only 3 minutes to
empty the sink when the drain is open
...
It can be emptied in 15 hrs
with the outlet pipe
...
The faucet
alone can fill the sink in 6 minutes, while it takes 8 minutes to empty it with
3
the drain
...
The sink
can be filled by a cold-water faucet in 3
...
If both faucets are open,
the sink is filled in 2
...
How long does it take to fill the sink with just
the hot-water faucet open?
1
20) A water tank is being filled by two inlet pipes
...
How long does it
take to fill the tank using only pipe B?
21) A tank can be emptied by any one of three caps
...
If all three working
8
together could empty the tank in 8 59 minutes, how long would the third take
to empty the tank?
1
1
22) One pipe can fill a cistern in 1 2 hours while a second pipe can fill it in 2 3 hrs
...
How long would it
take the third pipe alone to fill the tank?
23) Sam takes 6 hours longer than Susan to wax a floor
...
How long will it take each of them working
alone to wax the floor?
24) It takes Robert 9 hours longer than Paul to rapair a transmission
...
If they work together they can clean it in 5 minutes
...
Working together they can do
the job in 9 minutes
...
If they divide the job and work together it will take them 8 4
minutes to type 10 pages
...
If they work
together they can mow the lawn in 9 minutes
...
9
Quadratics - Simultaneous Products
Objective: Solve simultaneous product equations using substitution to
create a rational equation
...
When we do so we may end up with a quadratic equation to solve
...
If we have two products we will choose a variable to solve for first and
divide both sides of the equations by that variable or the factor containing the
variable
...
Example 495
...
We pick a variable to solve for, divide each side by that variable, or factor containing the variable
...
Quite often these problems will have two solutions
...
xy = − 35
(x + 6)(y − 2) = 5
y=
− 35
5
and y − 2 =
x
x+6
5
− 35
−2=
x+6
x
− 35x(x + 6)
5x(x + 6)
− 2x(x + 6) =
x
x+6
− 35(x + 6) − 2x(x + 6) = 5x
− 35x − 210 − 2x2 − 12x = 5x
− 2x2 − 47x − 210 = 5x
− 5x
− 5x
− 2x2 − 52x − 210 = 0
x2 + 26x + 105 = 0
(x + 5)(x + 21) = 0
x + 5 = 0 or x + 21 = 0
−5−5
− 21 − 21
x = − 5 or x = − 21
− 5y = − 35 or − 21y = − 35
−5 −5
− 21 − 21
5
y = 7 or y =
3
5
( − 5, 7) or − 21,
3
To solve for x, divide the first
equation by x, second by x + 6
Substitute
− 35
for y in the second equation
x
Multiply each term by LCD: x(x + 6)
Reduce fractions
Distribute
Combine like terms
Make equation equal zero
Divide each term by − 2
Factor
Set each factor equal to zero
Solve each equation
Substitute each solution into xy = − 35
Solve each equation
Our solutions for y
Our Solutions as ordered pairs
The processes used here will be used as we solve applications of quadratics
including distance problems and revenue problems
...
World View Note: William Horner, a British mathematician from the late 18th
century/early 19th century is credited with a method for solving simultaneous
equations, however, Chinese mathematician Chu Shih-chieh in 1303 solved these
equations with exponents as high as 14!
371
9
...
1)
xy = 72
(x + 2)(y − 4) = 128
2)
xy = 180
1
(x − 1)(y − 2 ) = 205
3)
xy = 150
(x − 6)(y + 1) = 64
4)
xy = 120
(x + 2)(y − 3)=120
5)
xy = 45
(x + 2)(y + 1) = 70
6)
xy = 65
(x − 8)(y + 2) = 35
7)
xy = 90
(x − 5)(y + 1) = 120
8)
xy = 48
(x − 6)(y + 3) = 60
9)
xy = 12
(x + 1)(y − 4) = 16
10)
xy = 60
(x + 5)(y + 3) = 150
12)
xy = 80
(x − 5)(y + 5) = 45
11)
xy = 45
(x − 5)(y + 3) = 160
372
9
...
A common application of quadratics comes from revenue and distance problems
...
Once they are set up, we will solve them in exactly the same way we solved the
simultaneous product equations
...
If we multiply the number of items by the price per
item we will get the total paid
...
Once we have the table filled out we will have our equations which we can solve
...
Example 497
...
After three fish die, he decides to sell the rest at
a profit of S5 per fish
...
How many fish did he buy to
begin with?
Number Price Total
Buy
n
p
56
Sell
Using our table, we don ′t know the number
he bought, or at what price, so we use varibles
n and p
...
Number Price Total
Buy
n
p
56
Sell n − 3 p + 5 60
When he sold, he sold 3 less (n − 3), for S5
more (p + 5)
...
A group of students together bought a couch for their dorm that cost S96
...
How many students were in the original group?
Number Price Total
Deal
n
p
96
Paid
S96 was paid, but we don ′t know the number
or the price agreed upon by each student
...
The
total here is still S96
...
For either of these we could have used either method or even factoring
...
Use the one that seems
easiest for the problem
...
The only difference is the variables are r and t (for rate and time), instead of
n and p (for number and price)
...
So for our distance problems our table becomes the following:
rate time distance
First
Second
Using this table and the exact same patterns as the revenue problems is shown in
the following example
...
375
Greg went to a conference in a city 120 miles away
...
How fast did he drive on the way to the conference?
rate time distance
There r
t
120
Back
We do not know rate, r, or time, t he traveled
on the way to the conference
...
rate time distance
There
r
t
120
Back r − 10 t + 2
120
Coming back he drove 10 mph slower (r − 10)
and took 2 hours longer (t + 2)
...
rt = 120
(r − 10)(t + 2) = 120
t=
120
120
and t + 2 =
r
r − 10
120
120
+2=
r − 10
r
120r(r − 10)
120r(r − 10)
+ 2r(r − 10) =
r
r − 10
120(r − 10) + 2r 2 − 20r = 120r
120r − 1200 + 2r 2 − 20r = 120r
2r 2 + 100r − 1200 = 120r
− 120r
− 120r
2
2r − 20r − 1200 = 0
r 2 − 10r − 600 = 0
(r − 30)(r + 20) = 0
r − 30 = 0 and r + 20 = 0
+ 30 + 30
− 20 − 20
r = 30 and r = − 20
30 mph
Equations are product of rate and time
We have simultaneous product equations
Solving for rate, divide by r and r − 10
Substitute
120
for t in the second equation
r
Multiply each term by LCD: r(r − 10)
Reduce each fraction
Distribute
Combine like terms
Make equation equal to zero
Divide each term by 2
Factor
Set each factor equal to zero
Solve each equation
Can ′t have a negative rate
Our Solution
World View Note: The world’s fastest man (at the time of printing) is
Jamaican Usain Bolt who set the record of running 100 m in 9
...
That is a speed of over 23 miles per hour!
Another type of simultaneous product distance problem is where a boat is traveling in a river with the current or against the current (or an airplane flying with
the wind or against the wind)
...
If a boat is traveling
376
upstream, the current will pull against it or decrease the rate by the speed of the
current
...
Example 500
...
If the current flows at 2 miles per hour,
how fast would the man row in still water?
8
rate time distance
down
t
30
up
8−t
30
Write total time above time column
We know the distance up and down is 30
...
Subtracting
8 − t becomes time upstream
rate time distance
down r + 2
t
30
up r − 2 8 − t
30
Downstream the current of 2 mph pushes
the boat (r + 2) and upstream the current
pulls the boat (r − 2)
(r + 2)t = 30
(r − 2)(8 − t) = 30
t=
30
30
and 8 − t =
r+2
r−2
8−
8(r + 2)(r − 2) −
30
30
=
r+2 r −2
30(r + 2)(r − 2) 30(r + 2)(r − 2)
=
r+2
r −2
8(r + 2)(r − 2) − 30(r − 2) = 30(r + 2)
8r2 − 32 − 30r + 60 = 30r + 60
8r 2 − 30r + 28 = 30r + 60
− 30r − 60 − 30r − 60
8r2 − 60r − 32 = 0
2r 2 − 15r − 8 = 0
(2r + 1)(r − 8) = 0
2r + 1 = 0 or r − 8 = 0
−1−1
+8+8
2r = − 1 or r = 8
2
2
1
r = − or r = 8
2
8 mph
Multiply rate by time to get equations
We have a simultaneous product
Solving for rate, divide by r + 2 or r − 2
Substitute
30
for t in second equation
r+2
Multiply each term by LCD: (r + 2)(r − 2)
Reduce fractions
Multiply and distribute
Make equation equal zero
Divide each term by 4
Factor
Set each factor equal to zero
Solve each equation
Can ′t have a negative rate
Our Solution
377
9
...
Had he bought 3 pieces more
for the same money, he would have paid S15 less for each piece
...
2) A number of men subscribed a certain amount to make up a deficit of S100
but 5 men failed to pay and thus increased the share of the others by S1 each
...
3) A merchant bought a number of barrels of apples for S120
...
How many barrels did he originally buy?
4) A dealer bought a number of sheep for S440
...
How
many sheep did he originally purchase?
5) A man bought a number of articles at equal cost for S500
...
How many articles did he buy?
6) A clothier bought a lot of suits for S750
...
How many suits did he buy?
7) A group of boys bought a boat for S450
...
50 more
...
If there had been 3 fewer
persons in the party, it would have cost each person S2 more than it did
...
The speed rates
range from 50 to 70 km/hr in the lower range and from 70 to 90 km/hr in the
higher range
...
At what rates should he drive if he
1
plans to complete the test in 3 2 hours?
10) A train traveled 240 kilometers at a certain speed
...
What was the rate of each
engine?
11) The rate of the current in a stream is 3 km/hr
...
The round trip required 1 hour and 20 minutes
...
He
completed the round trip in 8 hours
...
The one driving at the slower rate drives 70 kilometers down a speedway and
returns by the same route
...
Both drivers
1
leave at the same time, and the faster car returns 2 hour earlier than the
slower car
...
If the rate of
the current is 2 km/hr, how fast should he row?
15) An automobile goes to a place 72 miles away and then returns, the round
trip occupying 9 hours
...
Find the rate of speed in both going and returning
...
Returning, the rate was increased 10 miles an hour
...
17) The rate of a stream is 3 miles an hour
...
If the speed of a train
were increased 4 miles an hour, the trip would take 40 minutes less
...
How fast did it travel?
20) Mr
...
Recently a new freeway has opend up and, although the freeway route is 120
miles, he can drive 20 mph faster on average and takes 30 minutes less time to
make the trip
...
Jones rate on both the old route and on the
freeway?
21) If a train had traveled 5 miles an hour faster, it would have needed 1
less time to travel 150 miles
...
1
2
hours
22) A traveler having 18 miles to go, calculates that his usual rate would make
him one-half hour late for an appointment; he finds that in order to arrive on
time he must travel at a rate one-half mile an hour faster
...
11
Quadratics - Graphs of Quadratics
Objective: Graph quadratic equations using the vertex, x-intercepts,
and y-intercept
...
One way we can do that is to make a
table of values
...
y = x2 − 4x + 3
x y
0
1
2
3
4
y = (0)2 + 4(0) + 3 = 0 − 0 + 3 = 3
y = (1)2 − 4(1) + 3 = 1 − 4 + 3 = 0
y = (2)2 − 4(2) + 3 = 4 − 8 + 3 = − 1
y = (3)2 − 4(3) + 3 = 9 − 12 + 3 = 0
y = (4)2 − 4(4) + 3 = 16 − 16 + 3 = 3
x y
0 3
1 0
2 −1
3 0
4 3
Make a table of values
We will test 5 values to get an idea of shape
Plug 0 in for x and evaluate
Plug 1 in for x and evaluate
Plug 2 in for x and evaluate
Plug 3 in for x and evaluate
Plug 4 in for x and evaluate
Our completed table
...
Connect the dots with a smooth
curve
...
Quadratics have a graph that looks like a U shape that is called a parabola
...
She studied conic sections
...
380
The above method to graph a parabola works for any equation, however, it can be
very tedious to find all the correct points to get the correct bend and shape
...
These key points are described below
...
B
C
A
D
Points B and C: x-intercepts: Where
the graph crosses the horizontal x-axis
Point D: Vertex: The point where the
graph curves and changes directions
...
To graph the equation y = ax2 + b x + c, find the following points
1
...
x-intercepts: Found by making y = 0, this means solving 0 = ax2 + bx + c
3
...
−b
2a
to find x
...
y = x2 + 4x + 3
y=3
Find the key points
y = c is the y − intercept
0 = x2 + 4x + 3
0 = (x + 3)(x + 1)
x + 3 = 0 and x + 1 = 0
−3−3
−1−1
x = − 3 and x = − 1
To find x − intercept we solve the equation
Factor
Set each factor equal to zero
Solve each equation
Our x − intercepts
−4 −4
=
=−2
2(1)
2
y = ( − 2)2 + 4( − 2) + 3
y=4−8+3
y=−1
( − 2, − 1)
−b
2a
Plug this answer into equation to find y − coordinate
Evaluate
The y − coordinate
Vertex as a point
x=
To find the vertex, first use x =
381
Graph the y-intercept at 3, the xintercepts at − 3 and − 1, and the
vertex at ( − 2, − 1)
...
Our Solution
If the a in y = ax2 + bx + c is a negative value, the parabola will end up being an
upside-down U
...
Remember, if a is negative, then ax2 will also be
negative because we only square the x, not the a
...
y = − 3x2 + 12x − 9
y=−9
0 = − 3x2 + 12x − 9
0 = − 3(x2 − 4x + 3)
0 = − 3(x − 3)(x − 1)
x − 3 = 0 and x − 1 = 0
+3+3
+1+1
x = 3 and x = 1
− 12
− 12
=
=2
2( − 3)
−6
y = − 3(2)2 + 12(2) − 9
y = − 3(4) + 24 − 9
y = − 12 + 24 − 9
y=3
(2, 3)
x=
Find key points
y − intercept is y = c
To find x − intercept solve this equation
Factor out GCF first, then factor rest
Set each factor with a varaible equal to zero
Solve each equation
Our x − intercepts
−b
2a
Plug this value into equation to find y − coordinate
Evaluate
To find the vertex, first use x =
y − value of vertex
Vertex as a point
Graph the y-intercept at − 9, the xintercepts at 3 and 1, and the vertex at
(2, 3)
...
Our Solution
382
It is important to remember the graph of all quadratics is a parabola with the
same U shape (they could be upside-down)
...
Go
back and check your work to be sure they are correct!
Just as all quadratics (equation with y = x2) all have the same U-shape to them
and all linear equations (equations such as y = x) have the same line shape when
graphed, different equations have different shapes to them
...
Absolute Value
Cubic
y = |x|
y = x3
Quadratic
Exponential
y = x2
y = ax
Square Root
√
y= x
Logarithmic
y = logax
383
9
...
Use this
information to graph the quadratic
...
1 Function Notation
...
2 Operations on Functions
...
3 Inverse Functions
...
4 Exponential Functions
...
5 Logarithmic Functions
...
6 Application: Compound Interest
...
7 Trigonometric Functions
...
8 Inverse Trigonometric Functions
...
1
Functions - Function Notation
Objective: Idenfity functions and use correct notation to evaluate functions at numerical and variable values
...
An
equation gives the relationship between variables and numbers
...
Functions
have at most one output for any input
...
For this reason x is considered an input
variable and y is considered an output variable
...
A great way to visualize this definition is to look at the graphs of a few relationships
...
If the vertical line crosses the graph more than once, that means we have
too many possible y values
...
386
Example 504
...
Drawing a vertical line
through this graph will
cross the graph twice,
once at top and once at
bottom
...
Drawing a vertical line
through this graph will
cross the graph only
once, it is a function
...
If we have only one solution then it is a function
...
Is 3x2 − y = 5 a function?
− 3x2 − 3x2
− y = − 3x2 + 5
−1
−1 −1
y = 3x2 − 5
Yes!
Solve the relation for y
Subtract 3x2 from both sides
Divide each term by − 1
Only one solution for y
...
Is y 2 − x = 5 a function?
+x+x
y2 = x + 5
√
y2 = ± x + 5
√
y=± x+5
No!
Solve the relation for y
Add x to both sides
Square root of both sdies
Simplify
Two solutions for y (one + , one − )
Not a function
Once we know we have a function, often we will change the notation used to
emphasis the fact that it is a function
...
We read this notation “f of x”
...
It is important to point out that f (x) does not mean
f times x, it is mearly a notation that names the function with the first letter
(function f ) and then in parenthesis we are given information about what variables are in the function (variable x)
...
World View Note: The concept of a function was first introduced by Arab
mathematician Sharaf al-Din al-Tusi in the late 12th century
Once we know a relationship is a function, we may be interested in what values
can be put into the equations
...
When finding the domain, often it is easier
to consider what cannot happen in a given function, then exclude those values
...
Find the domain: f (x) =
3x − 1
x2 + x − 6
x2 + x − 6 0
(x + 3)(x − 2) 0
x + 3 0 and x − 2 0
+2+2
−3−3
x − 3, 2
With fractions, zero can ′t be in denominator
Solve by factoring
Set each factor not equal to zero
Solve each equation
Our Solution
The notation in the previous example tells us that x can be any value except for
− 3 and 2
...
Example 508
...
This means any number can be used for x
...
Find the domain: f (x) =
√
2x − 3
388
Square roots can ′t be negative
2x − 3 0
+3+3
2x 3
2 2
3
x
2
Set up an inequality
Solve
Our Solution
3
The notation in the above example states that our variable can be 2 or any
3
number larger than 2
...
Another use of function notation is to easily plug values into functions
...
This is shown in the following examples
...
f (x) = 3x2 − 4x; find f ( − 2)
f ( − 2) = 3( − 2)2 − 4( − 2)
f ( − 2) = 3(4) − 4( − 2)
f ( − 2) = 12 + 8
f ( − 2) = 20
Substitute − 2 in for x in the function
Evaluate, exponents first
Multiply
Add
Our Solution
Example 511
...
k(a) = 2|a + 4|; find k( − 7)
k( − 7) = 2| − 7 + 4|
k( − 7) = 2| − 3|
k( − 7) = 2(3)
Substitute − 7 in for a in the function
Add inside absolute values
Evaluate absolute value
Multiply
389
k( − 7) = 6
Our Solution
As the above examples show, the function can take many different forms, but the
pattern to evaluate the function is always the same, replace the variable with
what is in parenthesis and simplify
...
Often the expressions use the same variable, it is
important to remember each variable is replaced by whatever is in parenthesis
...
g (x) = x4 + 1; find g(3x)
g(3x) = (3x)4 + 1
g(3x) = 81x4 + 1
Replace x in the function with (3x)
Simplify exponet
Our Solution
Example 514
...
390
10
...
1) Which of the following is a function?
a)
b)
c)
d)
e) y = 3x − 7
f) y 2 − x2 = 1
g)
h) x2 + y 2 = 1
√
y +x=2
Specify the domain of each of the following funcitons
...
11) g(x) = 4x − 4; Find g(0)
12) g(n) = − 3 · 5−n; Find g(2)
13) f (x) = |3x + 1| + 1; Find f (0)
14) f (x) = x2 + 4; Find f ( − 9)
15) f (n) = − 2 − n − 2 + 1; Find f ( − 6)
16) f (n) = n − 3; Find f (10)
17) f (t) = 3t − 2; Find f ( − 2)
18) f (a) − 3a−1 − 3; Find f (2)
19) f (t) = |t + 3|; Find f (10)
20) w(x) = x2 + 4x; Find w( − 5)
21) w(n) = 4n + 3; Find w(2)
22) w(x) = − 4x + 3; Find w(6)
23) w(n) = 2n+2; Find w( − 2)
24) p(x) = − |x| + 1; Find p(5)
25) p(n) = − 3|n|; Find p(7)
26) k(a) = a + 3; Find k( − 1)
27) p(t) = − t3 + t; Find p(4)
28) k(x) = − 2 · 42x−2; Find k(2)
29) k(n) = |n − 1|; Find k(3)
30) p(t) = − 2 · 42t+1 + 1; Find p( − 2)
31) h(x) = x3 + 2; Find h( − 4x)
32) h(n) = 4n + 2; Find h(n + 2)
33) h(x) = 3x + 2; Find h( − 1 + x)
34) h(a) = − 3 · 2a+3; Find h( 4 )
35) h(t) = 2 − 3t − 1 + 2; Find h(n2)
36) h(x) = x2 + 1; Find h( 4 )
37) g(x) = x + 1; Find g(3x)
38) h(t) = t2 + t; Find h(t2)
39) g(x) = 5x; Find g( − 3 − x)
40) h(n) = 5n−1 + 1; Find h( 2 )
a
x
n
392
10
...
Several functions can work together in one larger function
...
The four basic operations on functions are adding, subtracting, multiplying, and dividing
...
Addition
Subtraction
Multiplication
Division
(f + g)(x) = f (x) + g(x)
(f − g)(x) = f (x) − g(x)
(f · g)(x) = f (x)g(x)
f
f (x)
(x) =
g
g(x)
When we do one of these four basic operations we can simply evaluate the two
functions at the value and then do the operation with both solutions
Example 515
...
Example 516
...
If there is no number to plug into the equations we will simply
use each equation, in parenthesis, and simplify the expression
...
f (x) = 2x − 4
g(x) = x2 − x + 5
Find (f − g)(x)
f (x) − g(x)
(2x − 4) − (x2 − x + 5)
2x − 4 − x2 + x − 5
− x2 + 3x − 9
Write subtraction problem of functions
Replace f (x) with (2x − 3) and g(x) with (x2 − x + 5)
Distribute the negative
Combine like terms
Our Solution
The parenthesis are very important when we are replacing f (x) and g(x) with a
variable
...
Example 518
...
Example 519
...
f (x) = 2x − 1
g(x) = x + 4
Find (f · g)(3x)
f (3x)g(3x)
[2(3x) − 1][(3x) + 4]
(6x − 1)(3x + 4)
18x2 + 24x − 3x − 4
18x2 + 21x − 4
Write as a product of functions
Replace x in f (x) and g(x) with 3x
Multiply our 2(3x)
FOIL
Combine like terms
Our Solution
The fifth operation of functions is called composition of functions
...
The notation used for composition
of functions is:
(f ◦ g)(x) = f (g(x))
To calculate a composition of function we will evaluate the inner function and
substitute the answer into the outer function
...
395
Example 521
...
In these problems
we will take the inside function and substitute into the outside function
...
f (x) = x2 − x
g(x) = x + 3
Find (f ◦ g)(x)
f (g(x))
f (x + 3)
2
(x + 3) − (x + 3)
(x2 + 6x + 9) − (x + 3)
x2 + 6x + 9 − x − 3
x2 + 5x + 6
Rewrite as a function in function
Replace g(x) with x + 3
Replace the variables in f with (x + 3)
Evaluate exponent
Distribute negative
Combine like terms
Our Solution
It is important to note that very rarely is (f ◦ g)(x) the same as (g ◦ f )(x) as the
following example will show, using the same equations, but compositing them in
the opposite direction
...
f (x) = x2 − x
g(x) = x + 3
Find (g ◦ f )(x)
g(f (x))
g(x2 − x)
(x2 − x) + 3
x2 − x + 3
Rewrite as a function in function
Replace f (x) with x2 − x
Replace the variable in g with (x2 − x)
Here the parenthesis don ′t change the expression
Our Solution
World View Note: The term “function” came from Gottfried Wihelm Leibniz, a
German mathematician from the late 17th century
...
2 Practice - Operations on Functions
Perform the indicated operations
...
3
Functions - Inverse Functions
Objective: Identify and find inverse functions
...
The result that we get
when we evaluate the function is called the output
...
To find this we use an inverse function
...
If a function is named f (x), the
inverse function will be named f −1(x) (read “f inverse of x”)
...
World View Note: The notation used for functions was first introduced by the
great Swiss mathematician, Leonhard Euler in the 18th century
...
If we had an input of 3, we could calculate f (3) = (3) + 5 = 8
...
If we plug this output into the inverse function we get f −1(8) = (8) −
5 = 3, which is the original input
...
It may
be difficult to determine just by looking at the functions if they are inverses
...
If f changes the variable x in some way,
then g undoes whatever f did, then we will be back at x again for our final solution
...
If it is anything but x the functions are not inverses
...
√
x3 − 4
Are f (x) = 3 3x + 4 and g(x) =
inverses?
3
Caculate composition
x3 − 4
3
x3 − 4
for variable in f
3
f (g(x))
f
Replace g(x) with
x3 − 4
3
Substitute
x3 − 4
+4
3
√
3
x3 − 4 + 4
√
3
x3
x
Yes, they are inverses!
3
3
Divide out the 3 ′s
Combine like terms
Take cubed root
Simplified to x!
Our Solution
Example 525
...
Are f (x) =
x+2
3x − 2
and g(x) =
inverses?
3 − 4x
4x + 1
Calculate composition
f (g(x))
f
3
4
Replace g(x) with
x+2
3 − 4x
Substitute
x+2
3 − 4x
x+2
3 − 4x
−2
+1
402
x+2
3 − 4x
x+2
for variable in f
3 − 4x
Distribute 3 and 4 into numerators
3x + 6
3 − 4x
4x + 8
3 − 4x
(3x + 6)(3 − 4x)
3 − 4x
(4x + 8)(3 − 4x)
3 − 4x
−2
+1
− 2(3 − 4x)
+ 1(3 − 4x)
3x + 6 − 2(3 − 4x)
4x + 8 + 1(3 − 4x)
3x + 6 − 6 + 8x
4x + 8 + 3 − 4x
11x
11
x
Yes, they are inverses
Multiply each term by LCD: 3 − 4x
Reduce fractions
Distribute
Combine like terms
Divide out 11
Simplified to x!
Our Solution
While the composition is useful to show two functions are inverses, a more
common problem is to find the inverse of a function
...
This means if we switch x and y in our function we will find
the inverse! This process is called the switch and solve strategy
...
Replace f (x) with y
2
...
Solve for y
4
...
Find the inverse of f (x) = (x + 4)3 − 2
y = (x + 4)3 − 2
x = (y + 4)3 − 2
+2
+2
403
Replace f (x) with y
Switch x and y
Solve for y
Add 2 to both sides
x + 2 = (y + 4)3
√
3
x+2 = y+4
−4
−4
√
3
x+2 −4= y
√
f −1(x) = 3 x + 2 − 4
Cube root both sides
Subtract 4 from both sides
Replace y with f −1(x)
Our Solution
Example 528
...
Be careful not to get them backwards
...
There
may be several ways to represent the same function so the switch and solve
strategy may not look the way we expect and can lead us to conclude two functions are not inverses when they are in fact inverses
...
3 Practice - Inverse Functions
State if the given functions are inverses
...
11) f (x) = (x − 2)5 + 3
4
√
12) g(x) = 3 x + 1 + 2
−3
14) f (x) = x − 3
13) g(x) = x + 2
15) f (x) =
− 2x − 2
x+2
16) g(x) =
9+x
3
17) f (x) =
10 − x
5
18) f (x) =
5x − 15
2
20) f (x) =
12 − 3x
4
19) g(x) = − (x − 1)3
21) f (x) = (x − 3)3
22) g(x) = 5
23) g(x) = x − 1
24) f (x) =
x
26) h(x) =
x−1
25) f (x) = x + 1
27) g(x) =
8 − 5x
4
29) g(x) = − 5x + 1
31) g(x) = − 1 + x3
33) h(x) =
4−
√
3
−x+2
2
4x
2
− 3 − 2x
x+3
x
x+2
28) g(x) =
−x+2
3
30) f (x) =
5x − 5
4
32) f (x) = 3 − 2x5
34) g(x) = (x − 1)3 + 2
−1
x+1
36) f (x) = x + 1
7 − 3x
x−2
38) f (x) = −
35) f (x) = x + 2
37) f (x) =
39) g(x) = − x
40) g(x) =
405
3x
4
− 2x + 1
3
10
...
As our study of algebra gets more advanced we begin to study more involved
functions
...
Here we will look at exponential functions and then we
will consider logarithmic functions in another lesson
...
(It is important
not to confuse exponential functions with polynomial functions where the variable
is in the base such as f (x) = x2)
...
According to the 2009 CIA World Factbook, the country with the
highest population growth rate is a tie between the United Arab Emirates (north
of Saudi Arabia) and Burundi (central Africa) at 3
...
There are 32 countries
with negative growth rates, the lowest being the Northern Mariana Islands (north
of Australia) at − 7
...
Solving exponetial equations cannot be done using the skill set we have seen in
the past
...
However, we may notice that 9 is 32
...
This is the process we will use to solve exponential functions
...
Example 529
...
As we do so, we will use various exponent properties to help
...
406
Example 530
...
Example 531
...
1
9
2x
= 37x−1
(3−2)2x = 37x−1
3−4x = 37x−1
− 4x = 7x − 1
− 7x − 7x
− 11x = − 1
− 11 − 11
1
x=
11
1
as 3−2 (negative exponet to flip)
9
Multiply exponents − 2 and 2x
Same base, set exponets equal
Subtract 7x from both sides
Rewrite
Divide by − 11
Our Solution
If we have several factors with the same base on one side of the equation we can
add the exponents using the property that states axa y = ax+ y
...
54x · 52x−1 = 53x+11
56x−1 = 53x+11
6x − 1 = 3x + 11
− 3x
− 3x
3x − 1 = 11
+1 +1
3x = 12
3 3
x=4
Add exponents on left, combing like terms
Same base, set exponents equal
Move variables to one sides
Subtract 3x from both sides
Add 1 to both sides
Divide both sides by 3
Our Solution
It may take a bit of practice to get use to knowing which base to use, but as we
practice we will get much quicker at knowing which base to use
...
Again, below are the properties we used to simplify
...
This is shown in the next example
...
16
2x−5
1
·
4
3x+1
1
= 32 ·
2
x+3
Write with a common base of 2
x+3
(24)2x−5 · (2−2)3x+1 = 25 · (2−1
28x−20 · 2−6x−2 = 25 · 2−x−3
22x−22 = 2−x+2
2x − 22 = − x + 2
+x
+x
3x − 22 = 2
+ 22 + 22
3x = 24
3 3
x=8
Multiply exponents, distributing as needed
Add exponents, combining like terms
Same base, set exponents equal
Move variables to one side
Add x to both sides
Add 22 to both sides
Divide both sides by 3
Our Solution
All the problems we have solved here we were able to write with a common base
...
To solve problems like
this we will need to use the inverse of an exponential function
...
408
10
...
1) 31−2n = 31−3n
2) 42x = 16
3) 42a = 1
4) 16−3p = 64−3p
1
6) 625−n−2 = 125
1
5) ( 25 )−k = 125−2k−2
1
1
7) 62m+1 = 36
8) 62r −3 = 6r −3
9) 6−3x = 36
10) 52n = 5−n
11) 64b = 25
12) 216−3v = 363v
1
13) ( 4 )x = 16
14) 27−2n−1 = 9
15) 43a = 43
16) 4−3v = 64
17) 363x = 2162x+1
18) 64x+2 = 16
19) 92n+3 = 243
20) 162k = 64
21) 33x−2 = 33x+1
22) 243 p = 27−3p
23) 3−2x = 33
24) 42n = 42−3n
25) 5m+2 = 5−m
26) 6252x = 25
1
27) ( 36 )b−1 = 216
28) 2162n = 36
29) 62−2x = 62
30) ( 4 )3v −2 = 641−v
1
1
1
216
6−2a
31) 4 · 2−3n−1 = 4
32)
33) 43k−3 · 42−2k = 16−k
34) 322p−2 · 8 p = ( 2 )2p
1
35) 9−2x · ( 243 )3x = 243−x
1
= 63a
1
36) 32m · 33m = 1
37) 64n−2 · 16n+2 = ( 4 )3n−1
38) 32−x · 33m = 1
39) 5−3n−3 · 52n = 1
40) 43r · 4−3r = 64
1
409
10
...
The inverse of an exponential function is a new function known as a logarithm
...
When working with radicals we found that
√
there were two ways to write radicals
...
Each form has its advantages, thus we need to be comfortable using both the
radical form and the rational exponent form
...
The first form we are very
familiar with, bx = a, where b is the base, a can be thought of as our answer, and
x is the exponent
...
The word “log” tells us that we are in this new form
...
b is still the base, a can still be thought of as our answer
...
Both
mean the same thing, both are still the same exponent problem, but just as roots
can be written in radical form or rational exponent form, both our forms have
their own advantages
...
This is what is shown in the next few examples
...
Write each exponential equation in logarithmic form
m3 = 5
logm5 = 3
Identify base, m, answer, 5, and exponent 3
Our Solution
72 = b
log7b = 2
Identify base, 7, answer, b, and exponent, 2
Our Solution
4
2
3
=
16
81
16
=4
3 81
log 2
2
16
Identify base, , answer, , and exponent 4
3
81
Our Solution
Example 536
...
We will first
evaluate logarithmic expressions
...
Example 537
...
Evaluate log1255
log1255 = x
125x = 5
(53)x = 5
53x = 5
3x = 1
3 3
1
x=
3
Set logarithm equal to x
Change to exponent form
Write as common base, 125 = 53
Multiply exponents
Same base, set exponents equal (5 = 51)
Solve
Divide both sides by 3
Our Solution
Example 539
...
Example 540
...
log2(3x + 5) = 4
24 = 3x + 5
16 = 3x + 5
−5
−5
11 = 3x
3 3
11
=x
3
Change to exponential form
Evaluate exponent
Solve
Subtract 5 from both sides
Divide both sides by 3
Our Solution
Example 542
...
Similar to square roots not writting the common index of 2 in the radical, we don’t write the common base of 10 in the logarithm
...
Example 543
...
In an
advanced algebra course logarithms will be studied in much greater detail
...
5 Practice - Logarithmic Functions
Rewrite each equation in exponential form
...
1
7) 80 = 1
8) 17−2 = 289
9) 152 = 225
1
11) 64 6 = 2
1
10) 144 2 = 12
12) 192 = 361
Evaluate each expression
...
23) log5 x = 1
24) log8 k = 3
25) log2 x = − 2
26) log n = 3
27) log11 k = 2
28) log4 p = 4
29) log9 (n + 9) = 4
30) log11 (x − 4) = − 1
31) log5 ( − 3m) = 3
32) log2 − 8r = 1
33) log11 (x + 5) = − 1
34) log7 − 3n = 4
35) log4 (6b + 4) = 0
36) log11 (10v + 1) = − 1
37) log5 ( − 10x + 4) = 4
38) log9 (7 − 6x) = − 2
39) log2 (10 − 5a) = 3
40) log8 (3k − 1) = 1
413
10
...
An application of exponential functions is compound interest
...
This money added to the balance is called interest
...
This idea of earning interest on interest is called compound interest
...
The next year
you will earn another 10% or S11, giving you a new balance of S121
...
10, giving you a new balance of S133
...
This pattern will continue each year until you close the account
...
The first way, as described above, is
compounded annually
...
But
interest can be compounded more often
...
When interest is compounded in any of these ways we can calculate the
balance after any amount of time using the following formula:
r nt
n
A = Final Amount
P = Principle (starting balance)
r = Interest rate (as a decimal)
n = number of compounds per year
t = time (in years)
Compound Interest Formula: A = P 1 +
Example 544
...
5% compounded quarterly, no payments required for the first five years, what will your balance be at
the end of those five years?
P = 25000, r = 0
...
065
A = 25000 1 +
4
A = 25000(1
...
01625)20
A = 25000(1
...
49
S34, 510
...
Example 545
...
5% compounded weekly for
4 years?
A = 3000, r = 0
...
065
3000 = P 1 +
52
3000 = P (1
...
00125)208
3000 = P (1
...
296719528
1
...
53 = P
S2313
...
296719528
Solution for P
Our Solution
It is interesting to compare equal investments that are made at several different
types of compounds
...
Example 546
...
03, n = 12, t = 7
12·7
0
...
0025)12·7
A = 4000(1
...
2333548)
A = 4933
...
42
Identify each variable
Plug each value into formula, evaluate parentheses
Multiply exponents
Evaluate exponent
Multiply
Our Solution
415
To investigate what happens to the balance if the compounds happen more often,
we will consider the same problem, this time with interest compounded daily
...
If S4000 is invested in an account paying 3% interest compounded daily, what is
the balance after 7 years?
P = 4000, r = 0
...
03
A = 4000 1 +
365
A = 4000(1
...
00008219 )2555
A = 4000(1
...
)
A = 4934
...
67
Identify each variable
Plug each value into formula, evaluate parenthesis
Multiply exponent
Evaluate exponent
Multiply
Our Solution
While this difference is not very large, it is a bit higher
...
Compound
Annually
Semi-Annually
Quarterly
Monthly
Weekly
Daily
Balance
S4919
...
02
S4930
...
42
S4934
...
67
As the table illustrates, the more often interest is compounded, the higher the
final balance will be
...
So once interest is paid into the account it will start
earning interest for the next compound and thus giving a higher final balance
...
This is when the interest is compounded continuously
...
Instead we will use the following formula:
Interest Compounded Continuously: A = Pert
A = Final Amount
P = Principle (starting balance)
e = a constant approximately 2
...
r = Interest rate (written as a decimal)
t = time (years)
416
The variable e is a constant similar in idea to pi (π) in that it goes on forever
without repeat or pattern, but just as pi (π) naturally occurs in several geometry
applications, so does e appear in many exponential applications, continuous
interest being one of them
...
World View Note: e first appeared in 1618 in Scottish mathematician’s
Napier’s work on logarithms
...
Some say he used e because his name begins
with E
...
Others say it is because
Euler’s work already had the letter a in use, so e would be the next value
...
Example 548
...
03, t = 7
A = 4000e0
...
21
A = 4000(1
...
71
S4934
...
21
Multiply
Our Solution
Albert Einstein once said that the most powerful force in the universe is compound interest
...
Example 549
...
16 in an account paying 12% interest compounded continuously
for 100 years, and that is all you have to leave your children as an inheritance,
what will the final balance be that they will receive?
P = 6
...
12, t = 100
A = 6
...
12·100
A = 6
...
16(162, 544
...
52
S1, 002, 569
...
16 investment grew to over one million dollars! That’s the power of compound interest!
417
10
...
S500 invested at 4% compounded annually for 10 years
...
S600 invested at 6% compounded annually for 6 years
...
S750 invested at 3% compounded annually for 8 years
...
S1500 invested at 4% compounded semiannually for 7 years
...
S900 invested at 6% compounded semiannually for 5 years
...
S950 invested at 4% compounded semiannually for 12 years
...
S2000 invested at 5% compounded quarterly for 6 years
...
S2250 invested at 4% compounded quarterly for 9 years
...
S3500 invested at 6% compounded quarterly for 12 years
...
All of the above compounded continuously
...
5 years?
6) What principal will amount to S1750 if invested at 3% interest compounded
quarterly for 5 years?
7) A thousand dollars is left in a bank savings account drawing 7% interest,
compounded quarterly for 10 years
...
What is the balance at the end of 10 years?
9) S1750 is invested in an account earning 13
...
What is the balance at the end of 9 years?
10) You lend out S5500 at 10% compounded monthly
...
If the bill
matured in 2 years, what was it worth at maturity?
12) You borrow S25000 at 12
...
If you are
unable to make any payments the first year, how much do you owe, excluding
penalties?
13) A savings institution advertises 7% annual interest, compounded daily, How
much more interest would you earn over the bank savings account or credit
union in problems 7 and 8?
14) An 8
...
If S2500 is deposited for 5 years,
what is the total accumulated?
15) You lend S100 at 10% continuous interest
...
7
Functions - Trigonometric Functions
Objective: Solve for a missing side of a right triangle using trigonometric ratios
...
We will discuss three of the functions here
...
To the right is a picture of a right triangle
...
In the picture,
angle A is the angle we will use to
name the other sides
...
The
side across from the angle A is called
the opposite side
...
It is important to remember that the opposite and adjacent sides are
named in relationship to the angle A or the angle we are using in a problem
...
The three trigonometric funtions are functions taken of angles
...
The
ratios are as describe below:
sinθ =
opposite
hypotenuse
cosθ =
adjacent
hypotenuse
tanθ =
opposite
adjacent
The “weird” variable θ is a greek letter, pronounced “theta” and is close in idea to
our letter “t”
...
Some
students remember the three ratios by remembering the word “SOH CAH TOA”
where each letter is the first word of: “Sine: Opposite over Hypotenuse; Cosine:
Adjacent over Hypotenuse; and Tangent: Opposite over Adjacent
...
World View Note: The word “sine” comes from a mistranslation of the Arab
word jayb
Example 550
...
6
8
10
3
adjacent
θ
α
First we will find the three ratios of θ
...
So we fill in the following:
opposite
8
6
4
sinθ = hypotenuse = 10 = 5
cosθ = hypotenuse = 10 = 5
opposite
6
3
tanθ = adjacent = 8 = 4
Now we will find the three ratios of α
...
So we fill in the following:
opposite
8
6
θ
10
Hypotenuse
4
adjacent
6
Opposite of θ
α
Adjacent of α
Adjacent of θ
Opposite of α
8
3
sinα = hypotenuse = 10 = 5
cosα = hypotenuse = 10 = 5
opposite
8
4
tanα = adjacent = 6 = 3
We can either use a trigonometry table or a calculator to find decimal values for
sine, cosine, or tangent of any angle
...
Using either a table or a calculator, we
can solve the next example
...
sin 42◦
0
...
213
Use calculator or table
Our Solution
cos 18◦
0
...
The trick will be to
determine which angle we are working with, naming the sides we are working
with, and deciding which trig function can be used with the sides we have
...
Find the measure of the missing side
...
The trig ratio that uses the opposite
and adjacent sides is tangent
...
tan25◦ =
4
x
0
...
466x = 4
0
...
466
x = 8
...
466
Our Solution
Example 553
...
We will be using the angle marked
70◦
...
x
9
70◦
cos70◦ =
x
9
0
...
08 = 1x
3
...
So we will
take the cosine of our angle
...
Our Solution
...
7 Practice - Trigonometric Functions
Find the value of each
...
1) cos 71◦
2) cos 23◦
3) sin 75◦
4) sin 50◦
Find the value of the trig function indicated
...
Round to the nearest tenth
...
3
◦
A
C
16)
A
x
A
63◦
x
7
...
2◦
20)
B
6
B
C
71
...
2◦
A
4
B
C
27)
28)
B
A
x
A
x
32◦
4
C
B
425
13
...
4
A
C
x
A
22◦
29◦
3
...
4
11
A
37
...
1
13
...
5◦
B
10
...
We used a special function, one of the trig functions, to take an angle of a triangle
and find the side length
...
Because this is the opposite operation, we will use the inverse
function of each of the trig ratios we saw before
...
sin−1
opposite
hypotenuse
= θ cos−1
adjacent
hypotenuse
= θ tan−1
opposite
adjacent
=θ
Just as with inverse functions, the − 1 is not an exponent, it is a notation to tell
us that these are inverse functions
...
We can
calculate inverse trig values using a table or a calculator (usually pressing shift or
2nd first)
...
sinA = 0
...
5) = A
30◦ = A
We don ′t know the angle so we use an inverse trig function
Evaluate using table or calculator
Our Solution
cosB = 0
...
667) = B
48◦ = B
We don ′t know the angle so we use an inverse trig function
Evaluate using table or calculator
Our Solution
tanC = 1
...
54) = C
57◦ = C
We don ′t know the angle so we use an inverse trig function
Evaluate using table or calculator
Our Solution
If we have two sides of a triangle, we can easily calculate their ratio as a decimal
and then use one of the inverse trig functions to find a missing angle
...
Find the indicated angle
...
θ
17
The trig function that uses opposite
and hypotenuse is the sine
12
12
17
−1
sin (0
...
Find the indicated angle
From the angle α, the given sides are
the opposite (5) and the adjacent (3)
3
5
α
The trig function that uses opposite
and adjacent is the tangent
As we are looking for an angle we will
use the inverse tangent
...
667)
59◦
tan−1
Tangent is opposite over adjacent
...
This is called solving a triangle
...
The angles of a triangle always add up to 180◦, because we have a right triangle,
90◦ are used up in the right angle, that means there are another 90◦ left in the
two acute angles
...
Another trick is on the sides of the angles
...
The
Pythagorean Theorem states that if c is the hypotenuse of the triangle, and a and
429
b are the other two sides (legs), then we can use the following formula, a2 + b2 = c2
to find a missing side
...
Example 557
...
We
can use these to find either other side
...
5
35◦
5
tan35◦ =
x
0
...
700x = 5
0
...
700
x = 7
...
12 = c2
25 + 50
...
41 = c2
8
...
700
The missing leg
...
7
5
The 5 is the opposite side, so we will
use the tangent to find the leg
...
1
In the previous example, once we found the leg to be 7
...
The angle sum and pythagorean theorem are just nice shortcuts to solve the
problem quicker
...
Solve the triangle
In this triangle we have two sides
...
3
9
Tangent uses opposite and adjacent
To find an angle we use the inverse
tangent
...
6◦
tan−1
Evaluate fraction
Evaluate tangent
The angle on the right side
90◦ − 71
...
4◦
Subtract angle from 90◦ to get other angle
The angle on the left side
a 2 + b 2 = c2
92 + 32 = c2
81 + 9 = c2
90 = c2
√
3 10 or 9
...
5
18
...
6◦
3
World View Note: Ancient Babylonian astronomers kept detailed records of the
starts, planets, and eclipses using trigonometric ratios as early as 1900 BC!
431
10
...
1) sin Z = 0
...
6293
3) sin Y = 0
...
6157
Find the measure of the indicated angle to the nearest degree
...
Round to the nearest tenth
...
7
16)
B
B
7
12
θ
A
4
C
A
17)
θ
C
13
18)
A
10
B
θ
5
...
3
A
19)
20)
C
A
9
...
2
22)
B
B
A
θ
5
C
15
4
θ
A
15
C
23)
24)
A
C
10
θ
9
6
θ
B
A
12
B
C
25)
26)
C
B
θ
9
B
4
θ
15
...
8
C
A
434
A
27)
28)
A
C
θ
4
14
θ
B
C
15
A
B
15
30)
29)
B
C
θ
7
7
θ
A
B
A
C
14
6
Solve each triangle
...
32)
31)
C
28
...
3
62◦
B
A
15
33)
C
34)
C
C
2
...
3
B
35)
36)
C
A
B
7
7
3
21◦
A
C
B
37)
38)
B
C
B
A
48
16
◦
10
...
8
A
437
Answers - Chapter 0
0
...
2
Answers - Fractions
1)
7
2
4)
8
3
2)
5
4
5)
3
2
3)
7
5
6)
5
4
438
7)
5
4
33) 3
8)
4
3
34) − 15
9)
3
2
35) − 10
59)
17
5
60) − 3
7
61)
8
3
36)
11)
5
2
38)
20
21
39)
2
9
40)
8
7
4
3
7
2
14)
4
3
15)
4
3
41) −
16)
3
2
42)
17)
6
5
18)
7
6
19)
3
2
20)
8
7
4
67) 1
25
21
68)
7
8
69)
5
27
19
20
2
70) − 5
40
9
145
56
46) −
1
10
71) −
47) −
45
7
72) − 15
29
34
7
48)
13
15
73)
49)
4
27
74) −
50)
32
65
75) − 8
51)
1
15
76) − 3
4
2
24) − 3
13
4
3
4
33
20
28)
2
3
66) − 3
21
26
3
2
45)
23) − 9
27)
65)
44) −
5
3
26)
7
64) − 6
43) −
21) 8
25) −
47
56
8
13)
22)
3
7
63)
37) − 7
33
20
62)
5
14
10)
12)
37
20
33
56
3
2
52) 1
5
77) − 24
53) − 1
54) −
29) 4
55)
78)
10
7
30)
5
56) 2
31)
1
2
57) 3
19
32) − 20
58) −
39
14
79) − 6
2
7
18
7
23
3
80)
1
10
81) 2
31
8
82)
439
62
21
0
...
4
Answers - Properties of Algebra
1) 7
19) 7
37) − 8x + 32
2) 29
20) 38
38) 24v + 27
3) 1
21) r + 1
39) 8n2 + 72n
4) 3
22) − 4x − 2
5) 23
23) 2n
40) 5 − 9a
6) 14
24) 11b + 7
7) 25
25) 15v
8) 46
26) 7x
9) 7
27) − 9x
10) 8
28) − 7a − 1
11) 5
29) k + 5
12) 10
30) − 3p
13) 1
31) − 5x − 9
14) 6
32) − 9 − 10n
15) 1
33) − m
16) 2
34) − 5 − r
17) 36
35) 10n + 3
52) 16x2 − 20x
18) 54
36) 5b
53) 14b + 90
41) − 7k 2 + 42k
42) 10x + 20x2
43) − 6 − 36x
44) − 2n − 2
45) 40m − 8m2
46) − 18p2 + 2p
47) − 36x + 9x2
48) 32n − 8
49) − 9b2 + 90b
50) − 4 − 28r
51) − 40n − 80n2
440
54) 60v − 7
55) − 3x + 8x2
56) − 89x + 81
57) − 68k 2 − 8k
64) 30r − 16r2
65) − 72n − 48 − 8n2
66) − 42b − 45 − 4b2
67) 79 − 79v
58) − 19 − 90a
68) − 8x + 22
59) − 34 − 49p
69) − 20n + 80n − 42
2
60) − 10x + 17
70) − 12 + 57a + 54a2
61) 10 − 4n
71) − 75 − 20k
62) − 30 + 9m
72) − 128x − 121
63) 12x + 60
73) 4n2 − 3n − 5
74) 2x2 − 6x − 3
75) 4p − 5
76) 3x2 + 7x − 7
77) − v 2 + 2v + 2
78) − 7b2 + 3b − 8
79) − 4k 2 + 12
80) a2 + 3a
81) 3x2 − 15
82) − n2 + 6
Answers - Chapter 1
1
...
2
Answers to Two-Step Equations
1) − 4
2) 7
441
3) − 14
16) − 15
29) − 6
4) − 2
17) 7
30) 6
5) 10
18) 12
6) − 12
19) 9
31) − 16
7) 0
20) 0
8) 12
21) 11
9) − 10
22) − 6
10) − 16
23) − 10
11) 14
24) 13
12) − 7
25) 1
37) − 12
13) 4
26) 4
38) 0
14) − 5
27) − 9
39) 12
15) 16
28) 15
40) − 9
32) − 4
33) 8
34) − 13
35) − 2
36) 10
1
...
4
Answers to Solving with Fractions
1)
3
4
11) 0
12)
4
2) − 3
22) − 1
4
3
23) − 2
3
3)
6
5
13) − 2
4)
1
6
14)
5) −
6)
1
7
26) − 2
17) 0
7) − 9
27) − 3
5
5
18) − 3
1
3
3
28) − 2
19) 1
9) − 2
10)
25) 16
4
3
16) 1
25
8
8) −
1
2
15) −
19
6
9
24) − 4
29)
30)
20) 1
3
2
21)
1
2
4
3
3
2
1
...
c = b − a
1
...
h = gi
22
...
x = g + f
3
...
y =
cm + cn
4
4
...
r =
k(a − 3)
5
cb
16
...
n = p − c
5
...
y = dm
7
...
D =
ds
S
9
...
m =
2E
v2
c−b
a
15
...
m =
p−q
2
24
...
k = qr + m
F
26
...
L = S − 2B
19
...
Ea = IR + Eg
L
21
...
v =
16t2 + h
t
28
...
Q2 =
Q1 + PQ1
P
30
...
a =
31
...
b =
32
...
y =
7 − 3x
2
46
...
b =
5a − 4
7
48
...
y =
4x − 8
5
50
...
t =
33
...
r =
d
t
V
35
...
h = πr2
5 + bw
a
at − s
w= b
40
...
x =
c − bx
x
42
...
y =
44
...
6
Answers to Absolute Value Equations
5
1) 8, − 8
14) 3, − 3
15) − 2, 0
2) 7, − 7
3) 1, − 1
6)
7) − 2, −
4
10
3
39
7
16
,−6
5
11) 7, −
12) −
29
3
1
,−1
3
13) − 9, 15
19) −
17 7
,
2 2
4
6
2
32) − 6, 5
21) − 6, − 8
1
33) 7, 5
25
3
23) 1, −
2
31) − 3 , − 7
20) − 5 , − 2
22) 6, −
13
,1
7
30) − 3, 5
18) − 4, 3
8) − 3, 9
10)
29) −
6
38
,−6
9
9) 3, −
2
17) − 7 , 0
29
4
16
3
28) 5 , 0
16) 0, − 2
4) 2, − 2
5) 6, −
27) 6, −
13
7
34) −
24) 7, − 21
22
2
, − 13
5
19
11
35) − 22 , − 38
25) − 2, 10
7
36) 0, −
26) − 5 , 1
12
5
1
...
5
12) cd = 28
=k
4)
r
s2
=k
5)
f
xy
=k
10)
13)
=3
444
t
u2
= 0
...
5
a
√
x2 y
= 0
...
5 hours
32) 1600 km
33) r = 36
25) 4
...
2 mph
26) 450 m
18) mn = 3
...
5 m
19) 6 k
28) 5
...
3 k
36) V = 100
...
3 cm
30) 100 N
22) 160 kg/cm3
31) 27 min
37) 6
...
25
1
...
9
Answers - Age Problems
1) 6, 16
4) 17, 40
2) 10, 40
5) 27, 31
3) 18, 38
6) 12, 48
445
7) 31, 36
19) 9, 18
31) 10, 28
8) 16, 32
20) 15, 20
32) 12,20
9) 12, 20
21) 50, 22
10) 40, 16
22) 12
11) 10, 6
23) 72, 16
12) 12, 8
24) 6
35) 84, 52
13) 26
25) 37, 46
36) 14, 42
14) 8
26) 15
15) 4
27) 45
16) 3
28) 14, 54
17) 10, 20
29) 8, 4
39) 38, 42
18) 14
30) 16, 32
40) 5
33) 141, 67
34) 16, 40
37) 10
38) 10, 6
1
...
1
446
Answers - Points and Lines
1) B(4, − 3) C(1, 2)
D( − 1, 4)
E( − 5, 0) F(2, − 3) G(1, 3)
H( − 1, − 4) I( − 2, − 1) J(0, 2)
K( − 4, 3)
2)
L
J
C
D
H
I
K
F
E G
3)
4)
5)
6)
7)
8)
9)
10)
12)
13)
11)
14)
447
15)
16)
18)
19)
21)
17)
22)
20)
2
...
3
448
Answers - Slope-Intercept
1
1) y = 2x + 5
29) y = − 2 x + 1
2) y = − 6x + 4
30) y = 5 x + 4
3) y = x − 4
37)
31)
6
4) y = − x − 2
3
5) y = − 4 x − 1
38)
1
6) y = − 4 x + 3
1
7) y = 3 x + 1
8) y =
32)
2
x+5
5
9) y = − x + 5
7
10) y = − 2 x − 5
11) y = x − 1
5
12) y = − 3 x − 3
39)
33)
13) y = − 4x
3
14) y = − 4 x + 2
1
37
15) y = − 10 x − 10
1
3
16) y = 10 x − 10
40)
34)
17) y = − 2x − 1
6
70
18) y = 11 x + 11
7
19) y = 3 x − 8
4
20) y = − 7 x + 4
41)
35)
21) x = − 8
1
22) y = 7 x + 6
23) y = − x − 1
5
24) y = 2 x
42)
36)
25) y = 4x
2
26) y = − 3 x + 1
27) y = − 4x + 3
28) x = 4
2
...
5
Answers - Parallel and Perpendicular Lines
1) 2
9) 0
2
2) − 3
10) 2
3) 4
17) x = 2
11) 3
4) −
9
6
5
12) − 4
19) y − 4 = 2 (x − 3)
13) − 3
10
3
5) 1
6)
7
18) y − 2 = 5 (x − 5)
20) y + 1 = − 4 (x − 1)
5
3
1
14) − 3
7) − 7
3
8) − 4
7
21) y − 3 = 5 (x − 2)
15) 2
3
16) − 8
22) y − 3 = − 3(x + 1)
450
1
23) x = 4
32) y − 5 = − 2 (x + 2)
7
24) y − 4 = 5 (x − 1)
33) y = − 2x + 5
3
x+5
5
25) y + 5 = − (x − 1)
34) y =
26) y + 2 = − 2(x − 1)
35) y = −
4
x−3
3
36) y = −
5
x−5
4
1
(x − 5)
5
27) y − 2 =
1
28) y − 3 = − (x − 1)
37) y = − 2 x − 3
1
29) y − 2 = − 4 (x − 4)
7
5
38) y = 2 x − 2
1
30) y + 5 = 3 (x + 3)
39) y = − 2 x − 2
31) y + 2 = − 3(x − 2)
40) y = 5 x − 1
3
41) y = x − 1
42) y = 2x + 1
43) y = 2
2
44) y = − 5 x + 1
45) y = − x + 3
5
46) y = − 2 x + 2
47) y = − 2x + 5
3
48) y = 4 x + 4
Answers - Chapter 3
3
...
} : R
36) n > 1: (1, ∞)
38) p
3: ( − ∞, 3]
3
...
} : R
31) − 1 < x
1: ( − 1, 1]
32) m > 4 or m
− 1 : ( − ∞, − 1]
(4, ∞)
3
...
1
Answers - Graphing
1) ( − 1, 2)
2) ( − 4, 3)
12) (4, − 4)
23) ( − 1, − 1)
13) (1, − 3)
24) (2, 3)
3) ( − 1, − 3)
14) ( − 1, 3)
5) No Solution
16) No Solution
6) ( − 2, − 2)
17) (2, − 2)
4) ( − 3, 1)
25) ( − 1, − 2)
15) (3, − 4)
26) ( − 4, − 3)
18) (4, 1)
27) No Solution
19) ( − 3, 4)
28) ( − 3, 1)
10) No Solution
21) (3, 2)
29) (4, − 2)
11) (3, − 4)
22) ( − 4, − 4)
30) (1, 4)
7) ( − 3, 1)
8) (4, 4)
9) ( − 3, − 1)
20) (2, − 1)
4
...
3
454
Answers - Addition/Elimination
1) ( − 2, 4)
12) (1, − 2)
25) ( − 1, − 2)
2) (2, 4)
13) (0, 4)
3) No solution
14) ( − 1, 0)
26) ( − 3, 0)
4) Infinite number of
solutions
15) (8, 2)
27) ( − 1, − 3)
16) (0, 3)
28) ( − 3, 0)
5) No solution
17) (4, 6)
6) Infinite number of
solutions
18) ( − 6, − 8)
29) ( − 8, 9)
7) No solution
20) (1, 2)
31) ( − 2, 1)
32) ( − 1, 1)
9) ( − 3, − 5)
21) (0, − 4)
22) (0, 1)
33) (0, 0)
10) ( − 3, 6)
23) ( − 2, 0)
11) ( − 2, − 9)
24) (2, − 2)
8) (2, − 2)
19) ( − 2, 3)
30) (1, 2)
34) Infinite number of
solutions
4
...
5
Answers - Value Problems
1) 33Q, 70D
455
26) (1, 2, 4)
27) ( − 25, 18, − 25)
2 3
2
28) ( 7 , 7 , − 7 )
2) 26 h, 8 n
19) 13 d, 19 q
3) 236 adult, 342 child
20) 28 q
4) 9d, 12q
21) 15 n, 20 d
5) 9, 18
22) 20 S1, 6 S5
6) 7q, 4h
23) 8 S20, 4 S10
7) 9, 18
24) 27
8) 25, 20
25) S12500 @ 12%
S14500 @ 13%
9) 203 adults, 226 child
10) 130 adults, 70
students
26) S20000 @ 5%
S30000 @ 7
...
5%
34) S7000 @ 9%
S5000 @ 7
...
25%
S5500 @ 6
...
25%;
S3000 @ 5
...
5%
S4000 @ 9%
11) 128 card, 75 no card
27) S2500 @ 10%
S6500 @ 12%
12) 73 hotdogs,
58 hamburgers
28) S12400 @ 6%
S5600 @ 9%
13) 135 students,
97 non-students
29) S4100 @ 9
...
5%
S9000 @ 6
...
6%
S4500 @ 6
...
8%;
S3500 @ 8
...
6
Answers - Mixture Problems
1) 2666
...
5
19) 20, 30
2) 2
11) 10
20) 100, 200
3) 30
12) 8
4) 1, 8
13) 9
...
25
18) 40, 110
26) 2, 3
21) 40, 20
22) 10, 5
23) 250, 250
24) 21, 49
456
27) 56, 144
33) 440, 160
39) 10
28) 1
...
5
34) 20
40) 30, 20
29) 30
35) 35, 63
30) 10
36) 3, 2
31) 75, 25
37) 1
...
1
Answers to Exponent Properties
1) 49
17) 42
31) 64
2) 47
18) 34
32) 2a
3) 24
19) 3
4) 36
20) 33
5) 12m2n
21) m
3
6) 12x
33)
34)
2
y3
512x24
y 5 x2
2
22)
xy 3
4
35) 64m12n12
8) x3 y 6
23)
4x2 y
3
36)
9) 312
24)
y2
4
37) 2x2 y
10) 412
25) 4x10 y 14
38) 2y 2
11) 48
26) 8u18v 6
39) 2q 7r 8 p
27) 2x17 y 16
40) 4x2 y 4z 2
28) 3uv
41) x4 y 16z 4
7) 8m6n3
12) 36
6 4
13) 4u v
14) x3 y 3
16
15) 16a
16) 16x4 y 4
29)
x2 y
6
42) 256q 4r 8
30)
4a2
3
43) 4y 4z
5
...
3
Answers to Scientific Notation
1) 8
...
662 × 10−6
29) 1
...
44 × 10−4
16) 5
...
1 × 10−2
17) 1
...
2 × 107
4) 1
...
353 × 108
5) 3
...
815 × 104
6) 1
...
836 × 10−1
7) 870000
21) 5
...
196 × 10−2
32) 2
...
715 × 1014
34) 8
...
149 × 106
22) 6
...
0009
23) 3
...
177 × 10−16
11) 2
25) 2
...
00006
26) 6
...
4 × 10−3
27) 2
...
372 × 103
14) 1
...
91 × 10−2
41) 1
...
939 × 109
37) 4
...
474 × 103
39) 3
...
2 × 106
5
...
5
Answers to Multiply Polynomials
1) 6p − 42
13) 15v 2 − 26v + 8
2) 32k 2 + 16k
14) 6a2 − 44a − 32
3) 12x + 6
4) 18n3 + 21n2
5) 20m5 + 20m4
6) 12r − 21
2
7) 32n + 80n + 48
2
8) 2x − 7x − 4
9) 56b2 − 19b − 15
15) 24x2 − 22x − 7
16) 20x2 − 29x + 6
17) 30x2 − 14xy − 4y 2
18) 16u2 + 10uv − 21v 2
19) 3x2 + 13xy + 12y 2
20) 40u2 − 34uv − 48v 2
10) 4r 2 + 40r+64
21) 56x2 + 61xy + 15y 2
11) 8x2 + 22x + 15
22) 5a2 − 7ab − 24b2
12) 7n2 + 43n − 42
23) 6r 3 − 43r 2 + 12r − 35
459
24) 16x3 + 44x2 + 44x + 40
32) 42u4 + 76u3v + 17u2v 2 − 18v 4
25) 12n3 − 20n2 + 38n − 20
33) 18x2 − 15x − 12
34) 10x2 − 55x + 60
26) 8b3 − 4b2 − 4b − 12
3
2
2
27) 36x − 24x y + 3xy + 12y
3
2
28) 21m + 4m n − 8n
35) 24x2 − 18x − 15
3
36) 16x2 − 44x − 12
3
37) 7x2 − 49x + 70
29) 48n4 − 16n3 + 64n2 − 6n + 36
38) 40x2 − 10x − 5
30) 14a4 + 30a3 − 13a2 − 12a + 3
39) 96x2 − 6
40) 36x2 + 108x + 81
31) 15k 4 + 24k 3 + 48k 2 + 27k + 18
5
...
7
Answers to Divide Polynomials
1
1
n2
1) 5x + 4 + 2x
2)
5x3
9
+ 5x2 +
3) 2n3 + 10 + 4n
4x
9
4)
3k 2
8
k
1
+2+4
460
31) 25 + 20r + 4r 2
32) m2 − 14m + 49
33) 4 + 20x + 25x2
37) n2 − 25
38) 49x2 + 98x + 49
x
5) 2x3 + 4x2 + 2
6)
5p3
4
+ 4p2 + 4p
3
3
20) n + 3 + n + 4
1
7) n2 + 5n + 5
8)
m2
3
5
34) k 2 − 3k − 9 − k − 1
2
+ 2m + 3
9) x − 10 +
9
x+8
10) r + 6 +
1
r −9
8
11) n + 8 − n + 5
5
12) b − 3 − b − 7
9
13) v + 8 − v − 10
14) x − 3 −
8
33) 3n2 − 9n − 10 − n + 6
19) x − 3 + 10x − 2
5
x+7
6
15) a + 4 − a − 8
2
16) x − 6 − x − 4
3
17) 5p + 4 + 9p + 4
1
18) 8k − 9 − 3k − 1
21) r − 1 + 4x + 3
1
35) x2 − 7x + 3 + x + 7
1
22) m + 4 + m − 1
3
36) n2 + 9n − 1 + 2n + 3
23) n + 2
4
4
37) p2 + 4p − 1 + 9p + 9
24) x − 4 + 2x + 3
5
25) 9b + 5 − 3b + 8
7
38) m2 − 8m + 7 − 8m + 7
5
26) v + 3 − 3v − 9
8
7
39) r 2 + 3r − 4 − r − 4
3
40) x2 + 3x − 7 + 2x + 6
27) x − 7 − 4x − 5
5
28) n − 7 − 4n + 5
29) a2 + 8a − 7 −
5
6
a+7
5
30) 8k 2 − 2k − 4 + k − 8
1
31) x2 − 4x − 10 − x + 4
41) 6n2 − 3n − 3 + 2n + 3
3
42) 6b2 + b + 9 + 4b − 7
32) x2 − 8x + 7
Answers - Chapter 6
6
...
2
Answers - Grouping
1) (8r 2 − 5)(5r − 1)
2) (5x2 − 8)(7x − 2)
2
11) (7x + 5)(y − 7)
12) (7r 2 + 3)(6r − 7)
21) (4u + 3)(8v − 5)
22) 2(u + 3)(2v + 7u)
3) (n − 3)(3n − 2)
13) (8x + 3)(4y + 5x)
5) (3b2 − 7)(5b + 7)
24) (4x − 5y 2)(6y − 5)
6) (6x2 + 5)(x − 8)
15) (8x + 1)(2y − 7)
16) (m + 5)(3n − 8)
7) (3x + 2)(x + 5)
17) (2x + 7y )(y − 4x)
25) (3u − 7)(v − 2u)
4) (2v 2 − 1)(7v + 5)
2
8) (7p2 + 5)(4p + 3)
9) (7x2 − 4)(5x − 4)
10)(7n2 − 5)(n + 3)
14) (3a + b2)(5b − 2)
2
18) (m − 5)(5n + 2)
19) (5x − y)(8y + 7)
23) (5x + 6)(2y + 5)
26) (7a − 2)(8b − 7)
27) (2x + 1)(8y − 3x)
20) (8x − 1)(y + 7)
6
...
4
Answers - Trinomials where a
1
1) (7x − 6)(x − 6)
15) (3x − 5)(x − 4)
29) (k − 4)(4k − 1)
2) (7n − 2)(n − 6)
16) (3u − 2v)(u + 5v)
30) (r − 1)(4r + 7)
3) (7b + 1)(b + 2)
17) (3x + 2y)(x + 5y)
31) (x + 2y)(4x + y)
4) (7v + 4)(v − 4)
18) (7x + 5y)(x − y)
32) 2(2m2 + 3mn + 3n2)
5) (5a + 7)(a − 4)
19) (5x − 7y)(x + 7y)
6) Prime
20) (5u − 4v)(u + 7v)
7) (2x − 1)(x − 2)
21) 3(2x + 1)(x − 7)
8) (3r + 2)(r − 2)
22) 2(5a + 3)(a − 6)
9) (2x + 5)(x + 7)
23) 3(7k + 6)(k − 5)
10) (7x − 6)(x + 5)
24) 3(7n − 6)(n + 3)
11) (2b − 3)(b + 1)
25) 2(7x − 2)(x − 4)
12) (5k − 6)(k − 4)
26) (r + 1)(4r − 3)
39) 4(x − 2y)(6x − y)
13) (5k + 3)(k + 2)
27) (x + 4)(6x + 5)
40) 2(3x + 2y)(2x + 7y)
14) (3r + 7)(r + 3)
28) (3p + 7)(2p − 1)
33) (m − 3n)(4m + 3n)
34) 2(2x2 − 3xy + 15y 2)
35) (x + 3y)(4x + y)
36) 3(3u + 4v)(2u − 3v)
37) 2(2x + 7y)(3x + 5y)
38) 4(x + 3y)(4x + 3y)
6
...
6
Answers - Factoring Strategy
1) 3(2a + 5y)(4z − 3h)
6) 5(4u − x)(v − 3u2)
2) (2x − 5)(x − 3)
7) n(5n − 3)(n + 2)
3) (5u − 4v)(u − v)
8) x(2x + 3y)(x + y)
4) 4(2x + 3y)2
9) 2(3u − 2)(9u2 + 6u + 4)
5) 2( − x + 4y)(x2 + 4xy + 16y 2)
10) 2(3 − 4x)(9 + 12x + 16x2)
464
11) n(n − 1)
27) (3x − 4)(9x2 + 12x + 16)
12) (5x + 3)(x − 5)
28) (4a + 3b)(4a − 3b)
13) (x − 3y)(x − y)
29) x(5x + 2)
14) 5(3u − 5v)2
30) 2(x − 2)(x − 3)
15) (3x + 5y)(3x − 5y)
31) 3k(k − 5)(k − 4)
16) (x − 3y)(x2 + 3xy + 9y 2)
32) 2(4x + 3y)(4x − 3y)
17) (m + 2n)(m − 2n)
33) (m − 4x)(n + 3)
18) 3(2a + n)(2b − 3)
34) (2k + 5)(k − 2)
19) 4(3b2 + 2x)(3c − 2d)
35) (4x − y)2
20) 3m(m + 2n)(m − 4n)
36) v(v + 1)
21) 2(4 + 3x)(16 − 12x + 9x2)
37) 3(3m + 4n)(3m − 4n)
22) (4m + 3n)(16m2 − 12mn + 9n2)
38) x2(x + 4)
23) 2x(x + 5y)(x − 2y)
39) 3x(3x − 5y)(x + 4y)
24) (3a + x2)(c + 5d2)
40) 3n2(3n − 1)
25) n(n + 2)(n + 5)
41) 2(m − 2n)(m + 5n)
26) (4m − n)(16m2 + 4mn + n2)
42) v 2(2u − 5v)(u − 3v)
6
...
1
Answers - Reduce Rational Expressions
1) 3
20)
7
8k
37)
3a − 5
5a + 2
1
3
21)
4
x
38)
9
p+2
22)
9x
2
39)
23)
3m − 4
10
2n − 1
9
24)
10
9n2(9n + 4)
40)
3x − 5
5(x + 2)
25)
10
2p + 1
41)
2(m + 2)
5m − 3
26)
1
9
42)
9r
5(r + 1)
9) − 2
27)
1
x+7
43)
10) 0, − 10
28)
7m + 3
9
2(x − 4)
3x − 4
29)
8x
7(x + 1)
44)
5b − 8
5b + 2
30)
7r + 8
8r
45)
7n − 4
4
31)
n+6
n−5
46)
5(v + 1)
3v + 1
32)
b+6
b+7
47)
(n − 1)2
6(n + 1)
33)
9
v − 10
48)
7x − 6
(3x + 4)(x + 1)
49)
7a + 9
2(3a − 2)
50)
2(2k + 1)
9(k − 1)
2)
1
3) − 5
4) undefined
5)
1
2
6) 6
7) − 10
8) 0, 2
5
11) 0
12) −
10
3
13) − 2
14) 0, −
1
2
15) − 8, 4
1
16) 0, 7
34)
3(x − 3)
5x + 4
18)
3
n
35)
2x − 7
5x − 7
19)
3
5a
36)
k −8
k+4
17)
7x
6
7
...
3
Answers - Least Common Denominators
1) 18
2) a2
3) ay
4) 20xy
5) 6a2c3
6) 12
7) 2x − 8
8) x2 − 2x − 3
9) x2 − x − 12
10) x2 − 11x + 30
467
11) 12a4b5
12) 25x3 y 5z
13) x (x − 3)
14) 4(x − 2)
15) (x + 2)(x − 4)
16) x(x − 7)(x + 1)
17) (x + 5)(x − 5)
18) (x − 3)2(x + 3)
19) (x + 1)(x + 2)(x + 3)
20) (x − 2)(x − 5)(x + 3)
21)
6a4
2b
,
10a3b2 10a3b2
22)
3x2 + 6x
2x − 8
,
(x − 4)(x + 2) (x − 4)(x + 2)
23)
x2 + 4x + 4
x2 − 6x + 9
,
(x − 3)(x + 2) (x − 3)(x + 2)
24)
5
2x − 12
− 3x
,
,
x(x − 6) x(x − 6) x(x − 6)
25)
x2 − 4x
3x2 + 12x
,
(x − 4)2(x + 4) (x − 4)2(x + 4)
26)
5x + 1
4x + 8
,
(x − 5)(x + 2) (x − 5) (x + 2)
27)
x2 + 7x + 6
2x2 − 9x − 18
2 , (x − 6)(x + 6)2
(x − 6)(x + 6)
28)
3x2 + 4x + 1
2x2 − 8x
,
(x − 4)(x + 3)(x + 1) (x − 4)(x + 3)(x + 1)
29)
4x
x2 + 4x + 4
,
(x − 3)(x + 2) (x − 3)(x + 2)
30)
3x2 + 15x
x2 − 4x + 4
5x − 20
, (x − 4)(x − 2)(x + 5) , (x − 4)(x − 2)(x + 5)
(x − 4)(x − 2)(x + 5)
7
...
5
Answers - Complex Fractions
1)
x
x−1
12)
2)
1−y
y
13)
3)
−a
a+2
14) 3x + 2
4)
5−a
a
15)
4b(a − b)
a
16)
x+2
x−1
17)
x−5
x+9
a−1
5) − a + 1
2a2 − 3a + 3
− 4a2 − 2a
x
3
b−2
25) − 2b + 3
26)
x+y
x−y
27)
a − 3b
a + 3b
6)
b3 + 2b − b − 2
7)
2
5
18) −
8)
4
5
19)
1
3x + 8
29) − y
20)
1
x+4
30) x2 − 1
21)
x−2
x+2
31)
y −x
xy
22)
x−7
x+5
32)
x2 − xy + y 2
y −x
8b
1
9) − 2
1
10) − 2
11)
x2 − x − 1
x2 + x + 1
(x − 3)(x + 5)
4x2 − 5x + 4
469
2x
28) − x2 + 1
2
33)
x2 + y 2
xy
35)
(1 − 3x)2
x2(x + 3)(x − 3)
34)
2x − 1
2x + 1
36)
x+y
xy
7
...
31
18) n =
61
3
32) 16
19) x =
38
3
33) 2
...
1 ft
4) x = 16
5) x =
3
2
6) n = 34
7) m = 21
16
7
30) x = − 1, 3
21) x = − 8, 5
35) 39
...
1 in
10) n = 25
23) m = − 7, 8
8) x =
79
8
11) b = −
12) r =
36
5
13) x =
5
2
14) n =
32
5
40
3
24) x = − 3, 9
37) T: 38, V: 57
25) p = − 7, − 2
38) J: 4 hr, S: 14 hr
26) n = − 6, 9
39) S8
27) n = − 1
28) n = − 4, − 1
40) C: 36 min,
K: 51 min
7
...
8
Answers - Dimensional Analysis
1) 12320 yd
16) 2,623,269,600 km/yr
2) 0
...
6 lb/in2
3) 0
...
51 km/hr2
4) 135,000 cm
19) 32
...
1 mi
20) 6
...
5 yd2
21)17280 pages/day; 103
...
435 km2
22) 2,365,200,000 beats/lifetime
8) 86,067,200 ft2
23) 1
...
58 cm3
25) 56 mph; 25 m/s
11) 0
...
15 yd3; 113 m3
12) 5
...
31 mph
28) 350,000 pages
14) 104
...
42 lb
Answers - Chapter 8
8
...
2
√
3
1) 5 5
√
2) 53 3
√
3) 53 6
√
4) 53 2
√
5) 53 7
√
6) 23 3
√
7) − 84 6
√
8) − 164 3
√
9) 124 7
√
10) 64 3
√
11) − 24 7
√
12) 154 3
√
4
13) 3 8a2
√
4
14) 2 4n3
Answers - Higher Roots
√
5
15) 2 7n3
√
5
16) − 2 3x3
√
17) 2p5 7
√
18) 2x6 4
√
19) − 67 7r
√
7
20) − 16b 3b
√
21) 4v 2 3 6v
√
22) 20a2 3 2
√
23) − 28n2 3 5
24) − 8n2
√
3
25) − 3xy 5x2
√
3
26) 4uv u2
√
27) − 2xy 3 4xy
√
3
28) 10ab ab2
√
29) 4xy 2 3 4x
472
√
30) 3xy 2 3 7
√
31) − 21xy 2 3 3y
32) − 8y 2 3 7x2 y 2
√
3
33) 10v 2 3u2v 2
√
34) − 403 6xy
√
3
35) − 12 3ab2
√
36) 9y 3 5x
37) − 18m2np2 3 2m2 p
38) − 12mpq 4 5p3
39) 18xy 4 8xy 3z 2
√
40) − 18ab2 4 5ac
√
4
41) 14hj 2k 2 8h2
42) − 18xz 4 4x3 yz 3
8
...
4
√
1) − 48 5
√
2) − 25 6
√
3) 6m 5
√
4) − 25r 2 2r
Answers - Multiply and Divide Radicals
√
5) 2x2 3 x
√
6) 6a2 3 5a
√
√
7) 2 3 + 2 6
√
√
8) 5 2 + 2 5
473
√
√
9) − 45 5 − 10 15
√
√
10) 45 5 + 10 15
√
√
11) 25n 10 + 10 5
√
√
12) 5 3 − 9 5v
√
13) − 2 − 4 2
√
14) 16 − 9 3
√
15) 15 − 11 5
√
√
√
16) 30 + 8 3 + 5 15 + 4 5
√
√
√
17) 6a + a 10 + 6a 6 + 2a 15
√
√
18) − 4p 10 + 50 p
√
19) 63 + 32 3
√
√
√
20) − 10 m + 25 2 + 2m − 5
√
22)
3
25
√
15
4
23)
1
20
21)
24) 2
√
29)
15
3
√
10
15
√
4 3
9
√
4 5
5
√
5 3xy
12y 2
30)
√
4 3x
16xy 2
25)
26)
27)
28)
31)
32)
33)
34)
35)
36)
37)
38)
√
6p
3
√
2 5n
5
√
3
10
5
√
3
15
4
√
3
10
8
√
4
8
8
√
4
5 10r 2
2
√
4
4n2
mn
8
...
6
Answers - Rational Exponents
√
5
1) ( m )3
2)
1
√
(4 10r )3
√
3) ( 7x )3
4)
−2
17)
18)
1
3a2
27)
m
35
8
7
n6
y 12
28)
5
x6
29)
20) 1
−4
1
8) (5a) 2
9) 4
1
5 3
y 4x2
1
3
n4
1
1
21) y 2
22)
30)
v2
y3
1
x4
7
u2
10) 2
11) 8
23)
1
1000
4
5
13) x 3 y 2
14)
1
2b 4
19) u2v 2
7
12)
26)
a2
11
1
2
7) n
3
25
3
5) (6x)
1
1 1
a2b2
16) 1
1
√
(3 6b )4
6) v
15)
24)
1
3
4
32)
17
2y 6
x3
y
7
10
3
x4
1
4
v3
4
31) xy 3
7 3
b4a4
25)
1
3y 12
33)
2
u2
3
v4
475
15
17
34) x 4 y 4
8
...
8
Answers - Complex Numbers
22) − 28 + 76i
2) − 4i
23) 44 + 8i
3) − 3 + 9i
24) 16 − 18i
41)
− 30i − 5
37
42)
1) 11 − 4i
48i − 56
85
6) 5 − 12i
27) 9i + 5
7) − 4 − 11i
28)
− 3i − 2
3
43) 9i
√
44) 3i 5
√
45) − 2 5
√
46) − 2 6
8) − 3 − 6i
29)
10i − 9
6
47)
9) − 8 − 2i
30)
4i + 2
3
48)
31)
3i − 6
4
12) 24
32)
5i + 9
9
13) 40
33) 10i + 1
14) 32
34) − 2i
15) − 49
35)
− 40i + 4
101
36)
9i − 45
26
37)
56 + 48i
85
38)
4 − 6i
13
39)
70 + 49i
149
40)
− 36 + 27i
50
4) − 1 − 6i
5) − 3 − 13i
10) 13 − 8i
11) 48
16) 28 − 21i
17) 11 + 60i
18) − 32 − 128i
19) 80 − 10i
20) 36 − 36i
21) 27 + 38i
25) − 3 + 11i
26) − 1 + 13i
√
1+i 3
2
√
2+i 2
2
49) 2 − i
50)
√
3 + 2i 2
2
51) i
52) − i
53) 1
54) 1
55) − 1
56) i
57) − 1
58) − i
Answers - Chapter 9
9
...
2
Answers - Solving with Exponents
√
1) ± 5 3
2) − 2
10)
√
−1±3 2
2
18)
√
√
13) − 7
3) ± 2 2
4) 3
19)
11) 65, − 63
12) 5
5) ± 2 6
14) −
6) − 3, 11
7) − 5
1
15)
11 5
,
2 2
21) −
191
64
3
34
, − 10
3
22) 3
23) −
5
17) − 8 , − 8
9) − 1
5
4
20) No Solution
11
5
,− 2
2
16) −
3
8) 5 , − 5
9
8
17
2
24) No Solutoin
9
...
− 7
46)
35) 7, 1
47)
36) 2, − 6
37)
38)
39)
√
√
− 6 + i 258 − 6 − i 258
,
6
6
√
√
− 6 + i 111 − 6 − i 111
,
3
3
√
√
5 + i 130 5 − i 130
,
5
5
40) 2, − 4
41)
42)
43)
44)
√
√
− 5 + i 87 − 5 − i 87
,
2
2
√
√
− 7 + 181 − 7 − 181
,
2
2
√
√
3 + i 271 3 − i 271
,
7
7
√
√
− 1 + 2i 6 − 1 − 2i 6
,
2
2
48)
49)
50)
51)
52)
53)
√
√
7 + i 139 7 − i 139
,
2
2
√
√
5 + 3i 7 5 − 3i 7
, 2
2
12
,−4
5
√
√
1 + i 511 1 − i 511
,
4
4
√
√
9 + 21 9 − 21
, 2
2
√
1 + i 163 1 − i163
, 2
2
√
√
− 5 + i 415 − 5 − i 415
,
8
8
√
√
11 + i 95 11 − i 95
,
6
6
√
√
5 + i 191 5 − i 191
,
2
2
54) 8, 7
5
55) 1, − 2
3
56) 3, − 2
9
...
5
Answers - Build Quadratics from Roots
NOTE: There are multiple answers for each problem
...
1) x2 − 7x + 10=0
2) x2 − 9x + 18 = 0
3) x2 − 22x + 40 = 0
2
4) x − 14x + 13 = 0
5) x2 − 8x + 16 = 0
6) x2 − 9x = 0
7) x2 = 0
8) x2 + 7x + 10 = 0
9) x2 − 7x − 44 = 0
15) 7x2 − 31x + 12 = 0
16) 9x2 − 20x + 4 = 0
17) 18x2 − 9x − 5 = 0
2
18) 6x − 7x − 5 = 0
19) 9x2 + 53x − 6 = 0
34) x2 + 4x + 20 = 0
22) x2 − 1 = 0
23) 25x2 − 1 = 0
25) x2 − 11 = 0
27) 16x2 − 3 = 0
28) x2 + 121 = 0
9
...
7
Answers - Rectangles
1) 6 m x 10 m
10) 1
...
25 in
4) 10 ft x 18 ft
13) 1
...
145 in
481
22) 23
...
5 ft
9
...
4
14) 12 hours
5) C = 4, J = 12
15) 16 hours
6) 1
...
6 hours
28) 12 and 36 min
26) 15 and 22
...
9
Answers - Simultaneous Product
1) (2, 36), ( − 18, − 4)
7) (45, 2), ( − 10, − 9)
9
2) ( − 9, − 20), ( − 40, − 2 )
8) (16, 3), ( − 6, − 8)
5
3) (10, 15), ( − 90, − 3 )
9) (1, 12), ( − 3, − 4)
10) (20, 3), (5, 12)
4) (8, 15), ( − 10, − 12)
5
5) (5, 9), (18, 2
...
10
Answers - Revenue and Distance
1) 12
9) 60 mph, 80 mph
2) S4
10) 60, 80
3) 24
11) 6 km/hr
4) 55
12) 200 km/hr
19) 45 mph
5) 20
13) 56, 76
20) 40 mph, 60 mph
6) 30
14) 3
...
11
482
21) 20 mph
22) 4 mph
Answers - Graphs of Quadratics
7)
1)
13)
(3,0)
(-2,0)
(0,-8)
(4,0)
(4,3)
(5,0)
(2,0)
8)
(2,3)
(-1,0)
(0, -3)
(1,0)
(3, 0)
(1, -4)
(6,0)
(0,-24)
(0,-45)
(1,-9)
2)
(4,8)
(3,0)
(-9,0)
14)
(1,0)
(-3,0)
(-1,-8)
(0,-6)
9)
3)
(2,9)
(0,10)
(1,0)
(5,0)
(0,5)
(-1,0)
15)
(5,0)
(-3,0) (-1,0)
(3,-8)
(-2,-3)
10)
4)
(0,16)
(2,0)
(1,0) (2,1)
(3,0)
(0,-3)
(4,0)
(3, -2)
(0,9)
16)
(0,45)
(-3,0)
11)
5)
(3,4)
(3,0)
(1,0)
(5,0)
17)
(0,-5)
(0,75)
(0,-18)
12)
6)
(3,8)
(1,0)
(-10,0)
(3,0)
(4,2)
(5,0)
(5,0)
(3,0) (5,0)
(4,-5)
18)
(0,15)
(0,-30)
(-3,0)
(-1,0)
(-2,-5)
483
19)
20)
(-6,5)
(-7,0)
(2,5)
(1,0)
(-5,0)
(0,-175)
(3,0)
(-15,0)
Answers - Chapter 10
10
...
yes b
...
no
d
...
yes f
...
yes h
...
2
Answers - Operations on Functions
1) 82
6) − 30
2) 20
7) − 3
3) 46
8) 140
4) 2
9) 1
5) 5
10) − 43
484
−2+n
2
+1
11) 100
36)
12) − 74
3n − 6
− n2 − 4n
37)
− x3 − 2x
− 3x + 4
13)
1
5
38) x4 − 4x2 − 3
14) 27
39)
− n2 − 2n
3
16) n − 2n
40)
32 + 23n − n3
8
18) − x3 + 2x2 − 3
42) 5
15) −
9
26
2
41) − 155
17) − x3 − 4x − 2
43) 21
19) − x2 − 8x + 2
44) 4
20) 2t2 − 8t
21) 4x3 + 25x2 + 25x
45) 103
22) − 2t3 − 15t2 − 25t
46) 12
47) − 50
23) x2 − 4x + 5
48) 112
24) 3x2 + 4x − 9
25)
49) 176
n2 + 5
3n + 5
50) 147
26) − 2x + 9
27)
51) 16x2 + 12x − 4
− 2a + 5
3a + 5
3
52) − 8a + 14
2
28) t + 3t − 3t + 5
53) − 8a + 2
3
29) n + 8n + 5
30)
54) t
4x + 2
x2 + 2x
55) 4x3
31) n6 − 9n4 + 20n2
56) − 2n2 − 12n − 16
32) 18n2 − 15n − 25
57) − 2x + 8
33) x + 3
58) 27t3 − 108t2 + 141t − 60
2
34) − 3
59) − 16t − 5
35) t4 + 8t2 + 2
60) 3x3 + 6x2 − 4
10
...
4
Answers - Exponential Functions
1) 0
15) 1
29) 0
2) − 1
16) − 1
30) No solution
3) 0
17) No solution
4) 0
18) − 3
4
1
3
31) 1
32) 3
5) − 4
19) − 4
5
20) − 4
3
33)
1
3
3
21) No solution
34)
2
3
6) − 4
7) − 2
22) 0
8) 0
9) −
24)
10) 0
11)
35) 0
3
23) − 2
2
3
36) 0
2
5
3
8
25) − 1
12) 0
37)
26)
5
6
38) − 1
1
4
1
13) − 2
27) − 2
14) − 6
28)
5
39) − 3
1
3
40) No solution
486
4x
3
10
...
6
Answers - Interest Rate Problems
1)
a
...
12; 745
...
1209
...
87
b
...
11; 859
...
1528
...
27
c
...
08; 953
...
2694
...
72
d
...
22; 1984
...
3219
...
99
2) 1640
...
60
3) 2868
...
66
4) 2227
...
98
5) 1726
...
12
6) 1507
...
19
i
...
17; 7190
...
43
13) 12
...
96
14) 3823
...
68
10
...
3256
14) 8
...
6
2) 0
...
1
29) 1
3) 0
...
8
30) 8
4) 0
...
2
31) 1
...
8
6)
8
15
19) 17
...
2
33) 5
...
3
34)2
21) 3
...
1
22) 10
...
2
4
5
24) 8
...
2
11) 16
...
5
38) 3
...
8
26) 24
...
1
13) 32
27) 4
...
2
9)
10)
36) 3
...
8
Answers - Inverse Trigonometric Functions
1) 29◦
11) 36◦
2) 39◦
12) 61
...
2◦
5) 24◦
15) 55
...
7◦
7) 15◦
17) 58◦
8) 18◦
18) 20
...
2◦
10) 35◦
20) 73
...
3◦
22) 45◦
32) m∠B = 22
...
2◦,
c = 16
...
4◦
33) m∠B = 22
...
5◦, c = 7
...
2◦
34) m∠A = 39◦, b = 7
...
9
25) 55◦
35) m∠B = 64
...
4◦, b = 6
...
5◦
36) m∠A = 69◦, b = 2
...
5
27) 47◦
37) m∠B = 38◦, b = 9
...
6
28) 15
...
4, c = 14
29) 30◦
39) m∠A = 45◦, b = 8, c = 11
...
1, c = 32
...
1◦, m∠A = 60
...
2
489
Title: INTERMEDIATE ALGEBRA (English)
Description: It's very helpful for new intermediate students.
Description: It's very helpful for new intermediate students.