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Title: Algebra Question 501
Description: Algebra Question From Learning Express Newyork. The Learning Express Skill Builder in Focus Writing Team is comprised of experts in test preparation, as well as educators and teachers who specialize in language arts and math.
Description: Algebra Question From Learning Express Newyork. The Learning Express Skill Builder in Focus Writing Team is comprised of experts in test preparation, as well as educators and teachers who specialize in language arts and math.
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501
Algebra Questions
501
Algebra Questions
2nd Edition
®
NEW YORK
Copyright © 2006 LearningExpress, LLC
...
Published in the United States by LearningExpress, LLC, New York
...
—2nd ed
...
cm
...
ed
...
1st ed
...
ISBN 1-57685-552-X
1
...
I
...
501 algebra
questions
...
LearningExpress (Organization)
...
Title: Five hundred one
algebra questions
...
Title: Five hundred and one algebra questions
...
A15 2006
512—dc22
2006040834
Printed in the United States of America
9 8 7 6 5 4 3 21
Second Edition
ISBN 1-57685-552-X
For more information or to place an order, contact LearningExpress at:
55 Broadway
8th Floor
New York, NY 10006
Or visit us at:
www
...
com
The LearningExpress Skill Builder in Focus Writing Team is
comprised of experts in test preparation, as well as educators and
teachers who specialize in language arts and math
...
James, New York
Colleen Schultz
Middle School Math Teacher, Grade 8
Vestal Central School District
Math Tutor
Vestal, New York
Contents
Introduction
ix
1
Working with Integers
1
2
Working with Algebraic Expressions
12
3
Combining Like Terms
24
4
Solving Basic Equations
41
5
Solving Multi-Step Equations
49
6
Solving Equations with Variables on Both Sides
of an Equation
58
7
Using Formulas to Solve Equations
72
8
Graphing Linear Equations
81
9
Solving Inequalities
110
10 Graphing Inequalities
119
11 Graphing Systems of Linear Equations
and Inequalities
142
12 Solving Systems of Equations Algebraically
172
13 Working with Exponents
186
Contents
14 Multiplying Polynomials
194
15 Factoring Polynomials
206
16 Using Factoring
215
17 Solving Quadratic Equations
229
18 Simplifying Radicals
242
19 Solving Radical Equations
250
20 Solving Equations with the Quadratic Formula
261
viii
Introduction
This book is designed to provide you with review and practice for algebra
success! It is not intended to teach common algebra topics
...
501 Algebra Questions is designed for many audiences
...
It can be used to supplement current instruction in a math class
...
If, at some point, you feel you need further explanation
about some of the algebra topics highlighted in this book, you can find them in
the LearningExpress publication Algebra Success in 20 Minutes a Day
...
The book is organized into 20 chapters with a variety of arithmetic,
algebra, and word problems
...
You may want to follow the sequence,
as each succeeding chapter builds on skills taught in previous chapters
...
Chapters are arranged using the same method
...
Second, there are
helpful tips on how to practice the problems in each chapter
...
In many books, you are given one model
problem and then asked to do many problems following that model
...
If you find
yourself getting stuck solving a problem, you can look at the answer explanation and
use it to help you understand the problem-solving process
...
The purpose of drills and practice is to make you proficient at solving problems
...
If, after
completing all the problems in a section, you feel that you need more practice, do
the problems over
...
You will probably want to have a calculator handy as you work through some of
the sections
...
If you have
difficulty factoring numbers, the multiplication chart on the next page may help
you
...
And don’t forget
to keep lots of scrap paper on hand
...
Make the commitment to improve your algebra skills
...
Why you do a math operation is as important
as how you do it
...
You can do it! When you achieve algebra success, you have laid the foundation for future challenges and success
...
When you find signed numbers with
addition and subtraction operations, you can simplify the task by changing
all subtraction to addition
...
For example, subtracting a three is the same as adding a negative
three
...
As
you go through the step-by-step answer explanations, you will begin to see
how this process of using only addition can help simplify your understanding of operations with signed numbers
...
After all, that’s the point of practice!
You work at the problems until the process becomes automatic
...
The Tips for Working with Integers section that follows gives you
some simple rules to follow as you solve problems with integers
...
That’s when you can consider them yours
...
You can use a memory device called a mnemonic
to help you remember a set of instructions
...
This nonsense word helps you remember to:
501 Algebra Questions
P
E
MD
AS
do operations inside Parentheses
evaluate terms with Exponents
do Multiplication and Division in order from left to right
Add and Subtract terms in order from left to right
Tips for Working with Integers
Addition
Signed numbers the same? Find the SUM and use the same sign
...
(The larger number is the one whose value without a positive or negative
sign is greatest
...
That is, you can add numbers in any order and
the result is the same
...
Subtraction
Change the operation sign to addition, change the sign of the number following the operation, then follow the rules for addition
...
Signs
different? Multiply or divide and give the result a negative sign
...
You can multiply terms in any order and the
result will be the same
...
Evaluate the following expressions
...
27 + −5
2
...
−15
− −7
4
...
8 + −4 − 12
6
...
−25
8
...
24 · −8 + 2
10
...
−15
+ 5 + −11
2
501 Algebra Questions
12
...
3 + −7 − 14 + 5
14
...
(−18 ÷ 2) − (6 · −3)
16
...
23 − (−4)2
18
...
21 + (11 + −8)3
20
...
A scuba diver descends 80 feet, rises 25 feet, descends 12 feet, and then
rises 52 feet where he will do a safety stop for five minutes before
surfacing
...
A digital thermometer records the daily high and low temperatures
...
The low was −12° C
...
A checkbook balance sheet shows an initial balance for the month of $300
...
Deposits were made into the account in the amounts of $84 and
$116
...
A gambler begins playing a slot machine with $10 in quarters in her coin
bucket
...
She
then plays 20 more quarters in the same machine before walking away
...
A glider is towed to an altitude of 2,000 feet above the ground before
being released by the tow plane
...
What is the glider’s
altitude now?
3
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
1
...
The sign of the larger term is positive, so the sign of
the result is positive
...
Change the subtraction sign to addition by
changing the sign of the number that follows it
...
Since the signs were negative, the result
is negative
...
Change the subtraction sign to addition by
changing the sign of the number that follows it
...
Give the result the sign of the larger term
...
Signs different? Subtract the value of the numbers
...
4
[27 − 5 = 22]
27 + −5 = +22
−18
+ −20 + (−16)
[18 + 20 + 16 = 54]
−18
+ −20 + −16 = −54
−18
− 20 − 16 = −54
−15
+7
[15 − 7 = 8]
−15 + 7 = −8
−15 − 7 = −8
[33 − 16 = 17]
33 + −16 = +17
501 Algebra Questions
5
...
With three terms, first group like terms and add
...
Substitute the result into the first expression
...
Give the result the sign of the larger term
...
First divide
...
Substitute the result into the expression
...
Give the result the sign of the term with the
larger value
...
First perform the multiplications
...
Signs different? Multiply the terms and give
the result a negative sign
...
Signs different? Subtract the value of the
numbers
...
Because all the operators are multiplication, you
could group any two terms and the result would
be the same
...
Signs the same? Multiply the terms and give
the result a positive sign
...
Signs different? Multiply the terms and give
the result a negative sign
...
Group the terms being multiplied and evaluate
...
Substitute
...
Give the result the sign of the term with the
larger value
...
Because all the operators are multiplication, you
could group any two terms and the result would be
the same
...
Signs the same? Multiply the terms and give the
result a positive sign
...
Signs the same? Multiply the terms and give the
result a positive sign
...
Because all the operators are addition, you could
group any two terms and the result would be the
same
...
Signs different? Subtract the value of the numbers
...
Substitute
...
The simplified result of the numeric expression
is as follows:
(−15 + 5) + −11
[15 − 5 = 10]
[(−15 + 5) = −10]
(−10) + −11
[10 + 11 = 21]
(−10) + −11 = −21
−15
+ 5 + −11 = −21
12
...
Signs different? Divide and give the result
a negative sign
...
Change the subtraction sign to addition
by changing the sign of the number that
follows it
...
The simplified result of the numeric
expression is as follows:
6
[49 ÷ 7 = 7]
[48 ÷ −4 = −12]
(7) − (−12)
7 + +12
7 + +12 = +19
(49 ÷ 7) − (48 ÷ −4) = +19
501 Algebra Questions
13
...
Now perform additions from left to right
...
[7 − 3 = 4]
[3 + −7 = −4]
(−4) + −14 + 5
(−4 + −14) + 5
Substitute
...
Signs the same? Add the value of the terms and
give the result the same sign
...
Signs different? Subtract the value of the
numbers and give the result the sign of the
higher value number
...
First evaluate the expressions within the
parentheses
...
Substitute the values into the original
expression
...
The simplified result of the numeric
expression is as follows:
7
[5 · 3 = 15]
[12 ÷ −4 = −3]
−(15)
+ (−3)
[15 + 3 = 18]
−(15) + (−3) = −18
−(5
· 3) + (12 ÷ −4) = −18
501 Algebra Questions
15
...
Signs different? Divide the value of the
terms and give the result a negative sign
...
Substitute the values into the original
expression
...
Signs different? Subtract the value of the
numbers and give the result the sign of the
higher value number
...
Evaluate the expressions within the parentheses
...
Substitute the value into the original expression
...
The simplified result of the numeric expression
is as follows:
[(−18 ÷ 2)]
[18 ÷ 2 = 9]
[(−18 ÷ 2 = −9)]
(6 · −3)
[6 · 3 = 18]
(6 · −3) = −18
(−9) − (−18)
(−9) + (+18)
[18 − 9 = 9]
(−9) + (+18) = +9
(−18 ÷ 2) − (6 · −3) = +9
(64 ÷ −16)
[64 ÷ 16 = 4]
(64 ÷ −16 = −4)
23 + (−4)
[23 − 4 = 19]
23 + (−4) = +19
23 + (64 ÷ −16) = +19
17
...
Signs the same? Multiply the terms and give
the result a positive sign
...
Change subtraction to addition and change
the sign of the term that follows
...
8
[4 · 4 = 16]
[(−4)2 = +16]
23 − (−4)2 = (8) − (+16)
8 + −16
[16 − 8 = 8]
8 + −16 = −8
501 Algebra Questions
The simplified result of the numeric expression
is as follows:
18
...
Change subtraction to addition and change
the sign of the term that follows
...
Substitute the values of the expressions in
parentheses into the original expression
...
Signs different? Multiply the value of the
terms and give the result a negative sign
...
Signs different? Subtract the value of the
numbers and give the result the sign of the
higher value number
...
First evaluate the expression within the parentheses
...
Substitute the value into the expression
...
Substitute the value into the expression
...
First evaluate the expressions within the
[32 + 6 = (9) + 6 = 15]
parentheses
...
Substitute values into the original expression
...
[−24 ÷ 8 = −3]
(15) ÷ (−3)
The simplified result of the numeric
expression is as follows:
(32 + 6) ÷ (−24 ÷ 8) = −5
9
[15 ÷ 3 = 5]
(15) ÷ (−3) = −5
501 Algebra Questions
21
...
Going up is in the
positive direction, while going down is in the negative direction
...
The resulting numerical expression would be
−80 + +25 + −12 + +52
as follows:
Because addition is commutative, you can
associate like-signed numbers
...
[−80 + −12 = −92]
[+25 + +52 = +77]
Substitute the values into the numerical
expression
...
[92 − 77 = 15]
−
The diver took his rest stop at 15 feet
...
You could simply figure that +5° C is 5° above zero and −11° C is 11°
below
...
Or you could find the difference between +5° and −11°
...
23
...
An expression to represent
the activity during the month
would be as follows:
300 + −25 + −82 + −213 + −97 + +84 + +116
Because addition is commutative, you can associate
like signed numbers
...
[300 + +84 + +116 = +500]
[(−25 + −82 + −213 + −97 = −417]
Substitute the values into
the revised expression
...
10
501 Algebra Questions
24
...
Four quarters per dollar gives you 4 · 10
= 40 quarters
...
In chronological
order, the expression would be as follows:
40 − 15 + 50 − 20
Change all operation signs to addition and
the sign of the number that follows
...
(40 + 50) + (−15 + −20)
Use the rules for adding integers with like
signs
...
(90) + (−35)
Signs different? Subtract the value of the
numbers and give the result the sign of the
higher value number
...
As in problem 21, ascending is a positive number while descending is a neg-
ative number
...
An expression that represents the problem
+2,000 + −450 + +1,750
is as follows:
Because addition is commutative, you can
associate like-signed numbers
...
[+2,000 + +1,750 = +3,750]
Substitute into the revised equation
...
[3,750 − 450 = 3,300]
The simplified result of the numeric
expression is as follows:
(+3,750) + −450 = +3,300
11
2
Working with
Algebraic Expressions
This chapter contains 25 algebraic expressions; each can contain up to
five variables
...
When given numerical values for the
variables, you can turn an algebraic expression into a numerical one
...
You will be evaluating expressions very much like the previous
numerical expressions
...
Work on developing a similar style throughout, and you will have one sure way of solving these kinds
of problems
...
Read through the Tips for Working with Algebraic Expressions
before you begin to solve the problems in this section
...
Use PEMDAS to perform operations in the proper order
...
501 Algebra Questions
Evaluate the following algebraic expressions when
a=3
b = −5
x=6
1
y = 2
z = −8
■
26
...
3x ÷ z
28
...
5ab + xy
30
...
7x ÷ 2yz
32
...
6y − 2ab
34
...
2(a2 + 2y) ÷ b
36
...
−2x
− b + az
38
...
5xy ÷ 2b
12
40
...
2b2 ÷ y
42
...
6y(z ÷ y) + 3ab
44
...
12ab ÷ y
x
46
...
10b3 − 4b2
48
...
z2 − 4a2y
50
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
Underlined expressions show the original algebraic expression as an equation with the expression
equal to its simplified result
...
Substitute the values for the variables into the expression
...
Substitute
...
Give the result the sign of the larger value
...
Substitute the values for the variables into the
expression
...
Substitute
...
1
(−24)
The value of the expression is as follows:
+4
4a + z = +4
3(6) ÷ (−8)
[3(6) = 18]
(18) ÷ (−8)
2
14
1
[18 ÷ 8 = 28 = 24]
1
3x ÷ z = −24 or −2
...
Substitute the values for the variables into the expression
...
Substitute
...
Signs the same? Add the value of the terms and give the
result the same sign
...
Substitute the values for the variables into the
expression
...
Signs different? Multiply the terms and give the
result a negative sign
...
Substitute the equivalent values into the original
expression
...
Give the result the sign of the larger value
...
Substitute the values for the variables into the
expression
...
Signs the same? Multiply the terms and give the
result a positive sign
...
Now evaluate the other term
...
Substitute the equivalent values into the original
expression
...
The simplified value of the expression is as follows:
31
...
PEMDAS: Multiply the terms in the
expression
...
Signs different? Divide and give the result a
negative sign
...
25]
(42) ÷ (−8) = −5
...
25
501 Algebra Questions
32
...
Group terms using order of operations
...
Signs different? Multiply and give the result
a negative sign
...
Signs different? Divide and give the result
a negative sign
...
1
(−5)(6) + (−8) ÷ (2)
1
(−5)(6) + {(−8) ÷ (2)}
[(−5)(6)]
[5 · 6 = 30]
[(−5)(6) = −30]
1
[(−8) ÷ (2)]
1
[8 ÷ 2]
1
2
[8 ÷ 2 = 8 · 1 = 8 · 2 = 16]
1
Substitute the equivalent values into the
original expression
...
The simplified value of the expression
is as follows:
33
...
Evaluate the terms on either side of the
subtraction sign
...
Positive times
negative is negative
...
Change the operator to addition and the sign
of the number that follows
...
Substitute the values for the variables into the
expression
...
Signs the same? Add the value of the terms and
give the result the same sign
...
Next you evaluate the term with the exponent
...
Substitute the equivalent values into the original
expression
...
Substitute the values for the variables into the
expression
...
The first term has an exponent
...
Evaluate the second term
...
Evaluate the first term
...
Signs different? Divide and give the result a
negative sign
...
Substitute the values for the variables into the
expression
...
Evaluate the second term
...
Signs different? Multiply and give the result a
negative sign
...
Change the subtraction to addition and the sign
of the number that follows
...
The simplified value of the expression is as
follows:
17
3(−13)2
[(−13)2 = −13 · −13]
[13 · 13 = +169]
3(169) = 507
a(b + z)2 = 507
1
2((3)2 + 2(2)) ÷ (−5)
1
[(3)2 + 2(2)]
[(3)2 = 3 · 3 = 9]
1
[2(2) = 1]
1
[(3)2 + 2(2) = 9 + 1 = 10]
2(10) ÷ (−5)
[2(10) = 20]
(20) ÷ (−5)
[20 ÷ 5 = 4]
(20) ÷ (−5) = −4
2(a2 + 2y) ÷ b = −4
1
(3)3 + 24(2) − 3(−5)
[(3)3 = 3 · 3 · 3 = 27]
1
[24(2) = 12]
[3(−5)]
[3 · 5 = 15]
[3(−5) = −15]
(27) + (12) − (−15)
(27) + (12) + (+15)
27 + 12 + 15 = 54
a3 + 24y − 3b = +54
501 Algebra Questions
37
...
Evaluate first and last terms
...
Substitute the equivalent values into the original
expression
...
Commutative property of addition allows
grouping of like signs
...
Substitute
...
The simplified value of the expression is as
follows:
38
...
PEMDAS: Evaluate the term with the exponent
first
...
PEMDAS: Evaluate terms with multiplication
next
...
Change the subtraction to addition and the
sign of the number that follows
...
All term signs are
positive, a result of addition +
...
Substitute the values for the variables into the
expression
...
Evaluate the first term by multiplying
...
18
−2(6)
− (−5) + (3)(−8)
[−2(6) = −2 · 6 = −12]
[(3)(−8) = 3 · −8 = −24]
(−12) − (−5) + (−24)
−12
+ (+5) + −24
(−12 + −24) + (+5)
[−12 + −24 = −36]
(−36) + (+5)
[36 − 5 = 31]
(−36) + (+5) = −31
−2x
− b + az = −31
5(−8)2 − 2(−8) + 2
[(−8)2 = (−8)(−8) = +64]
5(+64) − 2(−8) + 2
[5(+64) = 320]
[2(−8) = −16]
320 − (−16) + 2
320 + (+16) + 2
320 + 16 + 2 = 338
5z2 − 2z + 2 = 338
1
5(6)(2) ÷ 2(−5)
1
1
[5 · 6 · 2 = (5 · 6) · 2 = 15]
[2(−5) = −10]
501 Algebra Questions
Substitute the values into the original numerical
expression
...
[15 ÷ 10 = 12]
1
−
(15) ÷ ( 10) = −12
The simplified value of the expression is as
1
follows:
5xy ÷ 2b = −12 or −1
...
Substitute the values for the variables into the
expression
...
Evaluate the second term
...
Change the subtraction to addition and the sign of
the number that follows
...
The simplified value of the expression is as follows:
41
...
First, evaluate the term with the exponent
...
Signs different? Multiply and give the result a
negative sign
...
Substitute the values into the original numerical
expression
...
The simplified value of the expression is as
follows:
12
7(6) + (6) − (−8)
[7(6) = 7 · 6 = 42]
12
[(6) = 12 ÷ 6 = 2]
(42) + (2) − −8
42 + 2 + −8
42 + 2 + +8 = 52
12
7x + x − z = 52
1
2(−5)2 ÷ 2
[2(−5)2 = 2 · (−5)(−5)]
[{2 · (−5)} · (−5)]
[2 · (−5) = −10]
[(−10) · (−5) = +50]
1
(+50) ÷ 2
(+50) · 2 = 100
2b2 ÷ y = 100
42
...
(−5)(6)((−8) + 3)
First, evaluate the expression inside the parentheses
...
Substitute the result into the numerical expression
...
Negative times positive
equals negative
...
The simplified value of the expression is as follows:
19
[(−8) + 3]
[8 − 3 = 5]
[(−8) + 3 = −5]
(−5)(6)(−5)
[−5 · 6 = −30]
(−30) · −5 = +150
bx(z + 3) = +150
501 Algebra Questions
43
...
First evaluate the expression inside the
parentheses
...
Substitute the result into the numerical
expression
...
Evaluate the second term in the expression
...
Signs the same? Add the value of the terms
and give the result the same sign
...
Substitute the values for the variables into the
expression
...
Change the subtraction to addition and the
sign of the number that follows
...
Substitute the results into the numerical
expression
...
Substitute the result into the numerical
expression
...
The simplified value of the expression is as
follows:
20
1
1
6(2)(−8 ÷ 2) + 3(3)(−5)
1
2
[−8 ÷ 2 = −8 · 1 = −16]
1
6(2)(−16) + 3(3)(−5)
1
1
[6(2)(−16) = 6 · 2 · −16]
[3 · −16 = −48]
[3(3)(−5) = 3 · 3 · −5]
[9 · −5 = −45]
(−48) + (−45)
−48
+ −45 = −93
6y(z ÷ y) + 3ab = −93
2(−5)(6) ÷ ((−8) − (−5))
[(−8 − −5)]
[−8 + 5]
[8 − 5 = 3]
[−8 + +5 = −3]
2(−5)(6) ÷ (−3)
[2 · −5 · 6 = −60]
(−60) ÷ −3
[60 ÷ 3 = 20]
(−60) ÷ −3 = 20
2bx ÷ (z − b) = 20
501 Algebra Questions
45
...
Evaluate the first term
...
Signs different? Multiply the numbers and give
the result a negative sign
...
Division by a fraction is the same as multiplication
by its reciprocal
...
1
(−180) ÷ (2)
−180
[180 · 2 = 360]
−180 · 2 = −360
The simplified value of the expression is
as follows:
46
...
Evaluate the expression in the innermost
parentheses
...
Substitute the result into the numerical
expression
...
PEMDAS: Multiply before subtraction
...
Substitute the result into the numerical
expression
...
[2 − 3 = 3 − 3 = 0]
The simplified value of the expression is as
follows:
21
(6)
6
[(2 − 3) = 2 − 3]
6
1
(2){(0) − 4(3)}
[{0 − 4(3)} = 0 − 4 · 3]
[0 − 4 · 3 = 0 − 12]
[0 − 12 = 0 + −12 = −12]
1
1
(2){−12} = 2 · −12
1
[2 · 12 = 6]
1
2
· −12 = −6
x
y{(2 − 3) − 4a} = −6
501 Algebra Questions
47
...
Evaluate the first term
...
Evaluate the second term in the numerical
expression
...
Substitute the results into the numerical
expression
...
Same signs? Add the value of the terms
and give the result the same sign
...
Substitute the values for the variables into
the expression
...
Evaluate the first term
...
Evaluate the second term
...
Subtract
...
Multiply from left to right
...
Substitute the values for the variables into the
expression
...
Signs the same? Multiply and give the result
a positive sign
...
Multiply from left to right
...
Yes, you can just subtract
...
Substitute the values for the variables
into the expression
...
Change subtraction to addition and
the sign of the term that follows
...
PEMDAS: Evaluate terms with
exponents next
...
Multiply from left to right
...
The simplified value of the expression
is as follows:
23
1
(−8)2 − 4(3)2(2)
[(−8)2 = −8 · −8]
[−8 · −8 = 64]
1
1
[4(3)2(2) = 4 · 3 · 3 · 2]
[ 4 · 3 = 12]
[12 · 3 = 36]
1
[36 · 2 = 18]
(64) − (18)
64 − 18 = 46
z2 − 4a2y = 46
3(6)2(−5)(5(3) −3(−5))
[(5(3) − 3(−5)) = 5 · 3 − 3 · −5]
[5 · 3 − 3 · −5 = 15 − −15]
[15 + +15 = 30]
3(6)2(−5)(30)
[(6)2 = 6 · 6 = 36]
3(36)(−5)(30)
[3(36) = 108]
[(108) · (−5) = −540]
[(−540) · (30) = −16,200]
3(6)2(−5)(5(3) − 3(−5)) = −16,200
3x2b(5a −3b) = −16,200
3
Combining Like
Terms
In this chapter, you will practice simplifying algebraic expressions
...
You should know that
■
■
■
the numbers in front of the variable or variables are called
coefficients
...
like terms can have different coefficients, but the configuration
of the variables must be the same for the terms to be alike
...
You can think of an algebraic term as a series of factors with numbers,
and you can think of variables as factors
...
When you have terms that are alike, you can add
or subtract them as if they were signed numbers
...
This
501 Algebra Questions
strategy will continue to be shown in the answer explanations
...
You will also use the important commutative and associative properties of addition and multiplication
...
See the Tips for Combining Like Terms
...
Study the following
general and specific examples
...
Now look at the following examples:
Commutative Property of Addition
a+b=b+a
This equation reminds us that terms being combined by addition can change
their location (commute), but the value of the expression remains the same
...
25
501 Algebra Questions
Associative Property of Addition
(q + r) + s = q + (r + s)
This equation reminds us that when you are performing a series of additions
of terms, you can associate any term with any other and the result will be the
same
...
Identity Property of Addition
n+0=n
Identity Property of Multiplication
n·1=n
Term Equivalents
x=1·x
For purposes of combining like terms, a variable by itself is understood to
mean one of that term
...
a + −b = a − +b = a − b
Adding a negative term is the same as subtracting a positive term
...
Which one looks simpler? Of
course, it’s the last, a − b
...
Writing expressions
as simply as possible is always appreciated
...
26
501 Algebra Questions
Simplify the following expressions by combining like terms
...
5a + 2a + 7a
52
...
4x + 2y − x + 3y
54
...
7h + 6 + 2w − 3 + h
56
...
3(2a + 3b) + 7(a − b)
58
...
64 + 5(n − 8) + 12n − 24
60
...
−7(a
+ b) +12a − 16b
62
...
8s − 3r + 5(2r − s)
64
...
5(15 − 2j ) + 11(7j − 3)
66
...
8(x − 7) + x(2 − x)
68
...
2x − x(5 + y) + 3xy + 4(2x − y)
70
...
5(3x − y) + x(5 + 2y) − 4(3 + x)
72
...
9(2x − t) + 23xt + x(−4 + 5t)
74
...
8(2a − b − 3c) + 3(2a − b) − 4(6 − b)
27
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
51
...
Add like terms
...
Add like terms
...
Use the commutative property of addition to
move like terms together
...
Add like terms
...
The simplified result of the algebraic
expression is:
7a + 3a + 6b
(7a + 3a) + 6b
[(7a + 3a) = 10a]
(10a) + 6b
7a + 6b + 3a = 10a + 6b
53
...
Use the commutative property of
addition to move like terms together
...
Add like terms
...
The simplified algebraic
expression is:
4x + 2y + (−x) + 3y
4x + (−x) + 2y + 3y
(4x + −x) + (2y + 3y)
[4x + −x = +3x = 3x]
[2y + 3y = 5y]
(4x + −x) + (2y + 3y) = (3x) + (5y)
4x + 2y − x + 3y = 3x + 5y
28
501 Algebra Questions
54
...
Use the commutative property
for addition to put like terms
together
...
Add like terms
...
Rewrite addition of a negative
term as subtraction of a positive
term by changing addition to
subtraction and changing the
sign of the following term
...
Change subtraction to addition and change the
sign of the term that follows
...
Use the associative property for addition
...
Substitute the result into the expression
...
Use the distributive property of multiplication
on the first expression
...
Substitute the results into the expression
...
Use the associative property for addition
...
Substitute the results into the expression
...
Use the distributive property of multiplication
on the first term
...
[7(a − b) = 7 · a − 7 · b]
[7a − 7b]
(6a + 9b) + (7a − 7b)
6a + 9b + 7a − 7b
Substitute the results into the expression
...
Use the commutative property for addition
to put like terms together
...
Add like terms
...
Substitute the result into the expression
...
Use the distributive property of
multiplication on the first term
...
Substitute the result into the expression
...
Use the associative property for
addition
...
Substitute the result into the expression
...
Use the distributive property of
multiplication on the second term
...
Parentheses are no longer needed
...
30
[11(4m + 5) = 11 · 4m + 11 · 5]
[44m + 55]
[3(−3m + 8) = 3 · −3m + 3 · 8]
[−9m + 24]
(44m + 55) + (−9m + 24)
44m + 55 + −9m + 24
44m + −9m + 55 + 24
(44m + −9m) + (55 + 24)
[44m + −9m = 35m]
[55 + 24 = 79]
(35m) + (79)
35m + 79
[5(n − 8) = 5 · n − 5 · 8]
[5n − 40]
64 + (5n − 40) + 12n − 24
64 + 5n − 40 + 12n − 24
64 + 5n + −40 + 12n + −24
501 Algebra Questions
Use the commutative property for
addition to put like terms together
...
Add like terms
...
Substitute the results into the
expression
...
Use the distributive property of
multiplication on the first term
...
Substitute the results into the
expression
...
Use the commutative property for
addition to put like terms
together
...
Use the associative property for
addition
...
[4(x + y − 4) = 4 · x + 4 · y − 4 · 4]
[4 · x + 4 · y − 4 · 4 = 4x + 4y − 16]
[6(2 − 3y) = 6 · 2 − 6 · 3y = 12 − 18y]
(4x + 4y − 16) + (12 − 18y)
4x + 4y − 16 + 12 − 18y
4x + 4y + −18y + −16 + 12
4x + (4y + −18y) + (−16 + 12)
[4y + −18y = −14y]
[−16 + 12 = 12 + −16 = −4]
Substitute the results into the
expression
...
The simplified algebraic
expression is:
4x − 14y − 4
31
501 Algebra Questions
61
...
Substitute the results into the expression
...
Use the commutative property for addition
to put like terms together
...
Use the associative property for addition
...
Substitute the results into the expression
...
62
...
14 + 9(2w + 7) + −2(6 + −w)
Use the distributive property of
multiplication on the second term
...
[−2(6 + −w) = −2 · 6 + −2 · −w]
Notice the result of multiplication
for opposite and like-signed terms
...
14 + (18w + 63) + (−12 + +2w)
Parentheses are no longer needed
...
18w + +2w + 14 + 63 + −12
Use the associative property for
addition
...
[18w + +2w = 20w]
Add from left to right
...
(20w) + (65)
Parentheses are no longer needed
...
Change subtraction to addition and change
8s + −3r + 5(2r + −s)
the sign of the terms that follow
...
[5(2r + −s) = 5 · 2r + 5 · −s]
[5 · 2r + 5 · −s = 10r + −5s]
8s + −3r + (10r + −5s)
8s + −3r + 10r + −5s
Substitute the results into the expression
...
Use the commutative property of addition
to put like terms together
...
Add like terms
...
Parentheses are no longer needed
...
Change subtraction to addition and change
the sign of the terms that follow
...
Use the distributive property of
multiplication on the second term
...
Parentheses are no longer needed
...
Use the associative property for
addition
...
Add terms using the rules for terms
with the same signs
...
Parentheses are no longer needed
...
Use the commutative property
of addition
...
Either of the last two expressions is
correct, but the second is the simpler:
33
6(3m + −12) + −4(9m + 8)
[6(3m + −12) = 6 · 3m + 6 · −12]
[6 · 3m + 6 · −12 = 18m + −72]
[−4(9m + 8) = −4 · 9m + −4 · 8]
[−4 · 9m + −4 · 8 = −36m + −32]
(18m + −72) + (−36m + −32)
18m + −72 + −36m + −32
18m + −36m + −72 + −32
(18m + −36m) + (−72 + −32)
[18m + −36m = −18m]
[−72 + −32 = −104]
(−18m) + (−104)
−18m + −104
−18m
− +104
−104
+ −18m
−104
− (+18m)
−104
− 18m
501 Algebra Questions
65
...
5(15 + −2j ) + 11(7j + −3)
Use the distributive property of
multiplication on the first term
...
[11(7j + −3) = 11 · 7j + 11 · −3]
[11 · 7j + 11 · −3 = 77j + −33]
Substitute the results into the
expression
...
75 + −10j + 77j + −33
Use the commutative property of
addition to put like terms together
...
(77j + −10j ) + (75 + −33)
Add terms using the rules for terms
with different signs
...
[75 + −33 = +42]
Substitute the results into the
expression
...
67j + 42
66
...
Use the distributive property of
multiplication on the first term
...
Substitute the results into the expression
...
Use the commutative property of addition
to put like terms together
...
Add the first term using the rules for
terms with the same signs
...
Substitute the results into the expression
...
34
a(a + 4) + 3a2 + −2a + 10
[a(a + 4) = a · a + a · 4]
[a · a + a · 4 = a2 + 4a]
(a2 + 4a) + 3a2 + −2a + 10
a2 + 4a + 3a2 + −2a + 10
a2 + 3a2 + 4a + −2a + 10
(a2 + 3a2) + (4a + −2a) + 10
[a2 + 3a2 = 4a2]
[4a + −2a = 2a]
(4a2) + (2a) + 10
4a2 + 2a + 10
501 Algebra Questions
67
...
Use the distributive property of
multiplication on the first term
...
Substitute the results into the expression
...
Use the commutative property of addition
to put like terms together
...
Add terms in parentheses using the rules
for terms with the same signs
...
Use the commutative property of addition
to put the terms in exponential order
...
Change subtraction to addition and change
the sign of the terms that follow
...
Use the distributive property of
multiplication on the third term
...
Remove the parentheses
...
Use the associative property for addition
...
Add the second term using the rules for
terms with the same signs
...
Remove the parentheses
...
Change subtraction to addition and
the sign of the terms that follow
...
Use the rules for multiplying
signed terms
...
Use the rules for multiplying
signed terms
...
Remove the parentheses
...
Add the first set of terms using the
rules for terms with different signs
...
Substitute the results into the
expression
...
70
...
Use the distributive property of
multiplication on the first term
...
Use the distributive property of
multiplication on the third term
...
Substitute the results into the
last expression
...
Use the commutative property of
addition to move terms together
...
Combine like terms using addition
rules for signed numbers
...
36
2x + −x(5 + y) + 3xy + 4(2x + −y)
[−x(5 + y) = −x · 5 + −x · y]
[−x · 5 + −x · y = −5x + −xy]
[4(2x + −y) = 4 · 2x + 4 · −y]
[4 · 2x + 4 · −y = 8x + −4y]
2x + (−5x + −xy) + 3xy + (8x + −4y)
2x + −5x + −xy + 3xy + 8x + −4y
(2x + −5x + 8x) + (−xy + 3xy) + (−4y)
[2x + −5x + 8x = 5x]
[−xy + 3xy = 2xy]
(−5x) + (2xy) + (−4y)
5x + 2xy −4y
−7(c
+ −2d) + 21c + −3(d + −5)
[−7(c + −2d) = −7 · c + −7 · −2d ]
[−7 · c + −7 · −2d = −7c + +14d ]
[−3(d + −5) = −3 · d + −3 · −5]
[−3 · d + −3 · −5 = −3d + +15]
(−7c + +14d) + 21c + (−3d + +15)
−7c + +14d + 21c + −3d + +15
21c + −7c + +14d + −3d + +15
(21c + −7c) + (+14d + −3d) + +15
(14c) + (11d) + 15
14c + 11d + 15
501 Algebra Questions
71
...
Use the distributive property of
multiplication on the first term
...
Use the distributive property of
multiplication on the second term
...
Use the distributive property of
multiplication on the third term
...
Substitute the results into the
original expression
...
Use the commutative property of
addition to move like terms
together
...
Combine like terms using addition
rules for signed numbers
...
Remove the parentheses
...
Change subtraction to addition
and the sign of the terms
that follow
...
[6(m + −3n) = 6 · m + 6 · −3n]
Use the rules for multiplying
signed terms
...
[3m(n + 5) = 3m · n + 3m · 5]
Use the rules for multiplying
signed terms
...
[−2n(3 + −m) = −2n · 3 + −2n · −m]
Use the rules for multiplying
signed terms
...
(6m + −18n) + (3mn + 15m) + (−6n + +2mn)
Remove the parentheses
...
Use the associative
property for addition
...
(21m) + (5mn) + (−24n)
Adding a negative term is
the same as subtracting a
positive term
...
Change subtraction to addition and the
sign of the terms that follow
...
Use the rules for multiplying
signed terms
...
Use the rules for multiplying
signed terms
...
Remove the parentheses
...
Use the associative property for
addition
...
Adding a negative term is the same as
subtracting a positive term
...
Change subtraction to addition and the sign
of the terms that follow
...
Use the distributive property of
multiplication on the first term
...
Use the distributive property of
multiplication on the second term
...
Substitute the results into the expression
...
Use the commutative property of addition
...
Combine like terms using addition rules
for signed numbers
...
Substitute the results into the expression
...
Substitute into the expression
...
Use the commutative property of addition
...
Add like terms
...
Change subtraction to addition
and the sign of the terms
that follow
...
Use the rules for multiplying
signed terms
...
Use the rules for multiplying
signed terms
...
Use the rules for multiplying
signed terms
...
Remove the parentheses
...
Use the associative property
for addition
...
Adding a negative term
is the same as subtracting a
positive term
...
An equation is a mathematical statement where two
expressions are set equal to each other
...
Simply put, that
is what you have done when you have the variable on one side of the equal
sign and a number on the other
...
There will be
different solutions to similar problems to show a variety of methods for
solving equations
...
Look over the Tips
for Solving Basic Equations before you begin this chapter’s questions
...
The inverse of addition is subtraction
...
If you add or subtract an amount from
one side of the equation, you must do the same to the other
side to maintain the equality
...
If a variable is being
multiplied by a coefficient, you can eliminate the coefficient by
dividing both sides of the equation by that coefficient, leaving you with
just one of the variables
...
The inverse of division is multiplication
...
If b is a number ≠ 0, then b(b) = y
...
Just
remember that you must do it to both sides
...
76
...
x − 25 = 32
78
...
b − 15 = 71
80
...
s − −4 = −1
3
1
82
...
b − 2 = −3
1
84
...
m + 2(5 − 24) = −76
86
...
4x = −20
88
...
27b = 9
90
...
0
...
8
x
92
...
−4 = −12
2
94
...
5b = −56
42
501 Algebra Questions
7
96
...
This was 8 the suggested selling price of
the car
...
After putting 324 teddy bears into packing crates, there were 54 crates
filled with bears
...
Only 3% of turtle hatchlings will live to become breeding adults
...
This year, a farmer planted 300 acres of corn
...
5 times as many
acres as he planted last year
...
A business executive received a $6,000 bonus check from her company at
the end of the year
...
How much was her
annual salary before receiving the bonus?
43
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
76
...
Associate like terms
...
Zero is the identity element for addition
...
Add 25 to each side of the equation
...
Associate like terms
...
Zero is the identity element for addition
...
Subtract 17 from both sides of the equation
...
Change subtraction to addition and
change the sign of the term that follows
...
Zero is the identity element for addition
...
Change subtraction to addition and change
the sign of the term that follows
...
Associate like terms
...
Zero is the identity element for addition
...
Change subtraction to addition and change
the sign of the term that follows
...
Associate like terms
...
Associate like terms
...
Zero is the identity element for addition
...
Change subtraction to addition and change the
sign of the term that follows
...
Associate like terms
...
Subtracting zero is the same as adding zero
...
Add −14 to each side of the equation
...
Apply the rules for operating with
signed numbers
...
a = 42
83
...
5
1
a + (0) = 42
1
5
2
5
5
b + −2 = −3
2
5
Subtract −2 from both sides of the equation
...
Change subtraction to addition and change
the sign of the term that follows
...
Change the improper fraction to a
mixed number
...
Change subtraction to addition and change
the sign of the term that follows
...
Perform the operation in parentheses
...
Subtract 18 from both sides of the equation
...
Perform the operation in parentheses
...
1
c + −4(2 + −5) = 20
1
c + (−4 · 2 + −4 · −5) = 20
c + (−2 + +20) = 20
c + (18) = 20
c + 18 − 18 = 20 − 18
c + (18 − 18) = 20 − 18
c+0=2
c=2
85
...
m + (2 · 5 − 2 · 24) = −76
The order of operations is to multiply first
...
Associate like terms
...
m + (10 − 48) = −76
m + −38 + +38 = −76 + +38
m + (−38 + +38) = −76 + +38
m + (0) = −38
m = −38
86
...
Apply the rules for operating with signed numbers
...
Divide both sides of the equation by 4
...
Another look for this solution method is as follows:
88
...
Apply the rules for operating with signed numbers
...
Divide both sides of the equation by 27
...
90
...
Reduce fractions to their simplest form (common factor
of 15)
...
2c
0
...
Divide both sides of the equation by 0
...
5
...
2
5
...
2
c = 29
Divide
...
Multiply both sides of the equation by 7
...
93
...
−4
y
· −4 = −4 · −12
y = −4 · −12
Signs the same? Multiply and give the result a
positive sign
...
Divide both sides of the equation by 3
...
3
a = 54 · 2
a = 81
8
8
b
5
8
8
÷ 5 = −56 ÷ 5
8
b = −56 ÷ 5
95
...
Dividing by a fraction is the same as multiplying
by its reciprocal
...
Here’s one
...
Let x = the suggested selling price of the car
...
So your equation is:
Divide both sides of the equation by
7
8
...
47
7
8x
7
8x
= $21,000
7
7
÷ 8 = 21,000 ÷ 8
7
x = 21,000 ÷ 8
8
x = 21,000 · 7
x = $24,000
501 Algebra Questions
97
...
The first sentence tells you that the number of
packing crates times the number of bears in each
is equal to the total number of bears
...
Divide
...
Let t = the number of turtle hatchlings born
...
Three percent of the
turtles born is 1,200
...
03,
so the equation becomes
0
...
Divide both sides of the equation by 0
...
0
...
03 = 1,200 ÷ 0
...
03
Divide
...
Let c = the number of acres he planted last year
...
5 times c is 300
...
5
...
1
...
5c ÷ 1
...
5
c = 300 ÷ 1
...
Let d = her annual salary
...
Your
equation will be:
The numerical equivalent of 5% is
0
...
05
...
48
(5%)d = $6,000
0
...
0
...
05 = 6,000 ÷ 0
...
05
d = $120,000
5
Solving Multi-Step
Equations
Solving multi-step equations simply combines the work you have done in
the previous chapters
...
Tips for Solving Multi-Step Equations
■
■
■
There are at least two ways to show multiplication
...
In equations, this becomes confusing
...
Both
conventions will be used in the answers, so you should get used
to either one
...
Or it can be shown using a fraction
10
bar like this: 10 ÷ 2 = 2 = 5
...
Check your answers before looking at the answer solutions
...
If the
quantities you find are equal, your solution is correct
...
For
2
3
example, 9 ÷ 3 = 9(2)
...
Then write the equation based on the
information stated in the problem
...
101
...
13x + 21 = 60
103
...
5x − 6 = −26
x
105
...
7 − 5 = 1
107
...
4 = 4a + 20
109
...
0
...
25 = 1
2
111
...
9 = 4m − 3
113
...
4m − 14 = 50
2m
115
...
7m − 6 = −2
...
10s − 6 = 0
s
118
...
7 = 3
119
...
−55
= 25 − s
50
501 Algebra Questions
Solve the following word problems by letting a variable equal the unknown quantity, making an equation from the information given, and then solving the equation
...
A farmer is raising a hog that weighed 20 lbs
...
He
expects it to gain 12 pounds per month
...
How many months will it be before he will sell the animal?
122
...
50 less than twice Bill’s hourly wage
...
50 per
hour
...
At year’s end, a share of stock in Axon Corporation was worth $37
...
What
was the price of a share of Axon stock at the beginning of the year?
124
...
5 times her former salary by changing
jobs
...
What was her salary at her
previous employment?
2
125
...
If the number of girls participating
is 105, how many boys participate?
51
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
101
...
Associate like terms
...
Zero is the identity element for addition
...
102
...
Associate like terms
...
Zero is the identity element for addition
...
103
...
Change subtraction to addition and change
the sign of the term that follows
...
Perform numerical operations
...
Divide both sides of the equation by 3
...
Add 6 to each side of the equation
...
Associate like terms
...
Zero is the identity element for addition
...
52
4x + 7 − 7 = 11 − 7
4x + (7 − 7) = (11 − 7)
4x + (0) = (4)
4x = 4
4x ÷ 4 = 4 ÷ 4
x=1
13x + 21 − 21 = 60 − 21
13x + (21 − 21) = (60 − 21)
13x + (0) = (39)
13x = 39
13x ÷ 13 = 39 ÷ 13
x=3
3x − 8 + 8 = 16 + 8
3x + −8 + 8 = 16 + 8
3x + (−8 + 8) = 16 + 8
3x + (0) = 24
3x = 24
3x ÷ 3 = 24 ÷ 3
x=8
5x − 6 + 6 = −26 + 6
5x + −6 + 6 = −26 + 6
5x + (−6 + 6) = −26 + 6
5x + (0) = −20
5x = −20
5x ÷ 5 = −20 ÷ 5
x = −4
501 Algebra Questions
x
3
x
3
x
3
x
3
105
...
Associate like terms
...
106
...
Change subtraction to addition and change the
sign of the term that follows
...
Perform numerical operations
...
Multiply both sides of the equation by 7
...
Associate like terms
...
Zero is the identity element for addition
...
+ (4 − 4) = 10 − 4
=6
x
3(3) = 3(6)
x = 18
Zero is the identity element for addition
...
107
...
+ 4 − 4 = 10 − 4
39 + 9 = 3a − 9 + 9
39 + 9 = 3a + (−9 + 9)
39 + 9 = 3a + (−9 + 9)
48 = 3a + (0)
48 = 3a
48 ÷ 3 = 3a ÷ 3
16 = a
Associate like terms
...
Zero is the identity element for addition
...
−16
4
108
...
4a
= 4
−4 = a
Associate like terms
...
Zero is the identity element for addition
...
10a
10
109
...
53
2
= 10
1
a = 0
...
Subtract 0
...
Associate like terms
...
Zero is the identity element for addition
...
3a + 0
...
25 = 1 − 0
...
3a + (0
...
25) = 1 − 0
...
3a + (0) = 0
...
3a = 0
...
3
...
0
...
3
0
...
3
a = 2
...
Subtract 8 from both sides of the equation
...
Perform numerical operations
...
Multiply both sides of the equation by
2
the reciprocal of 3
...
Add 3 to both sides of the equation
...
9 + 3 = 4m + −3 + 3
Associate like terms
...
12 = 4m + (0)
Zero is the identity element for addition
...
4 3
= 3(4m)
16 = m
113
...
Associate like terms
...
Zero is the identity element for addition
...
Divide both sides of the equation by
Use the rules for operating with
signed numbers
...
114
...
Change subtraction to addition and change
the sign of the term that follows
...
54
41 − 41 − 2m = 65 − 41
(41 − 41) − 2m = 65 − 41
(0) − 2m = 24
−2m = 24
−2m
−2m
−2
= 24
= 2−24
m = −12
4m − 14 + 14 = 50 + 14
4m + −14 + 14 = 50 + 14
4m + (−14 + 14) = 50 + 14
501 Algebra Questions
Perform numerical operations
...
4m + (0) = 64
4m = 64
Divide both sides of the equation by 4
...
This equation presents a slightly different look
...
There are two methods for solving
...
Associate like terms
...
Zero is the identity element for addition
...
Use rules for multiplying whole numbers
and fractions
...
2m
2
Or you can recognize that
Then you would multiply by the reciprocal
of the coefficient
...
Add 6 to each side of the equation
...
5 2
()m
2 5
5
= (2)8
m = 20
7m − 6 + 6 = −2
...
Associate like terms
...
7m + −6 + 6 = −2
...
5 + 6
7m + (0) = 3
...
7m
7
117
...
Change subtraction to addition and change
the sign of the term that follows
...
Perform numerical operations
...
Express the answer in the simplest form
...
5
= 7
m = 0
...
6
501 Algebra Questions
s
4
s
4
s
4
118
...
7 from both sides of the equation
...
Perform numerical operations
...
7 − 2
...
7
+ (2
...
7) = 3 − 2
...
3
s
4(4) = 4(0
...
2
Multiply both sides of the equation by 4
...
Add 7 to each side of the equation
...
Associate like terms
...
8s + −7 + 7 = 41 + 7
8s + (−7 + 7) = 41 + 7
8s + (0) = 48
8s
8
48
= 8
s=6
Divide both sides of the equation by 8
...
−55
− 25 = 25 − 25 − s
Change subtraction to addition and change
−55 + −25 = 25 + −25 + −s
the sign of the term that follows
...
(−55 + −25) = (25 + −25) + −s
−80 = (0) + −s
Perform numerical operations
...
You are to solve for s, but the term remaining
is −s
...
Use the rules for operating with signed
numbers
...
Subtract 25 from both sides of the equation
...
Let x = the number of months
...
An equation that represents
these words would be
Subtract 20 from both sides of the equation
...
Perform numerical operations
...
12x + 20 − 20 = 200 − 20
12x + (20 − 20) = 200 − 20
12x + (0) = 180
12x
12
180
= 1
2
x = 15
The farmer would have to wait 15 months before selling his hog
...
122
...
Then 2x less
$1
...
The equation representing the last
statement would be
Add 1
...
56
2x − 1
...
50
...
50 + 1
...
50 + 1
...
2x = 14
...
2x
2
14
...
00
Bill’s hourly wage is $7
...
123
...
The statements tell us that if we multiply the share
price at the beginning of the year by 3 and then
subtract $8, it will equal $37
...
Perform numerical operations
...
3x − 8 + 8 = 37 + 8
3x = 45
3x
3
45
= 3
x = 15
One share of Axon costs $15 at the beginning of the year
...
124
...
The
statements tell us that $64,000 is
equal to 1
...
An algebraic equation to
represent this statement is
64,000 = 1
...
Subtract 4,000 from both sides
of the equation
...
5x + 4,000 − 4,000
Perform numerical operations
...
5x
Divide both sides of the
60,000
1
...
5
...
5
1
...
125
...
The question tells us
2
that 3 the number of boys plus 25 is equal
to the number of girls who participate
...
Perform numerical operations
...
2
x + 25 = 105
...
57
105 − 25
6
Solving Equations
with Variables on
Both Sides of an
Equation
If you have been solving the problems in this book with some success,
you will move easily into this chapter
...
Then check your answers with the solutions provided
...
There is often more than one way to
find a solution
...
Tips for Solving Equations with
Variables on Both Sides of the Equation
Use the distributive property of multiplication to expand and separate
terms
...
501 Algebra Questions
a(b − c) = ab − ac
−1(b + c) = −b − c
−1(b − c) = −b + c
−(b − c) = −b + c
The object is to isolate the variable on one side of the equation
...
There
are two instances that sometimes occur when solving equations
...
Stated another way, the solution is the null set—that is, a set containing no elements
...
In that instance, any value can make the equation true; therefore, there are an infinite number of solutions
...
Find the solutions for the following equations
...
11x + 7 = 3x − 9
127
...
5x + 3 + 6x = 10x + 9 − x
129
...
20x − 5 − 5x = 11x + 49
131
...
4 + 3x − 0
...
15 − 2x
132
...
2x = 10 − 0
...
2 + 6x − 0
...
1
134
...
3 + 5x − 0
...
2 − 3x
135
...
8x = 3
...
8x − 9
...
7(x + 2) + 1 = 3(x + 14) − 4x
137
...
13 − 8(x − 2) = 7(x + 4) + 46
139
...
2(2x + 19) − 9x = 9(13 − x) + 21
141
...
2x + 15x = 1 + 3x
5
143
...
6(2x + 2) = 3(x + 1)
145
...
7(0
...
3(3 − 0
...
10(x + 2) + 7(1 − x) = 3(x + 9)
147
...
5(2x + 3) − 9 = 14(x + 1)
149
...
0
...
5 = 0
...
9x
60
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
11x + 7 − 7 = 3x − 9 − 7
11x + (0) = 3x − 16
11x = 3x − 16
11x − 3x = 3x − 3x − 16
8x = −16
126
...
Simplify
...
Subtract 3x from both sides of the equation
...
8x
8
−16
= 8
1x = −2
x = −2
Divide both sides of the equation by 8
...
Solution
...
Substitute −2 for x
in the original equation
...
11(−2) + 7 = 3(−2) − 9
−22 + 7 = −6 − 9
−15 = −15
The result is a true statement, so this answer is a correct solution
...
Add 23 to both sides of the equation
...
Identity property of 0 for addition
...
Simplify
...
Simplify
...
Use the commutative property of addition with
like terms
...
Subtract 3 from both sides
...
Now subtract 9x from both sides of the
equation
...
of addition with like terms
...
Add 19 to both sides of the
equation
...
Subtract 3x from both sides
of the equation to isolate the
variable on one side of the
equation
...
Divide both sides of the
equation by 2
...
Let’s check this answer
...
Simplify by multiplying
factors
...
Add and subtract from left
to right
...
Simplify
...
Use the commutative property
5x + 6x + 3 = 10x − x + 9
10x − 5x + 27 − 46 = 3x + 32 − 19
5x − 19 = 3x + 13
5x + 19 − 19 = 3x + 19 + 13
5x = 3x + 32
5x − 3x = 3x − 3x + 32
2x = 32
2x
2
32
= 2
x = 16
10(16) + 27 − 5(16) − 46 = 32 + 3(16) − 19
(160) + 27 − (80) − 46 = 32 + (48) − 19
187 − 80 − 46 = 80 − 19
107 − 46 = 61
61 = 61
The previous statement is true; therefore, the solution is correct
...
Use the commutative property to move
20x − 3x − 11 = 9x + 43
like terms
...
Add 11 to both sides of the equation
...
Subtract 9x from both sides of the
equation
...
17x − 11 = 9x + 43
17x + 11 − 11 = 9x + 43 + 11
17x = 9x + 54
17x − 9x = 9x − 9x + 54
8x = 54
8x
8
131
...
Combine like terms on each side of
the equation
...
15 from both sides of
the equation
...
Identity property of addition
...
Simplify
...
54
= 8
x = 6
...
Simplify the expression
...
4 − 0
...
15 − 2x
3x + 0
...
15 − 2x
3x + 0
...
15 = 1
...
15 − 2x
3x + (0) = 1 − 2x
3x = 1 − 2x
3x + 2x = 1 − 2x + 2x
5x = 1
5x
5
1
= 5
1
x = 5
Simplify the expression
...
Use the commutative property with
like terms
...
Subtract 17 from both sides of
the equation
...
Add 0
...
Combine like terms on each side of
the equation
...
63
2x − 1
...
2x
0
...
2x
0
...
2x
0
...
2x
0
...
2x = 4 − 0
...
2x
1x = 4
x=4
501 Algebra Questions
133
...
6x + 2 − 0
...
1
Combine like terms and simplify the
expression
...
8 = 5x + 2
...
8 from both sides of the
equation
...
8 − 1
...
1 − 1
...
6x = 5x + 0
...
6x − 5x = 5x − 5x + 0
...
x = 0
...
Instead of adding 1
...
1
from both sides of the equation
...
8 − 2
...
1 − 2
...
6x − 0
...
6x − 6x − 0
...
3 = −x
equation
...
3
−1 = −1
the variable x, which is −1
...
Simplify
...
3 = x
The answer will be the same
...
134
...
Associate like terms on each side
of the equation
...
Remove the parentheses
...
Associate like terms
...
Subtract 0
...
Combine like terms on each side
of the equation
...
Divide both sides of the equation
by 5
...
64
3x + 0
...
25 = 1
...
4 − 0
...
15 − 2x
3x + (0
...
15 − 2x
3x + 0
...
15 − 2x
3x + 2x + 0
...
15 + 2x − 2x
(3x + 2x) + 0
...
15 + (2x − 2x)
5x + 0
...
15
5x + 0
...
15 = 1
...
15
5x + 0 = 1
5x = 1
5x
5
1
= 5
x = 0
...
Use the commutative property with
like terms
...
Simplify the expression
...
Associate like terms on each side
of the equation
...
Identity property of addition
...
8x to both sides of the
equation
...
Divide both sides of the equation
by 3
...
136
...
8x + 12 = 3
...
4 − 0
...
8x) + 12 = (3
...
4) − 0
...
2x + 12 = −6 − 0
...
2x + 12 − 12 = −6 − 12 − 0
...
2x + (12 − 12) = (−6 − 12) − 0
...
2x + (0) = (−18) − 0
...
2x = −18 − 0
...
2x + 0
...
8x − 0
...
Simplify the expression
...
Combine like terms on each side
of the equation
...
Combine like terms on each side of
the equation
...
Combine like terms on each side of
the equation
...
7(x) + 7(2) + 1 = 3(x) + 3(14) − 4x
7x + 14 + 1 = 3x + 42 − 4x
Simplify the expression
...
Use the distributive property of
4(4x) + 4(3) = 6x − 28
16x + 12 = 6x − 28
multiplication
...
Subtract 12 from both sides of the
equation
...
Subtract 6x from both sides of the
equation
...
16x + 12 − 12 = 6x − 28 − 12
16x = 6x − 40
16x − 6x = 6x − 6x − 40
10x = −40
10x
10
138
...
Simplify the expression
...
Combine like terms on each side of
the equation
...
Combine like terms on each side of
the equation
...
Combine like terms on each side
of the equation
...
Simplify the expression
...
Use the distributive property of
multiplication
...
Use the commutative property
with like terms
...
Add 11x to both sides of the
equation
...
Subtract 9 from both sides of the
equation
...
Simplify the expression
...
Divide both sides of the equation
by 21
...
140
...
Simplify the expression
...
Combine like terms on each side
of the equation
...
Combine like terms on each side
of the equation
...
Combine like terms on each
side of the equation
...
Simplify the expression
...
Simplify the expression
...
21x = −21
21x
21
−21
= 21
−
x= 1
2(2x) + 2(19) − 9x = 9(13) − 9(x) + 21
4x + 38 − 9x = 117 − 9x + 21
38 + 4x − 9x = 117 + 21 − 9x
38 − 5x = 138 − 9x
38 − 38 − 5x = 138 − 38 − 9x
−5x
= 100 − 9x
−5x
+ 9x = 100 − 9x + 9x
4x = 100
4x
4
100
= 4
x = 25
2(2(25) + 19) − 9(25) = 9(13 − (25)) + 21
2(50 + 19) − 225 = 9(−12) + 21
2(69) − 225 = −108 + 21
138 − 225 = −87
−87 = −87
The solution is correct
...
Use the distributive property of
multiplication
...
Combine like terms on each side
of the equation
...
Simplify the expression
...
Simplify the expression
...
Simplify the expression
...
Now let’s check the answer by
substituting the solution into the
original equation
...
Change whole numbers to
fractional equivalents
...
4
4
− 3 = −3 + 16
Add 43 to both sides of the
48
4
4
4
4
+ − = − + 16
equation
...
16 = 16
A true statement, so this solution is correct
...
Simplify the equation by adding like terms
...
Combine like terms on each side of the
equation
...
Remember that division by a fraction is the
same as multiplication by the reciprocal
5
of the fraction
...
Simplify the expression
...
Use the distributive property
on both sides
...
4
35x = 1 + 3x
4
35x − 3x = 1 + 3x − 3x
4
x
5
=1
5 4
(x)
4 5
5
x = 4
5
= 4(1)
5
5
(x) − (2) + 3x = 3(x) + 3(2)
2
2
5
x − 5 + 3x = 3x + 6 − 10
2
68
− 10
501 Algebra Questions
Use the commutative
property with like terms
...
In this case,
that would be a 2
...
Simplify the expressions
...
Add 10 to both sides of the
equation
...
Subtract 6x from both sides
of the equation
...
Divide both sides of the
equation by 5
...
5
x
2
+ 3x − 5 = 3x + 6 − 10
5
2(2x + 3x − 5) = 2(3x + 6 − 10)
5
2(2x) + 2(3x) − 2(5) = 2(3x) + 2(6) − 2(10)
5x + 6x − 10 = 6x + 12 − 20
11x − 10 = 6x − 8
11x − 10 + 10 = 6x − 8 + 10
11x = 6x + 2
11x − 6x = 6x − 6x + 2
5x = 2
5x
5
2
= 5
2
x = 5
144
...
Simplify the expression
...
Simplify the expression
...
3
3
Simplify the expression
...
145
...
Simplify the expressions
...
06x to both sides of
the equation
...
Add 0
...
Combine like terms on each
side of the equation
...
2
...
0
...
2x) − 0
...
3(3) − 0
...
2x)
0
...
7 = 0
...
06x
0
...
06x − 0
...
9 − 0
...
06x
0
...
7 = 0
...
2x − 0
...
7 = 0
...
7
0
...
6
0
...
2
1
...
2
x=8
69
501 Algebra Questions
146
...
10(x) + 10(2) + 7(1) − 7(x) = 3(x) + 3(9)
Simplify the expression
...
10x − 7x + 20 + 7 = 3x + 27
Combine like terms on each
side of the equation
...
3x + 27 − 27 = 3x + 27 − 27
Simplify the expression
...
x=x
The solutions for this equation are infinite
...
Use the distributive property of
multiplication
...
36 − 4x = 2x − 6x − 36
Combine like terms
...
36 + 4x − 4x = 4x − 4x − 36
Combine like terms on each side of
the equation
...
There is no solution for this equation
...
148
...
Simplify the expressions
...
Subtract 9x from both sides of the
equation
...
Subtract 14 from both sides of the
equation
...
Divide both sides of the equation by 4
...
70
5(2x) + 5(3) − 9 = 14(x) + 14
10x + 15 − 9 = 14x + 14
10x + 6 = 14x + 14
10x − 10x + 6 = 14x − 10x + 14
6 = 4x + 14
6 − 14 = 4x + 14 − 14
−8
−8
4
= 4x
4x
= 4
−2 = x
501 Algebra Questions
149
...
Simplify the expressions
...
Subtract 40 from both sides of
the equation
...
Subtract 11x from both sides of
the equation
...
Divide both sides of the equation
by −4
...
7(x) − 7(10) + 110 = 4(x) − 4(25) + 7x
7x − 70 + 110 = 4x − 100 + 7x
7x + 40 = 11x − 100
7x + 40 − 40 = 11x − 100 − 40
7x = 11x − 140
7x − 11x = 11x − 11x − 140
−4x
= −140
−4x
−
4
−140
= −4
x = 35
150
...
In this equation, many of the
terms are in decimal form
...
10[0
...
5] = 10[0
...
9x]
Use the distributive property
of multiplication
...
8(x + 20)] − 10[4
...
7(5 + x)] − 10[0
...
8(x + 20) − 45 = 7(5 + x) − 9x
Use the distributive
property again
...
8x + 160 − 45 = 35 + 7x − 9x
Combine like terms on
each side of the equation
...
8x + 115 − 115 = 35 − 115 − 2x
Combine like terms on each
side of the equation
...
8x + 2x = −80 − 2x + 2x
Combine like terms on each
side of the equation
...
10
10
−
Simplify the expression
...
Algebra is a useful skill to know
when faced with problems in these areas
...
These word problems require you to find an unknown value in
a formula
...
Tips for Using Formulas to Solve Equations
Given a formula with several variables, you will generally be given values
for all but one
...
It can be helpful to list each variable with its given value
...
Keep in mind the rules for order of operations
...
14) where r = radius and h = height
Surface area of a cylinder: S = 2πr (r + h) where r = radius of the base, h = height
of cylinder (let π = 3
...
Find the volume of a rectangular solid whose length is 12 cm, width is 8
cm, and height is 3 cm
...
A rectangular container has a volume of 98 ft3
...
5 ft, what is its height?
153
...
A flight from
Miami to Aruba takes 3
...
How far is it from Miami to Aruba?
154
...
How long would the trip take if they averaged 17 miles per hour
for the trip?
155
...
They planned to hike 6 hours per day and wanted to complete a
trail chapter that was 60 miles long
...
At an interest rate of 6%, how much interest would $12,000 earn over
2 years?
157
...
What was the interest rate?
158
...
How long will it take for a $3,000 savings account to double its value at a
simple interest rate of 10%?
160
...
What would be the height of a trapezoidal building if, at its base, it
measured 80 feet, its roofline measured 40 feet, and the surface area of one
side was 7,200 ft2?
162
...
What are the lengths of the bases, if the lower base is three times the
length of the upper?
73
501 Algebra Questions
163
...
What would that temperature be on the Fahrenheit scale?
164
...
He is
going to a country where the temperature is reported in Celsius
...
)
165
...
What
is its volume in cubic feet?
166
...
12 cubic centimeters
and whose height is 12 centimeters
...
The volume of a cylindrical aquarium tank is 13,565 cubic feet
...
The radius of the base of a cylinder is 20 cm
...
What is its surface area?
169
...
6 square inches
and whose radius is 3 inches?
170
...
The height of the
cylinder is three times the radius of the base of the cylinder
...
74
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
151
...
List the values for the variables
...
Simplify the expression
...
152
...
List the values for the variables
...
Simplify the expression
...
5
...
Include the units
...
Write the applicable formula
...
Substitute the values for the variables
...
Include the units
...
5
h=?
98 = (7)(3
...
5h
98
24
...
5h
=
24
...
5
D = (350)(3
...
Write the applicable formula
...
Substitute the values for the variables
...
Divide both sides of the equation by 17
...
Include the units
...
Write the applicable formula
...
Calculate the total number of hours
...
Simplify the expression
...
Simplify the expression
...
5 = r
Include the units
...
5 miles per hour = r
156
...
List the values for the variables
...
Simplify the expression
...
157
...
List the values for the variables
...
Simplify the expression
...
Simplify the expression
...
158
...
I = prt
I=?
p = 12,000
r = 6%
t=2
I = (12,000)(0
...
76
13,500r
=
13,5
00
0
...
Simplify the expression
...
04)(2)
1,000 = 0
...
08
...
Include the units
...
08
159
...
0
...
08
12,500 = p
$12,500 = p
I = prt
To double its value, the account would have
to earn $3,000 in interest
...
Substitute the given values into the formula
...
Divide both sides of the equation by 300
...
Include the units
...
10)t
3,000 = 300t
3,000
300
300t
=
300
10 = t
10 years = t
1
A = 2h(b1 + b2)
A=?
b1 = 14
b2 = 18
h=8
160
...
List the values for the variables
...
A = 2 · 8 · (14 + 18)
Simplify the expression
...
A = 2 · 256
A = 128
A = 128 cm2
1
1
1
161
...
List the values for the variables
...
A = 2h(b1 + b2)
A = 7,200
b1 = 40
b2 = 80
h=?
1
7,200 = 2 · h · (40 + 80)
1
Simplify the expression
...
Simplify the expression
...
7,200
6
0
77
60h
= 6
0
120 = h
120 ft = h
501 Algebra Questions
1
A = 2h(b1 + b2)
A = 240
b1 = x
b2 = 3x
h = 12
162
...
List the values for the variables
...
1
240 = 2 · 12 · (x + 3x)
1
Simplify the expression
...
Simplify the expression
...
240
24
24x
= 2
4
10 = x
b1 = 10
b2 = 3(10)
b2 = 30
5
C = 9(F − 32)
C = 40
F=?
163
...
List the values for the variables
...
Multiply both sides of the equation by 9
...
Add 160 to both sides of the equation
...
Divide both sides of the equation by 5
...
5
40 = 9(F − 32)
5
9(40) = 9[9(F − 32)]
360 = 5(F − 32)
360 = 5F − 5(32)
360 = 5F − 160
360 + 160 = 5F + 160 − 160
520 = 5F
520
5
5F
= 5
104 = F
5
C = 9(F − 32)
C=?
F = 50
164
...
List the values for the variables
...
C = 9(50 − 32)
Simplify the expression
...
Write the applicable formula
...
Substitute the given values into the formula
...
Include the units
...
Write the applicable formula
...
Substitute the given values into the formula
...
Divide both sides of the equation by 37
...
Simplify the expression
...
167
...
List the values for the variables
...
Simplify the expression
...
16
...
Include the units
...
Write the applicable formula
...
Substitute the given values into the formula
...
Include the units
...
14
V = (3
...
24
V = 50
...
12
r=?
h = 12
π = 3
...
12 = (3
...
12 = 37
...
12
37
...
68r2
=
37
...
14
13,565 = (3
...
16h
13,565
452
...
16h
=
452
...
14
S = 2(3
...
Write the applicable formula
...
6
r=3
h=?
π = 3
...
Substitute the given values into
the formula
...
Use the distributive property
of multiplication
...
Subtract 56
...
Simplify the expression
...
84
...
Include the units
...
6 = 2(3
...
6 = 18
...
6 = 18
...
84(h)
282
...
52 + 18
...
6 − 56
...
52 − 56
...
84
226
...
84h
226
...
84
8
...
8
4
12 = h
12 in = h
170
...
S = 2πr (r + h)
S = 2,512
r=x
h = 3x
π = 3
...
Substitute the given values into
the formula
...
Divide both sides of the
equation by 25
...
Simplify the expression
...
Substitute the value for x into
the values list
...
14)(x)(x + 3x)
2,512 = 6
...
12
25
...
12
2
100 = x
10 = x
r x = 10 feet
=
h = 3x = 3(10) = 30 feet
80
8
Graphing Linear
Equations
This chapter asks you to find solutions to linear equations by graphing
...
Every point on the line is a solution for the equation
...
You select a value for
x and solve for the y value
...
The slope and y-intercept method may require you to change an equation into the slope-intercept form
...
Written in this form, the m value
is a number that represents the slope of the solution graph and the b is a
number that represents the y-intercept
...
From one point to another, the slope is the rise
over the run
...
Another way of saying that is: The y-intercept is the place
where the value of x is 0
...
Use the b value to determine where the line crosses the y-axis
...
501 Algebra Questions
■
■
Use the value of m as the slope of the equation
...
If the value of m is a whole number, the slope is the whole
change in y
number over 1
...
If the value of m is negative, use a negative sign in only the numerator
–3
3
3
or the denominator, not both
...
Graph the following equations using the slope and y-intercept method
...
y = 2x + 3
172
...
y = −2x + 9
3
174
...
y = 2x − 3
176
...
y + 3x = −2
1
1
178
...
2x + 5y = 30
180
...
x − 3y = 12
182
...
−5x
7
− y = −2
184
...
0 = 3x + 2y
186
...
y − 0
...
3 y − 2x = 0
5
1
189
...
7x = 4y + 8
191
...
6y + 13x = 12
193
...
1x = 0
...
4
82
501 Algebra Questions
194
...
6y + 27x = −42
For the following problems, use the slope/y-intercept method to write an equation
that would enable you to draw a graphic solution for each problem
...
A glider has a 25:1 descent ratio when there are no updrafts to raise its
altitude
...
Write an equation to represent the glider’s descent
from an altitude of 250 feet
...
An Internet service provider charges $15 plus $0
...
Write an equation that would represent the monthly bill of a user
...
A scooter rental agency charges $20 per day plus $0
...
Write an equation to represent the cost of one
day’s rental
...
A dive resort rents scuba equipment at a weekly rate of $150 per week and
charges $8 per tank of compressed air used during the week of diving
...
200
...
Write an equation to represent
this ratio
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
Underlined expressions show the original algebraic expression as an equation with the expression
equal to its simplified result
...
The equation is in the proper
2
change in y
m = 2 = 1 =
change in x
The y-intercept is at the point (0,3)
...
b = 3
...
y
(1,5)
(0,3)
x
84
501 Algebra Questions
172
...
slope/y-intercept form
...
A change in y of 5 and in x of 1
gives the point
(0 + 1,−2 + 5) or (1,3)
...
The equation is in the proper
2
change in y
m = −2 = −1 =
chang
e in x
The y-intercept is at the point (0,9)
...
b = 9
...
y
(0,9)
(1,7)
x
86
501 Algebra Questions
174
...
slope/y-intercept form
...
A change in y of 3 and in x of 4
gives the point
(0 + 4,−1 + 3) or (4,2)
...
The equation is in the proper
5
change in y
m = 2 =
chang
e in x
The y-intercept is at the point (0,−3)
...
b = −3
...
y
(2,2)
x
(0,−3)
88
501 Algebra Questions
176
...
Add 2x to both sides of the
equation
...
The equation is in the proper
slope/y-intercept form
...
A change in y of 2 and in x of 1
gives the point
y + 2x − 2x = 2x + 4
y = 2x + 4
2
change in y
m = 2 = 1 =
chang
e in x
The y-intercept is at the point (0,4)
...
y
(1,6)
(0,4)
x
89
501 Algebra Questions
177
...
Subtract 3x from both sides
of the equation
...
The equation is in the proper
slope/y-intercept form
...
A change in y of −3 and in x of 1
gives the point
y + 3x − 3x = −3x − 2
y = −3x − 2
3
change in y
m = −3 = −1 =
chang
e in x
The y-intercept is at the point (0,−2)
...
y
x
(0,−2)
(1,−5)
90
501 Algebra Questions
178
...
1
Add 2x to both sides of the
equation
...
The equation is in the proper
slope/y-intercept form
...
A change in y of 1 and in x of 2
gives the point
change in y
1
m = 2 =
chang
e in x
1
The y-intercept is at the point (0,32)
...
y
(0,3 )
(2,4 )
x
91
501 Algebra Questions
179
...
Subtract 2x from both sides
of the equation
...
Divide both sides of the
equation by 5
...
−2
y = 5x + 6
The equation is in the proper
slope/y-intercept form
...
A change in y of −2 and in x of 5
gives the point
−2
change in y
m = 5 =
chang
e in x
The y-intercept is at the point (0,6)
...
y
(0,6)
(5,4)
x
92
501 Algebra Questions
180
...
Subtract 4x from both sides of
the equation
...
Divide both sides of the equation
by 2
...
The equation is in the proper
slope/y-intercept form
...
A change in y of −2 and in x of 1
gives the point
2
change in y
m = −2 = −1 =
chang
e in x
The y-intercept is at the point (0,5)
...
y
(0,5)
(1,3)
x
93
501 Algebra Questions
181
...
Add 3y to both sides
of the equation
...
Subtract 12 from both
sides of the equation
...
Divide both sides of the
equation by 3
...
x
3
(1)(x)
1
x
x
3 − 4 = y
1
x − 4 = y
3
1
=
(3)(
1) = 3 · 1 = 3x
The equation is equivalent
to the proper form
...
b = −4
...
(0 + 3,−4 + 1) or (3,−3)
...
Put the equation in the proper form
...
(3x + 9y)
3
3x
9y
+
3
3
−27
= 3
= −9
x + 3y = −9
Simplify the equation
...
Simplify the equation
...
x − x + 3y = −x − 9
3y = −x − 9
3y
3
(−x − 9)
= 3
−x
9
−x
−9
y = 3 − 3
Simplify the equation
...
b = −3
...
(0 + 3,−3 − 1) or (3,−4)
...
Put the equation in the proper form
...
7
5x − 5x − y = 5x − 2
7
Simplify the equation
...
−y
Simplify the equation
...
y = −5x + 2
7
= 5x − 2
−1(−y)
7
= −1(5x − 2)
7
5
change in y
m = −5 = −1 =
chang
e in x
1
1
b = 2 = 32
...
1
1
(0 + 1,32 − 5) or (1,−12)
...
Put the equation in the proper form
...
Simplify the equation
...
Simplify the equation
...
b = 2
...
(0 + 7,2 + 1) or (7,3)
...
Put the equation in the proper form
...
Simplify the equation
...
0 − 2y = 3x + 2y − 2y
−2y = 3x
−2y
−
2
3x
= −2
−3
y = 2x
Simplify the equation
...
There is no b showing in the
equation, so b = 0
...
(0 + 2,0 − 3) or (2,−3)
...
Put the equation in the proper form
...
(3x + 12y)
3
3x
12y
+
3
3
−18
= 3
= −6
x + 4y = −6
Simplify the equation
...
Simplify the equation
...
x − x + 4y = − x − 6
4y = −x − 6
4y
4
(−x − 6)
= 4
−x
6
y = 4 − 4
Simplify the equation
...
−1
change in y
m = 4 =
chang
e in x
1
b = −32
...
1
1
(0 + 4,−12 − 1) or (4,−22)
...
Put the equation in the proper form
...
6x to both sides of the
equation
...
The equation is in the proper
slope/y-intercept form
...
A change in y of 3 and in x of 5
gives the point
y + 0
...
6x = 0
...
6x − 2
6
3
change in y
m = 0
...
(0 + 5,−2 + 3) or (5,1)
...
Put the equation in the proper form
...
2
y
3
2
y
3
Simplify the equation
...
1
1
1
− 2x + 2x = 0 + 2x
1
= 2x
3 2
3 1
()y = (x)
2 3
2 2
3
1y = 4x
3
y = 4x
Simplify the equation
...
b = 0
...
(0 + 4,0 + 3) or (4,3)
...
Simplify the equation
...
So, if you multiply the whole
equation by the lowest common multiple of the denominators, you
will have whole numbers with coefficients
...
6(6x − 3y) = 6(2)
Use the distributive
5
1
property of multiplication
...
5x − 2y = 12
Subtract 5x from both
sides of the equation
...
Divide both sides of the
−2y
−5x
12
− = − + −2
equation by −2
...
The equation is in the proper
slope/y-intercept form
...
A change in y of 5 and in x of 2
gives the point
5
change in y
m = 2 =
chang
e in x
The y-intercept is at the point (0,−6)
...
y
x
(2,−1)
(0,−6)
102
501 Algebra Questions
190
...
Subtract 8 from both sides
of the equation
...
Exchange the terms on each
side of the equal sign
...
7x − 8 = 4y + 8 − 8
7x − 8 = 4y
4y = 7x − 8
4y
4
7x
8
= 4 − 4
7
y = 4x − 2
Simplify the equation
...
b = −2
...
(0 + 4,−2 + 7) or (4,5)
...
Exchange the terms on each side
5y = 20x − 15
of the equal sign
...
Simplify the equation
...
b = −3
...
(0 + 1,−3 + 4) or (1,1)
...
Put the equation in the proper form
...
Simplify the equation
...
6y + 13x − 13x = −13x + 12
6y = −13x + 12
6y
6
−13x
12
= 6 + 6
−13
y = 6x + 2
Simplify the equation
...
b = 2
...
(0 + 6,2 − 13) or (6,−11)
...
Once again, if it would be easier for you to operate with whole number
coefficients instead of decimals to start, you could multiply the whole
equation by 10
...
10(0
...
7y + 1
...
x = 7y + 14
Subtract 14 from both
sides of the equation
...
x − 14 = 7y
If a = b, then b = a
...
7
7
1
y = 7x − 2
Simplify the equation
...
b = −2
...
(0 + 7,−2 + 1) or (7,−1)
...
Exchange the terms on each side
17y = −34x + 85
of the equal sign
...
Simplify the equation
...
b = 5
...
(0 + 1,5 − 2) or (1,3)
...
Put the equation in the proper form
...
Simplify the equation
...
6y + 27x − 27x = −27x − 42
6y = −27x − 42
6y
6
–27x
42
= 6 − 6
−27
y = 6x − 7
Simplify the equation
...
The equation is in the proper
slope/y-intercept form
...
A change in y of 9 and in x of −2
gives the point
−9
y = 2x − 7
−9
9
change in y
m = 2 = −2 =
chang
e in x
The y-intercept is at the point (0,−7)
...
y
(−2,2)
x
(0,−7)
196
...
Forward is in the positive direction
...
Ascending is in the positive direction
...
The change in position of the glider is described by the slope
...
Slope =
chang
e in x =
25
108
501 Algebra Questions
The starting position for the purposes
of this graphic solution is at an altitude
of 250 ft or +250
...
A graph of this equation would have a slope of
−1 and the y-intercept would be at
25
b = 250
−1
y = 25x + 250
(0,250)
197
...
Let x = the hours of Internet use for the month
...
25 times the number of hours of use
...
25 or
The y-intercept would be at
y = 0
...
Let y = the cost of a scooter rental for one day
...
The problem tells us that the cost would be equal
to the daily charges plus the 0
...
Written as an equation, this would be
The graph would have a y-intercept at (0,20)
and the slope would be
y = 0
...
Let y = the total cost for equipment
...
The problem tells us that the cost would be equal to
the weekly charge for gear rental plus 8 times the
number of tanks used
...
y = 8x + 150
200
...
Let x = the number of chickadees that visited the feeder
...
109
y = 7x
9
Solving Inequalities
If you compare Chapter 9 to Chapter 6, you will find only a few differences between solving inequalities and solving equalities
...
Your goal is to isolate the variable on one side of the inequality, and the result will be your solution
...
Look at the following tips to see what makes solving inequalities different from solving equalities
...
Keep the symbol facing the same way when you
multiply or divide by a positive factor
...
When you have fractional coefficients or terms in an equality
or inequality, multiply both sides by the least common multiple
of the denominators and work with whole numbers
...
Solve the following inequalities
...
3x + 2 < 11
202
...
5x ≤ 18
204
...
8 − 6x < 50
206
...
2x + 0
...
79
208
...
3(5 − 4x) < x − 63
210
...
2(7x − 3) ≥ −2(5 + 3x)
212
...
−x
0
...
3x − 5 > x − 2
215
...
−4x
217
...
2x + 0
...
9 + x
1
2
219
...
−7(x
+ 3) < −4x
5
1
221
...
3(1 − 3x) ≥ −3 (x + 27)
223
...
11(1 − x) ≥ 3(3 − x) − 1
225
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
Underlined inequalities show the simplified result
...
Subtract 2 from both sides of the inequality
...
3x + 2 − 2 < 11 − 2
3x < 9
3x
3
9
< 3
x<3
Divide both sides of the inequality by 3
...
4x − 6 + 6 > 30 + 6
4x > 36
202
...
Simplify the inequality
...
Simplify
...
Multiply both sides of the inequality by 2
...
5 2
5
()x ≤ (18)
2 5
2
10
5 18
(10 )x ≤ 2(1)
(1)x ≤ 45
x ≤ 45
204
...
Simplify
...
Simplify the inequality
...
Subtract 8 from both sides of the inequality
...
Divide both sides of the inequality by −6 and
change the direction of the inequality sign
...
112
4x + 26 − 26 ≥ 90 − 26
4x ≥ 64
4x
4
64
≥ 4
x ≥ 16
8 − 8 − 6x < 50 − 8
−6x < 42
−6x
−
6
42
> −6
x > −7
501 Algebra Questions
206
...
Combine like terms on each side of
the inequality
...
5x − 9 + 9 ≤ −2 + 9
5x ≤ 7
5x
5
7
≤ 5
7
x ≤ 5
Simplify
...
Subtract 0
...
Combine like terms on each side of
the inequality
...
Simplify the inequality
...
29 − 0
...
79 − 0
...
50
2x
2
0
...
25
208
...
Simplify the expressions
...
Simplify
...
Use the distributive property of
multiplication
...
Add 12x to both sides of the inequality
...
Add 63 to both sides of the
inequality
...
Divide both sides of the inequality
by 13
...
113
−6(x + 1)
−
6
x+1≤
60
≤ −6
−10
x + 1 − 1 ≤ −10 − 1
x ≤ −11
3(5) − 3(4x) < x − 63
15 − 12x < x − 63
15 − 12x + 12x < x + 12x − 63
15 < 13x − 63
15 + 63 < 13x − 63 + 63
78 < 13x
78
13
13x
< 1
3
6
501 Algebra Questions
210
...
Simplify
...
Combine like terms on each side
of the inequality
...
Simplify the inequality
...
Simplify
...
Use the distributive property of
4(x) + 4(1) < 5(x) + 5(2)
4x + 4 < 5x + 10
4x + 4 − 4 < 5x + 10 − 4
4x < 5x + 6
4x − 5x < 5x − 5x + 6
−x < 6
−1(−x)
> −1(6)
x > −6
multiplication
...
Add 6x to both sides of the inequality
...
Add 6 to both sides of the inequality
...
Divide both sides of the inequality
by 20
...
x ≥ 20
Reduce the fraction to lowest terms
...
Use the distributive property of
multiplication
...
Add 1 to both sides of the inequality
...
Add 4x to both sides of the
inequality
...
Divide both sides of the
inequality by 20
...
213
...
3(0
...
3(20)
inequality by 0
...
Simplify the expressions
on both sides
...
Simplify the expressions
...
Add 5 to both sides of the inequality
...
Subtract 1x from both sides of the
inequality
...
Multiply both sides of the inequality
by 3
...
≥ −1(6)
> 3 (3)
x>9
215
...
Simplify
...
Combine like terms on each side of
the inequality
...
Combine like terms
...
Simplify
...
Use the distributive property of
multiplication
...
Combine like terms
...
Add x to both sides of the inequality
...
Subtract 6 from both sides of the
inequality
...
3x + 5 ≥ −2(x) − 2(10)
3x + 5 ≥ −2x − 20
3x + 5 − 5 ≥ −2x − 20 − 5
3x ≥ −2x − 25
3x + 2x ≥ 2x − 2x − 25
5x ≥ −25
≥ 5
x≥−5
−4x
+ 3(x) + 3(5) ≥ 3(x) + 3(2)
+ 3x + 15 ≥ 3x + 6
−
( 4x + 3x) + 15 ≥ 3x + 6
–x + 15 ≥ 3x + 6
x − x + 15 ≥ x + 3x + 6
15 ≥ 4x + 6
−4x
15 − 6 ≥ 4x + 6 − 6
9 ≥ 4x
Simplify
...
1
24
Divide both sides of the inequality by 4
...
You can simplify equations (and inequalities)
with fractions by multiplying them by a
common multiple of the denominators
...
Use the distributive property
of multiplication
...
Use the distributive property of
multiplication
...
Add 3 to both sides of the inequality
...
Add 3x to both sides of the equation
...
Divide both sides of the inequality by 7
...
Simplify the expressions
...
Subtract 0
...
Combine like terms on each side of the
inequality
...
Simplify
...
Simplify the expressions
...
Change the term to an improper fraction
...
Use the distributive property of
multiplication
...
Add 13 to both sides of the inequality
...
Subtract 2x from both sides of the
inequality
...
116
−3
3
4(x − 4) < 4(4(x + 2))
−3
3
4(x) − 4(4) < 4(4)(x + 2)
4x − 3 < −3(x + 2)
4x − 3 < −3(x) − 3(2)
4x − 3 < −3x − 6
4x − 3 + 3 < −3x − 6 + 3
4x < − 3x − 3
3x + 4x < 3x − 3x − 3
7x < −3
7x
7
−3
< 7
−3
x < 7
3
x
2
+ 0
...
1 ≥ 0
...
1 + x
3
x
2
≥ 0
...
8 +
2
1
x ≥ 0
...
8)
x−x
x ≥ 1
...
Use the distributive property of
multiplication
...
Add 21 to both sides of the inequality
...
Add 4x to both sides of the inequality
...
Divide both sides of the inequality by −3
and change the direction of the
inequality sign
...
221
...
Simplify the terms
...
1
Subtract 2x from both sides of the
inequality
...
Subtract 5 from both sides of the
inequality
...
4
Multiply both sides by 3
3
(the reciprocal of 4)
...
−7(x)
− 7(3) < −4x
− 21 < −4x
−7x − 21 + 21 < −4x + 21
−7x < −4x + 21
−7x + 4x < −4x + 4x + 21
−3x < 21
−7x
−3x
−
3
21
> −3
x > −7
5
5
1
1
(x) + (4) > (x) + (8)
4
4
2
2
5
1
x + 5 > x + 4 − 8
4
2
5
1
x + 5 > x − 4
4
2
1
1
−8
5
x
4
3
x
4
− 2x + 5 > 2x − 2x − 4
1
3
x
4
3
x
4
+5−5>4−5
+ 5 > −4
> −9
4 3
(x)
3 4
4
> 3(−9)
x > −12
222
...
Simplify terms
...
Combine like terms
...
Combine like terms
...
Simplify
...
Remove the inner brackets, use the
commutative property of addition and
combine like terms
...
Simplify terms
...
Combine like terms
...
Combine like terms
...
Simplify terms
...
Use the distributive property of
− 26 ≥ 26 − 26 + 3x
≥ 3x
multiplication
...
Use the commutative property
...
Subtract 8 from both sides of
the inequality
...
Add 11x to both sides
...
11(1) − 11(x) ≥ 3(3) − 3(x) − 1
11 − 11x ≥ 9 − 3x − 1
11 − 11x ≥ 9 − 1 − 3x
11 − 11x ≥ 8 − 3x
Divide both sides by 8
...
225
...
Simplify terms
...
Simplify terms
...
Combine like terms
...
Combine like terms
...
When there is only one variable, you use a number line
...
Tips for Graphing Inequalities
When using a number line to show the solution graph for an inequality, use
a solid circle on the number line as the endpoint when the inequality symbol is ≤ or ≥
...
A solid circle shows that the solution graph includes the
endpoint; an open circle shows that the solution graph does not include the
endpoint
...
Use the skills you have been practicing in the previous chapters to
transform the inequality into the slope/y-intercept form you used to graph
equalities with two variables
...
Draw
a solid line when the inequality symbol is ≤ or ≥
...
To complete the graph, shade the region
above the boundary line when the inequality symbol is > or ≥
...
501 Algebra Questions
A simple way to check your graphic solution is to pick a point on either side of
the boundary line and substitute the x and y values in your inequality
...
An
easy point to use, if it is not your y-intercept, is the origin (0,0)
...
226
...
x < −1
228
...
x < 5
230
...
(Use graph paper
...
y < x + 1
232
...
y < 4x − 5
1
234
...
2y − 3x < 8
236
...
3x − 4 ≥ 2y
3
238
...
0
...
x − y ≤ 7
y
2
241
...
−3y
+ 9x ≤ −6
243
...
5x > 0
...
9
244
...
3y + 4x < 9 − 2x
246
...
9y + 7 ≥ 2(x + 8)
x
248
...
2( y + 3) − x ≥ 6(1 − x)
250
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
The answers to these questions are the graphs
...
−10
−5
0
5
10
227
...
−10
−5
0
5
10
229
...
−10
−5
0
5
10
121
501 Algebra Questions
231
...
slope/y-intercept form
...
A change in y of 1 and in x of 1
gives the point
(0 + 1,1 + 1) or (1,2)
...
y
(1,2)
(0,1)
x
122
501 Algebra Questions
232
...
slope/y-intercept form
...
A change in y of −1 and in x of 1
gives the point
(0 + 1,2 − 1) or (1,1)
...
y
(0,2)
(1,1)
x
123
501 Algebra Questions
233
...
slope/y-intercept form
...
A change in y of 4 and in x of 1
gives the point
(0 + 1,−5 + 4) or (1,−1)
...
y
x
(1,−1)
(0,−5)
124
501 Algebra Questions
1
234
...
1
1
− 2x + y ≤ −2x + 3
1
y ≤ −2x + 3
Combine like terms
...
b = 3
...
(0 + 2,3 − 1) or (2,2)
...
y
(0,3)
(2,2)
x
125
501 Algebra Questions
235
...
Combine like terms
...
2y − 3x + 3x < 3x + 8
2y < 3x + 8
Simplify terms
...
b = 4
...
(0 + 2,4 + 3) or (2,7)
...
y
(2,7)
(0,4)
x
126
501 Algebra Questions
236
...
Combine like terms
...
3
change in y
m = 3 = 1 =
chang
e in x
b = 3
...
(0 + 1,3 + 3) or (1,6)
...
y
(1,6)
(0,3)
x
127
501 Algebra Questions
237
...
But for an inequality, the
direction of the inequality
symbol must change when
you change sides of the statement
...
Rewrite the
inequality with sides exchanged
and the symbol reversed
...
2
2 − 2
3
y ≤ 2x − 2
Simplify terms
...
3
change in y
m = 2 =
chang
e in x
b = −2
...
(0 + 2,−2 + 3) or (2,1)
...
y
(2,1)
x
(0,−2)
128
501 Algebra Questions
238
...
Combine like terms
...
Use the distributive property
of multiplication
...
The inequality is in the proper
slope/y-intercept form
...
A change in y of 4 and in x of 1
gives the point
The y-intercept is at the point (0,−8)
...
Draw a solid boundary line and shade above it
...
Subtract 3 from both sides
0
...
Combine like terms on each
side of the inequality
...
Combine like terms
...
5
...
5y − x > −3
0
...
5y > x − 3
0
...
5
3
> 0x
...
5
3
x
Simplify the expressions
...
5 − 0
...
The inequality is in the proper
slope/y-intercept form
...
A change in y of 2 and in x of 1
gives the point
The y-intercept is at the point (0,−6)
...
Draw a dotted boundary line and shade above it
...
Subtract x from both sides of the
inequality
...
Simplify the expression
...
Use the distributive property of
multiplication
...
Use the commutative property
of addition
...
b = −7
...
(0 + 1,−7 + 1) or (1,−6)
...
y
x
(1,−6)
(0,−7)
131
501 Algebra Questions
241
...
Use the distributive property of
multiplication
...
Use the commutative property
of addition
...
y
2
y
2
3(3) < 3(3 − x)
3(3) < 3(3) − 3(x)
y < 2 − 3x
y < −3x + 2
−3
change in y
m = −3 = 1 =
chang
e in x
b = 2
...
(0 + 1,2 − 3) or (1,−1)
...
y
(0,2)
x
(1,−1)
132
501 Algebra Questions
242
...
Combine like terms
...
−3y
−3y
+ 9x − 9x ≤ −6 − 9x
≤ −6 − 9x
−3y
−
3
≥ −63 − −3
Simplify the terms
...
The inequality is in the proper
slope/y-intercept form
...
A change in y of 3 and in x of 1
gives the point
The y-intercept is at the point (0,2)
...
Draw a solid boundary line and shade above it
...
Subtract 0
...
Combine like terms
...
5x from both sides
of the inequality
...
Divide both sides of the inequality
by −0
...
Simplify the expression
...
Subtracting a negative number is
the same as adding a positive
...
b = 3
...
5x − 0
...
3y − 0
...
9
0
...
3y > −0
...
5x − 0
...
3y > −0
...
9
−0
...
5x − 0
...
3y
0
...
3 < 0
...
3
–0
...
5x
0
...
3 < (
...
3 )
5
y < 3x − (−3)
5
y < 3x + 3
5
change in y
m = 3 =
chang
e in x
The y-intercept is at the point (0,3)
...
Draw a dotted boundary line and shade below it
...
Subtract 3x from both sides
of the inequality
...
Subtract y from both sides of the
inequality
...
Divide both sides of the inequality
by −2 and change the direction
of the inequality symbol
...
Simplify
...
−2
change in y
m = −2 = 1 =
change in x
b = 4
...
(0 + 1,4 − 2) or (1,2)
...
y
(0,4)
(1,2)
x
135
501 Algebra Questions
245
...
Combine like terms
...
3y + 4x − 4x < 9 − 2x − 4x
3y < 9 − 6x
Simplify the expressions
...
y < 3 − 2x
Use the commutative property
...
y < −2x + 3
3y
3
6x
< 93 − 3
6x
9
–2
change in y
m = 1 =
chang
e in x
b = 3
...
(0 + 1,3 − 2) = (1,1)
...
y
(0,3)
(1,1)
x
136
501 Algebra Questions
246
...
Add 3y to both sides of the
inequality
...
Add 12 to both sides of the
inequality
...
Divide both sides of the
inequality by 3
...
The inequality is in the proper
slope/y-intercept form
...
A change in y of −1 and in x of 1
gives the point
≤ −3x − 3y
3y − 12 ≤ −3x − 3y + 3y
3y − 12 ≤ −3x
3y − 12 + 12 ≤ −3x + 12
3y ≤ −3x + 12
3y
3
−
12
≤ 33x + 3
y ≤ –x + 4
−1
change in y
m = −1 = 1 =
chang
e in x
The y-intercept is at the point (0,4)
...
Draw a solid boundary line and shade below it
...
Use the distributive property
9y + 7 ≥ 2x + 16
of multiplication
...
Combine like terms
...
9y + 7 − 7 ≥ 2x + 16 − 7
9y ≥ 2x + 9
9y
9
9
≥ 29x + 9
2
y ≥ 9x + 1
Simplify the expressions
...
b = 1
...
(0 + 9,1 + 2) or (9,3)
...
y
(9,3)
(0,1)
x
138
501 Algebra Questions
248
...
Use the distributive property of
multiplication
...
Subtract x from both sides of
the inequality
...
Divide both sides of the
inequality by 3
...
The inequality is in the proper
slope/y-intercept form
...
A change in y of 8 and in x of 3
gives the point
8
change in y
m = 3 =
chang
e in x
The y-intercept is at the point (0,−5)
...
Draw a solid boundary line and shade below it
...
Use the distributive property
of multiplication
...
Add x to both sides of the
inequality
...
Subtract 6 from both sides of
the inequality
...
Combine like terms
...
2( y) + 2(3) − x ≥ 6(1) − 6(x)
2y + 6 − x ≥ 6 − 6x
2y + 6 − x + x ≥ 6 − 6x + x
2y + 6 ≥ 6 − 5x
2y + 6 − 6 ≥ 6 − 5x − 6
2y + 6 − 6 ≥ 6 − 6 − 5x
2y ≥ −5x
2y
2
−5x
≥ 2
−5
y ≥ 2x
Simplify terms
...
b = 0
...
(0 + 2,0 − 5) or (2,−5)
...
y
(0,0)
x
(2,−5)
140
501 Algebra Questions
250
...
Simplify terms
...
Combine like terms on each side
of the inequality
...
−28y
−28y
≥ 2x − 14( y) − 14(10)
≥ 2x − 14y − 140
−28y
+ 14y ≥ 2x − 14y + 14y − 140
−14y
≥ 2x − 140
Simplify the terms
...
b = 10
...
(0 + 7,10 − 1) or (7,9)
...
y
(0,10)
(7,9)
x
141
11
Graphing Systems
of Linear Equations
and Inequalities
This chapter will present 15 systems of equalities and ten systems of
inequalities as practice in finding solutions graphically
...
Graphing systems of linear equations on the same coordinate plane will
give you a solution that is common to both equations
...
The solution will be all the points on the line graph because
the equations coincide
...
In this case, the lines are parallel and
will not intersect
...
The graphic solution will either be the common areas of the graphs of the inequalities or
there will be no solution if the areas do not overlap
...
For equations, graph the lines and look for the point or points of intersection
...
For inequalities, graph the boundaries as the appropriate dotted or solid line and
shade the area for each inequality depending upon the inequality symbol present
...
When multiplying or dividing by a negative term, change the direction of the
inequality symbol for each operation
...
251
...
2y − x = 2
3x + y = 8
253
...
y − x = 5 − x
−4y
= 8 − 7x
255
...
2x + y = 4
3( y + 9) = 7x
257
...
4x − 5y = 5
5y = 20 − x
259
...
15y = 6(3x + 15)
y = 6(1 − x)
261
...
3(2x + 3y) = 63
27y = 9(x − 6)
143
501 Algebra Questions
263
...
3x + 4y = 12
6
y = 3 − 8x
265
...
266
...
6y < 5x − 30
2y < −x + 4
268
...
5y ≤ 8(x + 5)
5y ≤ 12(5 − x)
270
...
3y ≥ −2(x + 3)
3y ≤ 2(6 − x)
272
...
7( y − 5) < −5x
−3
1
< 4(2x − 3y)
7
274
...
5x − 2( y + 10) ≤ 0
2x + y ≤ −3
144
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
The graph is shown
...
Transform equations into
slope/y-intercept form
...
b = 4
...
y=x+4
1
change in y
m = 1 = 1 =
chang
e in x
The y-intercept is at the point (0,4)
...
b = 2
...
−1
change in y
m = −1 = 1 =
chang
e in x
The y-intercept is at the point (0,2)
...
y
(1,5)
(–1,3)
(1,1)
x
146
501 Algebra Questions
252
...
Add x to both sides
...
2y − x = 2
2y − x + x = x + 2
2y = x + 2
Divide both sides by 2
...
b = 1
...
y = 2x + 1
1
1
change in y
m = 2 =
chang
e in x
The y-intercept is at the point (0,1)
...
Simplify
...
b = 8
...
3x + y = 8
3x − 3x + y = −3x + 8
y = −3x + 8
−3
change in y
m = −3 = 1 =
chang
e in x
The y-intercept is at the point (0,8)
...
y
(0,8)
(1,5)
(0,1)
(2,2)
x
147
501 Algebra Questions
253
...
Use the distributive property
of multiplication
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go down
7 spaces and right 4 for (4,−14)
...
************
4y = x + 4
1
y = 4x + 1
Divide both sides by 4
...
b = 1
...
1
change in y
m = 4 =
chang
e in x
The y-intercept is at the point (0,1)
...
y
(4,2)
(0,1)
x
(−4,0)
(0,−7)
(4,−14)
148
501 Algebra Questions
254
...
Add x to both sides
...
The graph is a line parallel to
the x-axis through (0,5)
...
7
Simplify terms
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go up
7 spaces and right 4 for (4,5)
...
The solution is (4,5)
...
Transform equations into
slope/y-intercept form
...
The equation is in the proper
slope/y-intercept form
...
b = 7
...
6
−6
change in y
m = 2 = −2 =
chang
e in x
The y-intercept is at the point (0,7)
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go up
1 space and right 4 for (4,−3)
...
The solution for the system of equations is (−4,−5)
...
Transform equations into
slope/y-intercept form
...
Combine like terms on each side
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go down
2 spaces and right 1 for (1,2)
...
************
3( y + 9) = 7x
Use the distributive property of
multiplication
...
Simplify
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go up
7 spaces and right 3 for (3,−2)
...
The solution for the system of equations is (3,−2)
...
Transform equations into
slope/y-intercept form
...
b = 9
...
y=x+9
1
change in y
m = 1 = 1 =
chang
e in x
The y-intercept is at the point (0,9)
...
1
y = −4x + 4
Divide both sides by 4
...
b = 4
...
−1
change in y
m = 4 =
chang
e in x
The y-intercept is at the point (0,4)
...
y
(1,10)
(0,9)
(−4,5)
(0,4)
(4,3)
x
152
501 Algebra Questions
258
...
Subtract 4x from both sides
...
Use the commutative property
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go up
4 spaces and right 5 for (5,3)
...
************
5y = 20 − x
1
y = 4 − 5x
Divide both sides by 5
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go down
1 space and right 5 for (5,3)
...
The solution for the system of equations is (5,3)
...
Transform equations into
6y = 9(x − 6)
slope/y-intercept form
...
6y = 9x − 54
9
y = 6x − 9
Divide both sides by 6
...
b = −9
...
9
change in y
m = 6 =
chang
e in x
The y-intercept is at the point (0,−9)
...
Subtract 15x from both sides
...
6y + 15x = −6
6y + 15x − 15x = −15x − 6
6y = −15x − 6
Divide both sides by 6
...
b = −1
...
−15
y = 6x − 1
−15
−5
change in y
m = 6 = 2 =
chang
e in x
The y-intercept is at the point (0,−1)
...
y
x
(0,−1)
(2,−6)
(0,−9)
154
(6,0)
501 Algebra Questions
260
...
Use the distributive property
of multiplication
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go down
6 spaces and left 5 for (−5,0)
...
************
y = 6(1 − x)
Use the distributive property
of multiplication
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go down
6 spaces and right 1 for (1,0)
...
The solution for the system of equations is (0,6)
...
Transform equations into
slope/y-intercept form
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go up
2 spaces and right 1 for (1,4)
...
************
Use the distributive property
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go up
2 spaces and right 1 for (1,−8)
...
The slopes are the same, so the line graphs are parallel and do not
intersect
...
Transform equations into
slope/y-intercept form
...
Subtract 6x from both sides
...
Use the commutative property
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go down
2 spaces and right 3 for (3,5)
...
************
27y = 9(x − 6)
Use the distributive property
of multiplication
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go up
1 space and right 3 for (3,−1)
...
The solution for the system of equations is (9,1)
...
Transform equations into
slope/y-intercept form
...
x − 20 = 5y
5y = x − 20
Divide both sides by 5
...
b = −4
...
y = 5x − 4
1
change in y
1
m = 5 =
chang
e in x
The y-intercept is at the point (0,−4)
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go up
4 spaces and right 5 for (5,6)
...
The solution for the system of equations is (−10,−6)
...
Transform equations into
slope/y-intercept form
...
Simplify
...
The equation is in the proper
slope/y-intercept form
...
The slope tells you to go down
3 spaces and right 4 for (4,0)
...
************
6
y = 3 − 8x
Use the commutative property
...
b = 3
...
−6
y = 8x + 3
−6
change in y
m = 8 =
chang
e in x
The y-intercept is at the point (0,3)
...
y
(0,3)
(4,0)
x
(8,−3)
159
501 Algebra Questions
265
...
Use the distributive property
of multiplication
...
The equation is in the proper
slope/y-intercept form
...
}
b = −5
...
************
10
y = 16 x − 5
10
5
−5
change in y
m = 16 = 8 = −8 =
chang
e in x
The y-intercept is at the point (0,−5)
...
Simplify
...
The equation is in the proper
slope/y-intercept form
...
}
b = 7
...
17
y = 8x + 7
17
−17
The y-intercept is at the point (0,7)
...
y
(0,7)
(8,0)
(0,−5)
(−8,−10)
160
change in y
m = 8 = −8 =
chang
e in x
x
501 Algebra Questions
266
...
Add 3x to both sides
...
Divide both sides by 2
...
−3
...
Use a solid line for the border
and shade above the line
because the symbol is ≥
...
−5
change in y
m = 2 =
chang
e in x
{Use the negatives to keep the
coordinates near the origin
...
The slope tells you to go down
5 spaces and right 2 for (2,0)
...
The y-intercept is at the point (0,5)
...
y
(0,5)
(2,0)
(0,−3)
161
x
501 Algebra Questions
267
...
5
y < 6x − 5
Divide both sides by 6
...
b = −5
...
Use a dotted line for the border and
shade below it because the symbol is <
...
−1
change in y
m = 2 =
chang
e in x
The y-intercept is at the point (0,2)
...
The slope tells you to go down
1 space and right 2 for (2,1)
...
The solution for the system of inequalities is the double-shaded area on
the graph
...
Transform equations into
y−x≥6
y−x+x≥x+6
y≥x+6
slope/y-intercept form
...
Simplify
...
b = 6
...
Use a solid line for the border and
shade above it because the symbol is ≥
...
Simplify
...
y ≥ 11x − 2
−2
−2
change in y
m = 11 =
chang
e in x
The y-intercept is at the point (0,−2)
...
The slope tells you to go down
2 spaces and right 11 for (11,−4)
...
The solution for the system of inequalities is the double-shaded area
on the graph
...
Transform equations into
slope/y-intercept form
...
3x + 4y = 12
Divide both sides by 5
...
}
b = 8
...
The slope tells you to go down
8 spaces and left 5 for (−5,0)
...
************
5y ≤ 12(5 − x)
Use the distributive property
of multiplication
...
5y ≤ 60 − 12x
5y ≤ −12x + 60
−12
y ≤ 5x + 12
Divide both sides by 5
...
b = 12
...
Use a solid line for the border and shade
below the line because the symbol is ≤
...
(0,12)
y
(0,8)
(−5,0)
(5,0)
x
164
501 Algebra Questions
270
...
Use the distributive property
of multiplication
...
Simplify the inequality
...
y > 10 x + 3
3
3
change in y
m = 10 =
chang
e in x
The y-intercept is at the point (0,3)
...
The slope tells you to go up
3 spaces and right 10 for (10,6)
...
************
4x + y < 4x + 5
Subtract 4x from both sides
...
y < 0x + 5
With a slope of zero, the line is
parallel to the x-axis
...
Use a dotted line for the border and
shade below it because the symbol is <
...
y
(10,6)
(0,5)
(0,3)
x
165
501 Algebra Questions
271
...
Use the distributive property
of multiplication
...
y ≥ 3x − 2
−2
−2
change in y
m = 3 =
chang
e in x
The y-intercept is at the point (0,−2)
...
The slope tells you to go down
2 spaces and right 3 for (3,−4)
...
************
Use the distributive property
of multiplication
...
3y ≤ 12 − 2x
3y ≤ −2x + 12
Divide both sides by 3
...
}
b = 4
...
The slope tells you to go down
2 spaces and right 3 for (3,2)
...
The solution for the system of inequalities is the double-shaded area
on the graph
...
Transform equations into
9( y − 4) < 4x
slope/y-intercept form
...
Add 36 to both sides
...
y < 9x + 4
4
4
change in y
m = 9 =
chang
e in x
The y-intercept is at the point (0,4)
...
The slope tells you to go up
4 spaces and right 9 for (9,8)
...
************
Use the distributive property
of multiplication
...
Change
the direction of the symbol
when dividing by a negative
...
b = −2
...
Use a dotted line for the border and shade
above the line because the symbol is >
...
y
(9,8)
(0,4)
x
(0,−2)
(9,−4)
167
501 Algebra Questions
273
...
Use the distributive property
of multiplication
...
Simplify
...
y < 7x + 5
−5
−5
change in y
m = 7 =
chang
e in x
The y-intercept is at the point (0,5)
...
The slope tells you to go down
5 spaces and right 7 for (7,0)
...
************
−3
Multiply both sides of the
inequality by 4
...
1
< 4(2x − 3y)
1
Add 3y to both sides
...
Add 12 to both sides
...
4(−3) < 4(4)(2x − 3y)
−12 < 1(2x − 3y)
−12 < 2x −3y
−12 + 3y < 2x − 3y + 3y
−12 + 3y < 2x
−12 + 12 + 3y < 2x + 12
3y < 2x + 12
Divide both sides by 3
...
The slope tells you to go up
2 spaces and right 3 for (3,6)
...
change in y
m = 3 =
chang
e in x
The y-intercept is at the point (0,4)
...
168
501 Algebra Questions
y
(3,6)
(0,5)
(0,4)
(7,0)
x
Graph for question 273
274
...
Use the distributive property
of multiplication
...
7
y > 7 − 4x
−7
y > 4x + 7
−7
change in y
m = 4 =
chang
e in x
The y-intercept is at the point (0,7)
...
The slope tells you to go down
7 spaces and right 4 for (4,0)
...
************
3( y + 5) > 7x
Use the distributive property
of multiplication
...
Simplify the inequality
...
y > 3x − 5
b = −5
...
7
7
169
change in y
501 Algebra Questions
The slope tells you to go up 7 spaces
and right 3 for (3,2)
...
The solution for the system of inequalities is the double-shaded area
on the graph
...
Transform equations into
slope/y-intercept form
...
Subtract 5x from both sides
...
Simplify the inequality
...
Change
the direction of the symbol
when dividing by a negative
...
The slope tells you to go up
5 spaces and right 2 for (2,−5)
...
170
501 Algebra Questions
Use a solid line for the border and shade
above it because the symbol is ≥
...
Simplify the inequality
...
b = −3
...
Use a solid line for the border and shade
below the line because the symbol is ≤
...
y
x
(0,−3)
(1,−5)
(2,−5)
(0,−10)
Graph for question 275
171
12
Solving Systems
of Equations
Algebraically
There is a faster way to solve systems of equations than graphing and
finding the solution point
...
The two methods you will practice here are called the elimination method and the substitution method
...
In the elimination method, you will transform one or both of the two
equations in the system so that when you add the two equations together,
one of the variables will be eliminated
...
When you find a numerical value for the
remaining variable, you just substitute the found value into one of the equations and solve for the other variable
...
Then you will eliminate
the variable by substituting into the other equation and solve
...
One method is not better than the other
...
Practice will help you decide
...
Then transform the equation or equations so
that you will get the result you want
...
Use the elimination method to solve the following systems of equations
...
x + y = 4
2x − y = −1
277
...
7x + 3y = 11
2x + y = 3
279
...
5x + 5y = 28
3x − y = 13
280
...
2x + 2y = 11
2x − y = 17
282
...
6y + 3x = 30
2y + 6x = 0
284
...
3x + y = 20
x
3
+ 10 = y
286
...
3x − 5y = −21
2(2y − x) = 16
1
288
...
289
...
x + y = 3
3x + 101 = 7y
291
...
6
y + 21x = 8
...
8x − y = 0
10x + y = 27
x
293
...
y + 3x = 0
y − 3x = 24
295
...
2x + y = 2 − 5y
x−y=5
2x
y
297
...
x + 6y = 11
x − 3 = 2y
299
...
2(2 − x) = 3y − 2
3x + 9 = 4(5 − y)
174
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
Be aware that you may have used a different method
of elimination to arrive at the correct answer
...
Add the equations
...
Now solve for x
...
Simplify terms
...
Subtract 1 from both sides
...
The solution for the system of equations is (1,3)
...
We could add the equations together if we had
a −3x in the second equation
...
Simplify
...
Identity element of addition
...
Substitute the value of y into one of the equations
in the system and solve for x
...
Subtract 4 from both sides
...
Multiply both sides by −1
...
175
3(–x + 2y = 1)
−3x + 6y = 3
3x + 4y = 17
0 + 10y = 20
10y = 20
y=2
−x
+ 2(2) = 1
+4=1
−x + 4 − 4 = 1 − 4
−x = −3
x=3
−x
501 Algebra Questions
278
...
Multiply the equation by −3
...
Add the first equation to the transformed second
...
2(2) + y = 3
4+y=3
4−4+y=3−4
y = −1
Subtract 4 from both sides
...
The solution for the system of equations is (2,−1)
...
If you multiply the second equation by 5
and add the equations together, you
can eliminate the y
...
Simplify terms
...
Additive identity
...
5
...
Simplify
...
Simplify
...
Simplify
...
5(3x − y) = 5(13)
5(3x) − 5( y) = 5(13)
15x − 5y = 65
0
...
5x + 0 = 93
15
...
See what the first equation looks like after
distributing the multiplication on the left
...
Multiply the second equation by −3 and add
equations to eliminate y
...
Simplify terms
...
Additive identity
...
176
3x + 3y = 18
−3(5x
+ y) = −3(−2)
− 3( y) = −3(−2)
−15x − 3y = 6
3x + 3y = 18
−12x + 0 = 24
−12x = 24
x = −2
−3(5x)
501 Algebra Questions
Substitute the value of x into one of the equations
in the system and solve for y
...
Simplify
...
281
...
Use the distributive property of multiplication
...
Add the first equation to the second
...
42x = 45
Multiply the equation by 2 to simplify the fraction
...
Divide both sides by 9
...
Subtract 20 from both sides
...
Multiply the equation by −1
...
2(42x = 45)
9x = 90
x = 10
1
2(10) − y = 17
20 − 20 − y = 17 − 20
−y = −3
y=3
282
...
Add 15 to both sides
...
Subtract y from both sides
...
Multiply the second equation by 8 and add
the first equation to the second
...
Add the first equation to the second
...
Divide both sides by 29
...
Simplify
...
177
3x − 15 + 15 = y + 15
3x = y + 15
3x − y = y − y + 15
3x − y = 15
8(3x − y) = 15
24x − 8y = 120
5x + 8y = 25
29x + 0 = 145
29x = 145
x=5
3(5) − 15 = y
15 − 15 = y
0=y
501 Algebra Questions
283
...
Additive identity
...
Substitute the value of x into one of the
equations in the system and solve for y
...
Simplify
...
The solution for the system of equations is (−2,6)
...
Transform the first equation into
familiar form (ax + by = c)
...
Simplify
...
Subtract x from both sides
...
Multiply equation by 3
...
Add the transformed first equation
...
Divide both sides by 13
...
Divide both sides by 3
...
178
−3(2y
+ 6x = 0)
− 18x = 0
6y + 3x = 30
0 − 15x = 30
x = −2
−6y
2y + 6(−2) = 0
2y − 12 = 0
2y − 12 + 12 = 0 + 12
2y = 12
y=6
3x = 5 − 7y
3x + 7y = 5 − 7y + 7y
3x + 7y = 5
2y = x − 6
−x + 2y = x − x − 6
−x + 2y = −6
3(−x + 2y = −6)
3(−x) + 3 (2y) = 3(−6)
−3x + 6y = −18
3x + 7y = 5
0 + 13y = −13
y = −1
3x = 5 − 7(−1)
3x = 5 + 7
3x = 12
x=4
501 Algebra Questions
285
...
Multiply the equation by 3
...
Simplify terms
...
Simplify
...
Simplify
...
Use the distributive property
...
Add the transformed second equation to the first
...
Divide both sides by 10
...
Simplify terms
...
286
...
Use the distributive property
...
Second equation
...
Simplify terms
...
Additive identity
...
Substitute the value of y into one of the
equations in the system and solve for x
...
Subtract 28 from both sides
...
Divide both sides by 3
...
179
3(2x + 7y = 45)
3(2x) + 3(7y) = 3(45)
6x + 21y = 135
−2(3x + 4y = 22)
−2(3x) − 2(4y) = −2(22)
−6x − 8y = −44
6x + 21y = 135
0 + 13y = 91
y=7
3x + 4(7) = 22
3x + 28 = 22
3x + 28 − 28 = 22 − 28
3x = −6
x = −2
501 Algebra Questions
287
...
Use the distributive property of multiplication
...
Commutative property of addition
...
Distributive property
...
Distributive property
...
Add the first equation to the second
...
Substitute the value of y into one of the
equations in the system and solve for x
...
Combine like terms on each side
...
The solution for the system of equations is (−2,3)
...
Transform the second equation by multiplying it by 3
...
3(2x − 3y = 21)
Use the distributive property of multiplication
...
3(2x) − 3(3y) = 3(21)
6x − y = 63
Add the first equation to the second
...
1
1
64x
1
64
1
Divide both sides by 64
...
Substitute the value of x into one of the equations
in the system and solve for y
...
Simplify
...
180
=
75
1
64
x = 12
1
4(12)
+ y = 12
3 − 3 + y = 12 − 3
y=9
501 Algebra Questions
289
...
Substitute 5x for y in the second equation
and then solve for x
...
Divide both sides by 29
...
The solution for the system of equations is (3,15)
...
Transform the first equation so that the value of x
is expressed in terms of y
...
Simplify
...
Use the distributive property of multiplication
...
Add like terms
...
Combine like terms
...
Substitute the value of y into one of the equations
in the system and solve for x
...
Combine like terms on each side
...
x+y−y=3−y
x=3−y
3(3 − y) + 101 = 7y
9 − 3y + 101 = 7y
9 + 101 − 3y = 7y
110 − 3y + 3y = 7y + 3y
110 = 10y
11 = y
x + (11) = 3
x + 11 − 11 = 3 − 11
x = −8
291
...
5x + y = 3
...
5x − 5x + y = 3
...
y = 3
...
(3
...
4
Combine like terms
...
6 + 16x = 8
...
6 from both sides
...
6 − 3
...
4 − 3
...
16x = 4
...
x = 0
...
5(0
...
6
Simplify terms
...
5 + y = 3
...
5 from both sides
...
5 − 1
...
6 − 1
...
y = 2
...
3,2
...
181
501 Algebra Questions
292
...
Add y to both sides of the equation
...
Substitute the value of y into the second equation
...
Divide both sides by 18
...
Simplify terms
...
Simplify
...
3
8(2) − y = 0
12 − y = 0
12 − y + y = 0 + y
12 = y
293
...
Multiply the equation by 3
...
Simplify
...
Use the distributive property of multiplication
...
Combine like terms
...
Combine like terms on each side
...
Substitute the value of y into one of the
equations in the system and solve for x
...
Combine like terms
...
The solution for the system of equations is (36,10)
...
Express y in terms of x in the first equation
...
Combine like terms and simplify
...
Combine like terms
...
Substitute the found value for x into
one of the equations and solve for y
...
Add 12 to both sides
...
182
x
3((3) = y + 2)
x
3(3) = 3y + 6
x = 3y + 6
2(3y + 6) − 4y = 32
6y + 12 − 4y = 32
6y − 4y + 12 = 32
2y + 12 − 12 = 32 − 12
2y = 20
y = 10
2x − 4(10) = 32
2x − 40 + 40 = 32 + 40
2x = 72
x = 36
y + 3x = 0
y + 3x − 3x = 0 − 3x
y = −3x
(−3x) − 3x = 24
= 24
x = −4
−6x
y + 3(−4) = 0
y − 12 = 0
y = 12
501 Algebra Questions
295
...
Subtract 5x
from both sides of the equation
...
Substitute the value of y into the second
equation in the system and solve for x
...
5x − 5x + y = 20 − 5x
y = 20 − 5x
1
3x = 2(20 − 5x) + 1
5
3x = 10 − 2x + 1
5
3x = 11 − 2x
Combine like terms
...
3x + 2x = 11 + 2x − 2x
Combine like terms
...
=
11
1
52
Simplify terms
...
5(2) + y = 20
Simplify
...
y = 10
The solution for the system of equations is (2,10)
...
Transform the second equation so that the
value of x is expressed in terms of y
...
Combine like terms on each side
...
Use the distributive property
of multiplication
...
Combine like terms on each side
...
Combine like terms on each side
...
Substitute the found value for y into one
of the equations and solve for x
...
x−y+y=5+y
x=5+y
2(5 + y) + y = 2 − 5y
10 + 2y + y = 2 − 5y
10 + 2y + y + 5y = 2 − 5y + 5y
10 + 8y = 2
10 − 10 + 8y = 2 − 10
8y = −8
y = −1
x − (−1) = 5
x+1=5
x=4
−
The solution for the system of equations is (4, 1)
...
Transform the first equation by eliminating
the denominators
...
10(10 + 5) = 10(1)
Use the distributive property of multiplication
...
10(10) + 10(5) = 10
2x + 2y = 10
183
2x
2x
y
y
501 Algebra Questions
2x + 2y
2
10
= 2
x+y=5
Divide both sides by 2
...
Now express x in terms of y
...
Simplify
...
Use the distributive property of multiplication
...
Add y to both sides
...
Subtract 12 from both sides
...
Substitute the value of y into one of the equations
in the system and solve for x
...
Combine like terms on each side
...
The solution for the system of equations is (2,3)
...
Transform the second equation so that the value
of x is expressed in terms of y
...
Combine like terms on each side
...
Use the commutative property of addition
...
Subtract 3 from both sides
...
Divide both sides by 8
...
Subtract 6 from both sides
...
The solution for the system of equations is (5,1)
...
Transform the second equation so that the
value of y is expressed in terms of x
...
Combine like terms on each side
...
Use the distributive property of multiplication
...
Combine like terms on each side
...
184
y + 10 − 10 = 3x − 10
y = 3x − 10
4(3x − 10) + 31 = 3x
4(3x) − 4 (10) + 31 = 3x
12x − 40 + 31 = 3x
12x − 9 = 3x
12x − 9 + 9 = 3x + 9
501 Algebra Questions
Combine like terms on each side
...
Combine like terms on each side
...
Substitute the value of x into the second
equation and solve for y
...
Combine like terms on each side
...
12x = 3x + 9
12x − 3x = 3x − 3x + 9
9x = 9
x=1
y + 10 = 3(1)
y + 10 − 10 = 3 − 10
y = −7
300
...
Use the distributive property
of multiplication
...
Combine like terms on each side
...
Substitute the value of x into the second
equation and solve for y
...
3x + 9 = 20 − 4y
3x + 9 − 9 = 20 − 9 − 4y
3x = 11 − 4y
11 − 4y
x = 3
4 − 2x = 3y − 2
11 − 4y
4 − 2(3) = 3y − 2
22 − 8y
Multiply the numerator by the factor 2
...
3(4 − (3)) = 3(3y − 2)
Use the distributive property
22 − 8y
of multiplication
...
12 − (22 − 8y) = 9y − 6
Simplify the second term and the − sign
...
Add 6 to both sides
...
−4 + 8y − 8y = 9y − 8y
Subtract 8y from both sides
...
Substitute the value of y into the first
equation in the system and solve for x
...
4 − 2x = −12 − 2
Subtract 4 from both sides
...
−
Divide both sides by 2
...
185
13
Working with
Exponents
In this chapter, you will practice adding, subtracting, multiplying, and
dividing expressions that contain variables with exponents
...
Tips for Working with Exponents
Add and subtract like terms:
3n + 5n = 8n, or 5x2y − 3x2y = 2x2y
...
1
2x−1 = 2(x1 ) = x
2
1
n−3 = n3
When the result of a division leaves an exponent of zero, the term raised to the
power of zero equals 1:
z0 = 1
r2
3r2 = 3r 0 = 3(1) = 3
When a variable with an exponent is raised to a power, you multiply the exponent
to form the new term:
(b2)3 = b2 · b2 · b2 = b(2 + 2 + 2) = b6
(2x2y)2 = 2x2y · 2x2y = 2 · 2 · x2 · x2 · y · y = 22x2 + 2y1 + 1 = 4x4y2
Remember order of operations: PEMDAS
...
187
501 Algebra Questions
Simplify the following expressions:
301
...
5ab4 − ab4
303
...
5c2 + 3c − 2c2 + 4 − 7c
305
...
(5n2)(2n5 − 2n3 + 3n7)
307
...
(5a2 · 3ab) + 2a3b
2
8xy
309
...
(4a2)3 + (2a3)2 − 11a6
3
(3x)
311
...
3s
3
313
...
(3xy5)2 − 11x2y2(4y4)2
2
2
3
2(3x y) (xy)
315
...
x5y3
317
...
x5 − x8
319
...
− a2
(2x)2
x4
x
(2x2y)4
321
...
8ax2
23
(a x)
323
...
(2ab2 x)2
4
9y2
325
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
Underlined expressions show the simplified result
...
Add like terms
...
Subtract like terms
...
Use the commutative property
9mn3 + 2mn3 + 8mn
11mn3 + 8mn
of addition
...
304
...
Combine like terms
...
Use the commutative property
3x2 + 7x2 + 4ax − 2ax − 8a2 + 7a2
10x2 + 2ax − a2
of addition
...
(5n2)(2n5) − (5n2)(2n3) + (5n2)(3n7)
306
...
Use the commutative property
of multiplication
...
(10 · n2 + 5) − (10 · n2 + 3) + (15 · n2 + 7)
10n7 − 10n5 + 15n9
Show expression in decreasing
exponential order
...
Use the commutative property
of multiplication
...
(30x2y2) + 7x2y2
Combine like terms
...
Use the commutative property
of multiplication
...
15a3b + 2a3b
Combine like terms
...
Divide numerical terms
...
8xy2
2xy
= x
y
4xy2
4xy2
xy
= 4x1 − 1y2 − 1
4x0y1 = 4(1)y = 4y
Simplify
...
Terms within parentheses are the base of
(4a2)(4a2)(4a2)
+ (2a3)(2a3) − 11a6
Use the distributive property of multiplication
...
64(a2 + 2 + 2) + 4(a3 + 3) − 11a6
Simplify
...
57a6
Another way of solving this problem is to
multiply the exponents of each factor
inside the parentheses by the exponent
outside of the parentheses
...
43a6 + 22a6 − 11a6
Simplify terms
...
57a6
the exponent outside the parentheses
...
In the denominator, multiply the exponents of each
factor inside the parentheses by the exponent
outside of the parentheses
...
When similar factors, or bases, are being divided,
you subtract the exponent in the denominator
from the exponent in the numerator
...
312
...
When similar factors, or bases, are multiplied, add
the exponents of the variables
...
190
33x3
x2 • x4
3
3 x3
27x3
2
x + 4 = x6
27x3 − 6 = 27x−3
27x
27x−3 = 3
12 · 2s2 · s4
3s3
12 · 2s2 + 4
3s3
8s6
3
s
501 Algebra Questions
When similar factors, or bases, are being divided,
subtract the exponent in the denominator from the
exponent in the numerator
...
Divide out the common factor of 7 in the
numerator and denominator
...
1 · 7a3b5
4 · 7ab2
1a3b5
4ab2
1a3b5
=
4ab2
1a3 – 1b5 – 2
= 4
1a2b3
4
Simplify exponents
...
Either expression is an acceptable answer
...
Multiply the exponents of each factor inside
the parentheses by the exponent outside
the parentheses
...
When similar factors, or bases, are multiplied,
add the exponents of the variables
...
Combine like terms
...
Multiply the exponents of each factor inside the
2(32x4y2)(x3y3)
2
3(x y2)
parentheses by the exponent outside the parentheses
...
When similar factors, or bases, are multiplied, add the
exponents of the variables
...
When similar factors, or bases, are being divided,
subtract the exponent in the denominator from the
exponent in the numerator
...
2 · 9 · x4 + 3y2 + 3
2
3x y2
6x7y5
2
x y2
6x7 − 2y5 − 2
6x5y3
316
...
2x2y5
5
x y3
= 2x2 − 5y5 − 3
Simplify the operations in the exponents
...
2x−3y2 = x
3
317
...
Use the commutative property of multiplication
...
When similar factors, or bases, are being
multiplied, add the exponents
...
108a3x3 − 2
108a3x
318
...
Simplify the operations in the exponents
...
A base with a negative exponent in the numerator
is equivalent to the same variable or base in the
denominator with the inverse sign for the exponent
...
Multiply the exponents of each factor inside the
53a6x9y3
125a6x9y3
parentheses by the exponent outside the parentheses
...
320
...
+
−
2
4
ax
2x
x
Evaluate the numerical coefficients and divide
3
5
2
6x
3x
9a x2
out common numerical factors in the terms
...
6x3 − 2 + 3x5 − 4 − 9a2 − 2x2
−1
Simplify the operations in the exponents
...
Combine like terms
...
Multiply the exponents of each factor
inside the parentheses by the exponent
outside the parentheses
...
Simplify the operations in the exponents
...
Add like terms
...
Multiply the exponents of each factor inside the
parentheses by the exponent outside the parentheses
...
Simplify the operations in the exponents
...
8a 5x 1 =
a5x1 = a5x
323
...
Evaluate the numerical coefficients
...
Combine like terms in the expression
...
Multiply the exponents of each factor inside the parentheses
–
42b2x 2
22a2b4x2
by the exponent outside the parentheses
...
Simplify the numerical factors in the numerator and the
denominator
...
–
4b2x 2
a2b4x2
4b2 – 4x–2 – 2
a2
4b−2x−4
a2
Simplify the operations in the exponents
...
4
a2b2x4
325
...
(2xy)2 · (x)2 +
32y2
Repeat the previous step
...
(4x2y2)(x2 ) + 9y2
The last term is equivalent to 1
...
(4x2y2)(x2 ) + 1
Use the commutative property of multiplication
...
Divide out the common factor of x2 in the
numerator and denominator
...
Specifically, you will practice solving problems multiplying
a monomial (one term) and a polynomial, multiplying binomials (expressions with two terms), and multiplying a trinomial and a binomial
...
a(b + c + d + e) = ab + ac + ad + ae
501 Algebra Questions
When multiplying a binomial by a binomial,
you use the mnemonic FOIL to remind
you of the order with which you multiply
terms in the binomials
...
Multiply the first terms
of each binomial
...
O is for outer
...
([a] + b)(c + [d]) gives the term ad
...
Multiply the inner terms
of each binomial
...
L is for last
...
(a + [b])(c + [d]) gives the term bd
...
ac + ad + bc + bd
Multiplying a trinomial by a binomial is relatively easy
...
Multiply
each term in the trinomial by the first and then the second term in the binomial
...
(a + b)(c + d + e) = (ac + ad + ae) + (bc + bd + be)
Multiply the following polynomials
...
x(5x + 3y − 7)
327
...
4bc(3b2c + 7b − 9c + 2bc2 − 8)
329
...
4x(9x3 + x2 − x4 + x
2 )
331
...
(x − 4)(x − 9)
333
...
(x + 2)(x − 3y)
335
...
(5x + 7)(5x − 7)
x
337
...
(3x2 + y2)(x2 − 2y2)
339
...
(2x2 + y2)(x2 − y2)
341
...
(2x − 3)(x3 + 3x2 − 4x)
343
...
(3y − 7)(6y2 − 3y + 7)
345
...
(x + 2)(2x + 1)(x − 1)
347
...
(2n − 3)(2n + 3)(n + 4)
349
...
(3x2 + 4)(x − 3)(3x2 − 4)
196
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
326
...
Simplify terms
...
Multiply each term in the trinomial by 2a
...
328
...
Simplify terms
...
Multiply each term in the
polynomial by 3mn
...
x(5x) + x(3y) − x(7)
5x2 + 3xy − 7x
2a(5a2) − 2a(7a) + 2a(9)
10a3 − 14a2 + 18a
4bc(3b2c) + 4bc(7b) − 4bc(9c) + 4bc(2bc2) − 4bc(8)
12b3c2 + 28b2c − 36bc2 + 8b2c3 − 32bc
3mn(−4m) + 3mn(6n) + 3mn(7mn2) − 3mn(3m2n)
+ 18mn2 + 21m2n3 − 9m3n2
−12m2n
197
501 Algebra Questions
330
...
4x(6x − 1)
12x
5
Simplify terms
...
36x4 + x
2 − 4x +
x2
When similar factors, or bases, are
being divided, subtract the exponent
in the denominator from the
exponent in the numerator
...
36x4 + 12x−1 − 4x5 + 24x0 − 4x−1
Use the associative property
of addition
...
36x4 + 8x−1 − 4x5 + 24x0
A base with a negative exponent
in the numerator is equivalent
to the same variable or base in
the denominator with the
8
inverse sign for the exponent
...
36x4 + x − 4x5 + 24(1)
−4x5
Simplify and put in order
...
Use FOIL to multiply binomials
...
Multiply the outer terms in each binomial
...
Multiply the last terms in each binomial
...
Combine like terms
...
Use FOIL to multiply binomials
...
Multiply the outer terms in each binomial
...
Multiply the last terms in each binomial
...
Combine like terms
...
Use FOIL to multiply binomials
...
Multiply the outer terms in each binomial
...
Multiply the last terms in each binomial
...
Combine like terms
...
Use FOIL to multiply binomials
...
Multiply the outer terms in each binomial
...
Multiply the last terms in each binomial
...
335
...
Multiply the first terms in each binomial
...
Multiply the inner terms in each binomial
...
Add the products of FOIL together
...
336
...
Multiply the first terms in each binomial
...
Multiply the inner terms in each binomial
...
Add the products of FOIL together
...
199
([2x] + 1)([3x] − 7)
6x2
([2x] + 1)(3x − [7])
−14x
(2x + [1])([3x] − 7)
+3x
(2x + [1])(3x − [7])
−7
6x2 − 14x + 3x − 7
6x2 − 11x − 7
([x] + 2)([x] − 3y)
x2
([x] + 2)(x − [3y])
−3xy
(x + [2])([x] − 3y)
+2x
(x + [2])(x − [3y])
−6y
2
x − 3xy + 2x − 6y
([7x] + 2y)([2x] − 4y)
14x2
([7x] + 2y)(2x − [4y])
−28xy
(7x + [2y])([2x] − 4y)
+4xy
(7x + [2y])(2x − [4y])
−8y2
2
14x − 28xy + 4xy − 8y2
14x2 − 24xy − 8y2
([5x] + 7)([5x] − 7)
25x2
([5x] + 7)(5x − [7])
−35x
(5x + [7])([5x] − 7)
+35x
(5x + [7] (5x − [7])
−49
25x2 − 35x + 35x − 49
25x2 − 49
501 Algebra Questions
337
...
Multiply the first terms in each binomial
...
Multiply the inner terms in each binomial
...
Add the products of FOIL together
...
338
...
Multiply the first terms in each binomial
...
Multiply the inner terms in each binomial
...
Add the products of FOIL together
...
339
...
Multiply the first terms in each binomial
...
Multiply the inner terms in each binomial
...
Add the products of FOIL together
...
340
...
Multiply the first terms in each binomial
...
Multiply the inner terms in each binomial
...
Add the products of FOIL together
...
200
x
([28x] + 7)([7] − 11)
4x2
x
([28x] + 7)(7 − [11])
−308x
x
(28x + [7])([7] − 11)
+x
x
(28x + [7])(7 − [11])
−77
4x2 − 308x + x − 77
4x2 − 307x − 77
([3x2] + y2)([x2] − 2y2)
3x4
([3x2] + y2)(x2 − [2y2])
−6x2y2
2
(3x + [y2])([x2] − 2y2
+x2y2
(3x2 + [y2])(x2 − [2y2])
−2y4
4
2
3x − 6x y2 + x2y2 − 2y4
3x4 − 5x2y2 − 2y4
([4] + 2x2)([9] − 3x)
+36
([4] + 2x2)(9 − [3x])
−12x
(4 + [2x2])([9] − 3x)
+18x2
(4 + [2x2])(9 − [3x])
−3x3
36 − 12x + 18x2 − 3x3
−3x3
+ 18x2 − 12x + 36
([2x2] + y2)([x2] − y2)
2x4
([2x2] + y2)(x2 − [y2])
−2x2y2
2
(2x + [y2])([x2] − y2)
+x2y2
(2x2 + [y2])(x2 − [y2])
−y4
4
2
2x − 2x y2 + x2y2 − y4
2x4 − x2y2 − y4
501 Algebra Questions
341
...
Simplify terms
...
[x(3x2 − 5x + 2)]
[x(3x2) − x(5x) + x(2)]
[3x3 − 5x2 + 2x]
[2(3x2 − 5x + 2)]
[2(3x2) − 2(5x) + 2(2)]
[6x2 − 10x + 4]
Simplify terms
...
[3x3 − 5x2 + 2x] + [6x2 − 10x + 4]
Use the commutative property of addition
...
3x3 + x2 − 8x + 4
342
...
Use the distributive property of
multiplication
...
Multiply the trinomial by the second
term in the binomial, −3
...
Simplify terms
...
Use the commutative property
of addition
...
343
...
Use the distributive property of
multiplication
...
Multiply the trinomial by the
second term in the binomial, b
...
Simplify terms
...
Use the commutative property of
addition
...
[2x(x3 + 3x2 − 4x)]
[2x(x3) + 2x(3x2) − 2x(4x)]
[2x4 + 6x3 − 8x2]
[−3(x3 + 3x2 − 4x)]
[−3(x3) − 3(3x2) − 3(−4x)]
[−3x3 − 9x2 + 12x]
[2x4 + 6x3 − 8x2] + [−3x3 − 9x2 + 12x)]
2x4 + 6x3 − 3x3 + 8x2 − 9x2 + 12x
2x4 + 3x3 − 17x2 + 12x
[4a(5a2 + 2ab − b2)]
[4a(5a2) + 4a(2ab) − 4a(b2)]
[20a3 + 8a2b − 4ab2]
[b(5a2 + 2ab − b2)]
[b(5a2) + b(2ab) − b(b2)]
[5a2b + 2ab2 – b3]
[20a3 + 8a2b − 4ab2] + [5a2b + 2ab2 − b3)]
20a3 + 8a2b + 5a2b − 4ab2 + 2ab2 − b3
20a3 + 13a2b − 2ab2 − b3
201
501 Algebra Questions
344
...
Use the distributive property of
multiplication
...
Multiply the trinomial by the
second term in the binomial, −7
...
Simplify terms
...
Use the commutative property
of addition
...
[3y(6y2 − 3y + 7)]
[3y(6y2) − 3y(3y) + 3y(7)]
[18y3 − 9y2 + 21y]
[−7(6y2 − 3y + 7)]
[−7(6y2) − 7(−3y) − 7(7)]
[−42y2 + 21y − 49]
[18y3 − 9y2 + 21y] + [−42y2 + 21y − 49]
18y3 − 9y2 − 42y2 + 21y + 21y − 49
18y3 − 51y2 + 42y − 49
345
...
Use the distributive property
of multiplication
...
Multiply the trinomial by the second
term in the binomial, 2
...
Simplify terms
...
Use the commutative property
of addition
...
[3x(3x2 − 2x − 5)]
[3x(3x2) − 3x(2x) − 3x(5)]
[9x3 − 6x2 − 15x]
[2(3x2 − 2x − 5)]
[2(3x2) − 2(2x) − 2(5)]
[6x2 − 4x − 10]
[9x3 − 6x2 − 15x] + [6x2 − 4x − 10]
9x3 – 6x2 + 6x2 − 15x − 4x − 10
9x3 − 19x − 10
346
...
Multiply the first terms in each binomial
...
([x] + 2)(2x + [1])
+x
Multiply the inner terms in each
binomial
...
(x + [2])(2x + [1])
+2
2
Add the products of FOIL together
...
2x2 + 5x + 2
202
501 Algebra Questions
Multiply the resulting trinomial by the
last binomial in the original
expression
...
Use the distributive property
of multiplication
...
Multiply the trinomial by the second
term in the binomial, −1
...
Add the results of multiplying by the
terms in the binomial together
...
Combine like terms
...
Multiply the first two parenthetical terms
in the expression using FOIL
...
Multiply the outer terms in each
binomial
...
Multiply the last terms in each
binomial
...
15a + 6a − 20a − 8
Combine like terms
...
(a + 3)(15a2 − 14a − 8)
Multiply the trinomial by the first
term in the binomial, a
...
[a(15a2) − a(14a) – a(8)]
Simplify terms
...
[3(15a2) − 3(14a) − 3(8)]
Use the distributive property of
multiplication
...
[15a3 − 14a2 − 8a] + [45a2 − 42a − 24]
Use the commutative property
of addition
...
15a3 + 31a2 − 50a − 24
348
...
Multiply the first terms in each binomial
...
Multiply the inner terms in each binomial
...
Add the products of FOIL together
...
Now we again have two binomials
...
Multiply the first terms in each binomial
...
Multiply the inner terms in each binomial
...
Add the products of FOIL together
...
([2n] − 3)([2n] + 3)
4n2
([2n] − 3)(2n + [3])
+6n
(2n − [3])([2n] + 3)
−6n
(2n − [3])(2n + [3])
−9
2
4n + 6n − 6n − 9
4n2 − 9
(n + 4)(4n2 − 9)
([n] + 4)([4n2] − 9)
4n3
([n] + 4)(4n2 − [9])
−9n
(n + [4])([4n2] − 9)
+16n2
(n + [4])(4n2 − [9])
−36
3
4n − 9n + 16n2 − 36
4n3 + 16n2 − 9n − 36
349
...
Use the distributive property of
multiplication
...
Multiply the trinomial by the second term in the binomial, −7
...
Simplify terms
...
Use the commutative property
of addition
...
Multiply the first two parenthetical terms
in the expression using FOIL
...
Multiply the outer terms in each binomial
...
Multiply the last terms in each binomial
...
Multiply the resulting trinomial by the last
binomial in the original expression
...
Use the distributive property
of multiplication
...
Multiply the trinomial by the second
term in the binomial, –4
...
Add the results of multiplying by the
terms in the binomial together
...
Combine like terms
...
You can
use three different techniques to factor polynomials
...
In the second,
you will factor polynomials that are the difference of two perfect squares
...
The trinomial expressions in this chapter will be in the form of x2 ± ax ± b, where a and b are
whole numbers
...
Complete explanations of the solutions will follow
...
Put that factor outside a set of
parentheses and the polynomial inside with the factor removed from
each term, e
...
2x2 + 8 = 2(x2 + 4)
Factoring using the difference of two perfect squares: Polynomials
in the form x2 − y2 can be factored into two terms: (x + y)(x − y)
...
The factors of the first term go in the first
position in the parentheses and the factors of the third term go in the second
position in each factor, e
...
x2 + 2x + 1 = (x + 1)(x + 1)
...
351
...
3a2x + 9ax
353
...
4a2 − 25
355
...
7x4y2 − 35x2y2 + 14x2y4
357
...
9r2 − 49
359
...
x2 + 5x + 6
361
...
b2 − 100
363
...
x2 − 3x − 18
365
...
b2 − 4b − 21
367
...
x2 + 10x + 25
369
...
x2 + 20x + 99
371
...
h2 − 12h + 11
373
...
v4 − 13v2 − 48
375
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
Underlined expressions show the original algebraic expression as an equation with the expression
equal to its simplified result
...
The terms have a common factor of 3
...
9a + 15 = 3(3a + 5)
352
...
Factor 3ax out of each term and write the
expression in factored form
...
Both terms in the polynomial are perfect
squares
...
x2 − 16 = (x + 4)(x − 4)
2
Check using FOIL
...
Both terms in the polynomial are perfect
squares
...
Use
the form for factoring the difference of
two perfect squares and put the roots
of each factor in the proper place
...
(2a + 5)(2a − 5) = 4a − 10a + 10a − 25 = 4a2 − 25
355
...
Factor
7n out of each term and write the expression
in factored form
...
The terms have a common factor of 7x2y2
...
208
7x2y2(x2 − 5 + 2y2)
501 Algebra Questions
357
...
The factors of x2 are x and x, and the
factors of 2 are 1 and 2
...
(x + 2)(x + 1) = x2 + x + 2x + 2 = x2 + 3x + 2
The factors are correct
...
Both terms in the polynomial are perfect squares
...
Use the form for
factoring the difference of two perfect
squares and put the roots of each factor
in the proper place
...
This expression can be factored using the
trinomial method
...
You want the result of the O and I of
the FOIL method for multiplying factors to
add up to −2x
...
Place the factors (2)(4) into the trinomial
factor form and check using
FOIL
...
(x + 2)(x − 4) = x2 − 4x + 2x − 8 = x2 − 2x − 8
The factors of the trinomial are now correct
...
This expression can be factored using the
trinomial method
...
Since the numerical term of the polynomial is positive, the signs in the factor form
for trinomials will be the same because only
two like signs multiplied together will result
in a positive
...
In order to add up to 5x, the
result of multiplying the Inner and Outer
terms of the trinomial factors will have to be
positive
...
Check using FOIL
...
This expression can be factored using the
trinomial method
...
You want the result of the O and I of
the FOIL method for multiplying factors
to add up to +1x
...
Place the factors (2)(3) into
the trinomial factor form and check
using FOIL
...
362
...
b2 = (b)2 and 100 = 102
...
b2 − 100 = (b + 10)(b − 10)
2
Check using FOIL
...
This expression can be factored using the
trinomial method
...
You want the result of the
O and I of the FOIL method for multiplying
factors to add up to +7x
...
Since
all signs are positive, use positive signs
in the factored form
for the trinomial
...
This is not just luck
...
364
...
The factors of x2 are
x and x, and the factors of 18 are (1)(18) or
(2)(9) or (3)(6)
...
The sum of the results of multiplying
the Outer and Inner terms of the trinomial
factors needs to add up to a −3x
...
(x + 3)(x − 6) = x2 − 6x + 3x − 18 = x2 − 3x − 18
210
501 Algebra Questions
365
...
The factors of b2 are b
and b, and the factors of 8 are (1)(8) or
(2)(4)
...
The signs within the parentheses of the factorization of the trinomial
must be the same to result in a positive
numerical term in the trinomial
...
How can you get 6 from
adding the two of the factors of 8? Right!
Use the (2)(4)
...
(b − 2)(b − 4) = b2 − 4b − 2b + 8 = b2 − 6b + 8
366
...
The factors of b2 are
b and b, and the factors of 21 are (1)(21) or
(3)(7)
...
Only the product
of a positive and a negative numerical term
will result in −21
...
Check using FOIL
...
This expression can be factored using the
trinomial method
...
You want the result of the
O and I of the FOIL method for multiplying
factors to add up to +11a
...
Since the signs in the factors
must be one positive and one negative, use the
factors 12 and 1 in the trinomial factors form
...
(a + 12)(a − 1) = a2 − 1a + 12a − 12 = a2 + 11a − 12
211
501 Algebra Questions
368
...
The factors of x2 are
x and x, and the factors of 25 are (1)(25)
or (5)(5)
...
The sum of the results
of multiplying the Outer and Inner terms of
the trinomial factors needs to add up to a
+10x
...
(x + 5)(x + 5) = x2 + 5x + 5x + 25 = x2 + 10x + 25
369
...
36y2 = (6y2)2 and 4z2 = (2z)2
Use the form for factoring the difference of
two perfect squares and put the roots of
each factor in the proper place
...
(6y + 2z)(6y − 2z) = 36y − 12y z + 12y2z − 4z2 = 36y4 − 4z2
370
...
The factors of x2 are
x and x, and the factors of 99 are (1)(99)
or (3)(33) or (9)(11)
...
The
sum of the results of multiplying the Outer
and Inner terms of the trinomial factors
needs to add up to a +20x
...
Check using FOIL
...
This expression can be factored using the
trinomial method
...
The sign of the numerical
term is positive, so the signs in the factors
of our trinomial factorization must be the
same
...
This leads one to believe that the signs in
the trinomial factors will both be negative
...
Check using FOIL
...
The factors of h2 are h and h, and the factors
of 11 are (1)(11)
...
The sign of the first-degree term (the
variable to the power of 1) is negative
...
Check
your answer
...
This expression can be factored using the
trinomial method
...
The sign of the numerical
term is positive, so the signs in the factors of
our trinomial factorization must be the same
...
The only factors of 18
that can be added or subtracted in any way to
equal 11 are 2 and 9
...
Check your answer
using FOIL
...
This expression can be factored using the
trinomial method
...
Only
the product of a positive and a negative
numerical term will result in −48
...
Use 3 and
16 and a positive and negative sign in the terms
of the trinomial factors
...
(v2 + 3)(v2 − 16) = v4 − 16v2 + 3v2 − 48 = v4 − 13v − 48
You may notice that one of the two factors of
the trinomial expression can itself be factored
...
Factor (v2 − 16) using the form for
factoring the difference of
two perfect squares
...
213
501 Algebra Questions
375
...
The sign of the numerical term is positive, so
the signs in the factors of our trinomial
factorization must be the same
...
This leads one to
believe that the signs in the trinomial factors
will both be negative
...
Use them and
two negative signs in the trinomial factor form
...
(x − 2)(x − 18) = x2 − 18x − 2x + 36 = x2 − 20x + 36
214
16
Using Factoring
This chapter will present polynomial expressions for you to factor
...
In
this chapter, the coefficients of the second-degree terms will often be whole
numbers greater than 1
...
In some cases, you will find that you can
factor using more than one of the three methods of factoring polynomials
on the given expression
...
Then analyze the remaining expression to determine if
other factoring methods can be used
...
Greatest common factor method
2
...
Trinomial method
When presented with a polynomial with a coefficient greater than 1 for
the second-degree term, use the trinomial factor form (ax ± ( ))(bx ± ( ))
where a · b = the coefficient of the second-degree term
...
501 Algebra Questions
After choosing terms to try in the trinomial factors form, use FOIL to check
your guesses for the trinomial factors
...
Tips for Using Factoring
When factoring a trinomial expression, first determine the signs that will be used
in the two factors
...
Then list the factors of the numerical term
...
Be systematic in your attempts to be sure you try all possible choices
...
Factor the following expressions
...
2x2 + 7x + 6
377
...
5x2 − 14x − 3
379
...
9x2 + 34x − 8
381
...
4a2 − 16a − 9
383
...
6a2 − 5a − 6
385
...
6x2 + 15x − 36
387
...
2a6 + a3 − 21
389
...
8x2 − 6x − 9
216
501 Algebra Questions
391
...
9x3 − 4x
393
...
4x4 − 37x2 + 9
395
...
4xy3 + 6xy2 − 10xy
397
...
3c2 + 19c − 40
399
...
4x4 + 2x2 − 30
217
501 Algebra Questions
Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original
expression
...
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression
...
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within
are to be multiplied
...
Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first
and work outward
...
376
...
(ax + ( ))(bx + ( ))
The factors of the second-degree term are 2x2 = (2x)(x)
...
You want to get 7x from adding the result of the Outer and Inner
multiplications when using FOIL
...
(2x + (1))(x + (6))
(2x + (6))(x + (1))
(2x + (2))(x + (3))
(2x + (3))(x + (2))
Now just consider the results of the Outer and Inner products of the terms
for each guess
...
(2x + (1))(x + (6)) will result in Outer product plus Inner product:
2x(6) + (1)x = 12x + x = 13x
...
(2x + (2))(x + (3)) will result in Outer product plus Inner product:
2x(3) + (2)x = 6x + 2x = 8x
...
Place the factors in the trinomial factor form so that the product of the
outer terms (2x)(2) = 4x and the product of the inner terms (3)(x) = 3x
...
(2x + (3))(x + (2))
Check using FOIL
...
The factors check out
...
Both signs in the trinomial are positive, so use positive signs in the
trinomial factor form
...
The factors of the numerical term 12 = (1)(12) = (2)(6) = (3)(4)
...
Place the factors in the trinomial
factor form so that the product of the outer terms (3x)(3) = 9x and the
product of the inner terms (4)(x) = 4x
...
(3x + 4)(x + 3)
Check using FOIL
...
3x2 + 9x + 4x + 12 = 3x2 + 13x + 12
The factors check out
...
Both signs in the trinomial are negative
...
(ax + ( ))(bx − ( ))
The factors of the second-degree term are 5x2 = (5x)(x)
...
When you multiply the Outer and Inner terms of the trinomial factors,
the results must add up to be −14x
...
Adding (−15x) + (+1x) = −14x
...
(5x + (1))(x − (3))
Check using FOIL
...
5x2 − 15x + 1x − 3 = 5x2 − 14x − 3
The factors check out
...
Both signs in the trinomial are positive, so use positive signs in the
trinomial factor form
...
The factors of the numerical term 4 = (1)(4) or 4 = (2)(2)
To get 15x from adding the result of the Outer and Inner multiplications
when using FOIL, place the factors in the trinomial factor form so that
the product of the outer terms (3x)(1) = 3x and the product of the
inner terms (4)(3x) = 12x
...
(3x + 4)(3x + 1)
Check using FOIL
...
(3x + 2)(3x + 1) = 9x2 + 3x + 6x + 2 = 9x2 + 9x + 2
380
...
So the signs in the trinomial
factor form will have to be + and −
...
(ax + ( ))(bx − ( ))
The factors of the second-degree term are 9x2 = (9x)(x) or 9x2 = (3x)(3x)
...
Let’s try putting in
factors in the trinomial factor form and see what we get
...
No, that doesn’t work
...
Changing position of the signs would help but not
with these factors because the term would be +71x
...
(9x − 2)(x + 4)
2
Check using FOIL
...
Be persistent and learn from your
mistakes
...
The three terms have a common factor of 3
...
Now factor the trinomial in the
parentheses, and don’t forget to include the factor 3 when you are done
...
So the signs in the trinomial
factor form will have to be + and −, because that is the only way to get a
negative sign when multiplying the Last terms when
checking with FOIL
...
The sign of the second term is negative
...
The factors of the
numerical term 6 are (1)(6) or (2)(3)
...
(x + 2)(x − 3)
Check using FOIL
...
(x + 2)(x − 3) = x2 − 3x + 2x − 6 = x2 − x − 6
Include the common factor of 3 so that
3(x + 2)(x − 3) = 3(x2 − x − 6) = 3x2 − 3x − 18
...
Both signs in the trinomial expression are negative
...
(ax + ( ))(bx − ( ))
The factors of the second-degree term 4a2 = (4a)(a) or 4a2 = (2a)(2a)
...
The coefficient of the first-degree term is 2 less than 18
...
Use this
information to place factors within the trinomial factor form
...
First—−(2a)(2a) = 4a2
Outer—−(2a)(−9) = −18a
Inner—−(1)(2a) = 2a
Last—−(1)(−9) = −9
The result of multiplying the factors is
(2a + 1)(2a − 9) = 4a2 − 18a + 2a − 9 = 4a2 − 16a − 9
...
Both signs in the trinomial expression are negative
...
(ax + ( ))(bx − ( ))
The factors of the second-degree term 6a2 = (6a)(a) or (2a)(3a)
...
We can predict that 13 = 18 − 5
...
The remaining
factors (a)(5) = 5a
...
(6a + 5)(a − 3)
Check using FOIL
...
The factors check out
...
Both signs in the trinomial expression are negative
...
(ax + ( ))(bx − ( ))
The factors of the second-degree term 6a2 = (6a)(a) or 6a2 = (2a)(3a)
...
The trinomial looks balanced with a 6 on each end and a 5 in the middle
...
(3a + 2)(2a − 3)
Check using FOIL
...
Didn’t that work out nicely? A sense of balance can be useful
...
This expression is the difference between two perfect squares
...
However, there is a greatest common factor that could be factored out first
to leave 4(4y2 − 25)
...
4(4y2 − 25) = 4(2y + 5)(2y − 5)
The first factorization is equivalent to the second because you
can factor out two from each of the factors
...
When factoring
polynomials, watch for the greatest common factors first
...
The terms of the trinomial have a greatest common factor of 3
...
You need
only factor the trinomial within the parentheses
...
So the signs in the trinomial
factor form will have to be + and −
...
The factors of the numerical term 12 are (1)(12) or (2)(6) or (3)(4)
...
It’s clear that
8x − 3x = 5x
...
(2x − 3)(x + 4)
Check using FOIL
...
Now include the greatest common factor of 3 for the final solution
...
Each term in the polynomial has a common factor of 2b
...
So the signs in the trinomial
factor form will have to be + and − because that is the only way to get a
negative sign when multiplying the Last terms when checking with
FOIL
...
The factors of the numerical term 21 are (1)(21) or (3)(7)
...
Place these factors in
the trinomial factor form so that the result of the Outer and Inner
products when using FOIL to multiply are +14c and −3c
...
First—−(2c)(c) = 2c2
Outer—−(2c)(7) = 14c
Inner—−(−3)(c) = −3c
Last—−(−3)(7) = −21
The product of the factors is
(2c − 3)(c + 7) = 2c2 + 14c − 3c − 21 = 2c2 + 11c − 21
...
2b(2c − 3)(c + 7) = 2b(2c2 + 11c − 21 = 4bc2 + 22bc − 42b
388
...
Then the expression becomes
2(a3)2 + (a3) − 21
...
The
sign of the numerical term is negative
...
(ax + ( ))(bx − ( ))
The second-degree term 2(a3)2 = 2(a3)(a3)
...
The factors
(a3)(7) = 7(a3) and the factors (2(a3))(3) = 6(a3)
...
Place these factors in the trinomial factor form so that the first
degree term is 1(a3)
...
First—−(a3)(2a3) = 2a6
Outer—−(a3)(7) = 7a3
Inner—−(−3)(2a3) = −6a3
Last—−(−3)(7) = −21
The product of the factors is
(a3 − 3)(2a3 + 7) = 2a6 + 7a3 − 6a3 − 21 = 2a6 + a3 − 21
...
223
501 Algebra Questions
389
...
Factoring 3x out results in the expression 3x(2a2 − 13a − 24)
...
The sign of the numerical term is negative
...
(ax + ( ))(bx − ( ))
The factors of the term 2a2 = (2a)(a)
...
The factors (2a)(8) = 16a, and the related factors (a)(3) = 3a
...
Place these numbers in the trinomial factor
form, and check the expression using FOIL
...
Now include the greatest common factor if 3x
...
The numerical term of the trinomial has a negative sign, so the signs
within the factors of the trinomial will be a + and −
...
The numerical term of the trinomial 9 has factors of (1)(9) or (3)(3)
...
The
numbers 2(3) = 6, and the corresponding 4(3) = 12
...
So use the second-degree term factors (2x)(4x)
and the numerical factors (3)(3)
...
First—−(2x)(4x) = 8x2
Outer—−(2x)(3) = 6x
Inner—−(−3)(4x) = −12x
Last—−(−3)(3) = −9
The product of the factors
(2x − 3)(4x + 3) = 8x2 + 6x − 12x − 9 = 8x2 − 6x − 9
...
The numerical term of the trinomial has a negative sign, so the signs
within the factors of the trinomial will be a + and −
...
The numerical term of the trinomial 2 has factors of (1)(2)
...
The last of these is equal to the desired −9c, which gives the
factoring (5c + 1)(c − 2)
...
(5c + 1)(c − 2)
Check using FOIL
...
392
...
Factoring
x out of the expression results in x(9x2 − 4)
...
Factor that expression using the form for the difference of two
perfect squares
...
2
9x = (3x)2
4 = 22
Using the form, the factorization of the difference of two perfect squares is
(3x − 2)(3x + 2)
...
First—−(3x)(3x) = 9x2
Outer—−(3x)(2) = 6x
Inner—−(−2)(3x) = −6x
Last—−(−2)(2) = −4
Include the greatest common factor x in the complete factorization
...
The terms in the trinomial expression are all positive, so the signs in the
trinomial factor form will be positive
...
The numerical term 63 has the factors (1)(63) or (3)(21) or (7)(9)
...
Let’s look at the possibilities using the 2r and 4r
...
2r(3) + 4r (21) = 87r is still too much
...
Getting closer
...
Nope
...
Bingo!
(2r + 7)(4r + 9)
Check using FOIL
...
8r2 + 18r + 28r + 63 = 8r2 + 46r + 63
The factors check out
...
When you think of x4 = (x2)2, you can see that the expression is a trinomial
that is easy to factor
...
The sign of the first-degree term is negative, so you
will use two − signs
...
The numerical term 9 has (1)(9) or (3)(3) as factors
...
Use these factors in the trinomial factor form
...
Now you need to notice that the factors of the original trinomial
expression are both factorable
...
Use the factor form for the difference of two perfect squares for each
factor of the trinomial
...
(4x2 − 1)(x2 − 9) = (2x + 1)(2x − 1)(x + 3)(x − 3)
395
...
(ax + ( ))(bx − ( ))
This expression has a nice balance to it with 12 at the extremities and a
modest 7 in the middle
...
Use FOIL to check
...
Those are the right terms but the wrong signs
...
(4d − 3)(3d + 4)
Multiply the factors using FOIL
...
396
...
When factored
out, the expression becomes 2xy(2y2 + 3y − 5)
...
The last sign is negative, so the signs within the factor form will be a +
and −
...
The numerical term 5 has factors (5)(1)
...
(2y + 5)( y − 1)
Multiply using FOIL
...
Now include the
greatest common factor to complete the factorization of the original
expression
...
The terms of the trinomial have a greatest common factor of 2a
...
The
expression within the parentheses is a trinomial and can be factored
...
Only a (+)(–) = (–)
...
The numerical term 33 has (1)(33) or (3)(11) as factors
...
(2x + 3)(x − 11)
Check using FOIL
...
Now include the
greatest common factor of the original expression to get the complete
factorization of the original expression
...
The signs within the terms of the factor form will be + and − because the
numerical term has a negative sign
...
The numerical term 40 has (1)(40) or (2)(20) or (4)(10) or (5)(8) as factors
...
Using trial and error, you can determine that 3c(8) = 24c,
and c(5) = 5c, and 24c − 5c = 19c
...
(3c − 5)(c + 8) = 3c2 + 24c − 5c − 40 = 3c2 + 19c − 40
The complete factorization of the original expression is (3c − 5)(c + 8)
...
The signs within the terms of the factor form will be + and − because the
numerical term has a negative sign
...
The numerical term 84 has (1)(84) or (2)(42) or (3)(28) or (4)(21) or (6)(14)
or (7)(12) as factors
...
2a(12) = 24a, and a(7) = 7a, and 24a − 7a = 17a
...
Place them in position so you get the result that you want
...
400
...
If you think of the variable as x2, you
can see that the expression is in the trinomial form
...
The trinomial you will be factoring
looks like this: 4(x2)2 + 2(x2) − 30
...
(ax + ( ))(bx − ( ))
The term (4x2)2 can be factored as (4x2)2 = x2(4x2) or (2x2)(2x2)
...
The factors (4x2)(3) = 12x2 and x2(10) = 10x2 will give you 12x2 − 10x2 =
2x2 when you perform the Inner and Outer multiplications and combine
like terms using FOIL with the terms in the trinomial factor form
...
Check using FOIL
...
However, the first factor has a greatest common factor of 2
...
Did you notice that you could have used the greatest common factor
method to factor out a 2 from each term in the original polynomial? If
you did, you would have had to factor the trinomial expression
2x4 + x2 − 15 and multiply the result by the factor 2 to equal the original
expression
...
228
17
Solving Quadratic
Equations
This chapter will give you practice in finding solutions to quadratic equations
...
While there are several methods for solving quadratic equations, solutions for all the equations presented here can
be found by factoring
...
Use these methods to factor the
equations that have been transformed into quadratic equations
...
There will be two solutions
for each quadratic equation
...
The solutions will be the same for equations with or without the
numerical factors
...
401
...
n2 − 169 = 0
403
...
y2 − 15y + 56 = 0
405
...
4x2 = 49
407
...
2n2 + 20n + 42 = 0
409
...
100r2 = 144
411
...
7a2 − 21a − 28 = 0
413
...
2x2 + 9x = –10
415
...
9x2 + 12x = −4
417
...
8b2 + 10b = 42
419
...
6b2 + 20b = −9b − 20
421
...
7x2 = 52x − 21
423
...
12r2 = 192 − 40r
425
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
The solutions are underlined
...
The expression is the difference of two perfect squares
...
Subtract 5 from both sides of the equation
...
Let the second factor equal zero
...
Combine like terms on each side
...
(x + 5)(x − 5) = 0
...
The expression is the difference of two perfect
squares
...
Subtract 13 from both sides of the equation
...
Let the second factor equal zero
...
Combine like terms on each side
...
231
(n + 13)(n − 13) = 0
...
Factor the trinomial expression using the trinomial
factor form
...
(a + 4)(a + 8) = 0
(a + 4) = 0
a+4−4=0−4
Combine like terms on each side
...
(a + 8) = 0
Subtract 8 from both sides
...
a = −8
The solutions for the quadratic equation a2 + 12a + 32 = 0 are a = −4
and a = −8
...
Factor the trinomial expression using the trinomial
factor form
...
y−8+8=0+8
Combine like terms on each side
...
( y − 7) = 0
Add 7 to both sides
...
y=7
2
The solutions for the equation y − 15y + 56 = 0 are y = 8 and y = 7
...
Factor the trinomial expression using the trinomial
factor form
...
(b + 10)(b − 9) = 0
(b + 10) = 0
b + 10 − 10 = 0 − 10
Combine like terms on each side
...
(b − 9) = 0
Add 9 to both sides
...
b=9
The solutions for the quadratic equation b2 + b − 90 = 0 are b = −10
and b = 9
...
Transform the equation so that all terms are on one
side and are equal to zero
...
Combine like terms on each side
...
The equation factors into
Applying the zero product property (if (a)(b) = 0,
then a = 0 or b = 0 or both = 0), the first factor
or the second factor or both must equal zero
...
Combine like terms on each side
...
232
4x2 − 49 = 49 − 49
4x2 − 49 = 0
(2x + 7)(2x − 7) = 0
...
Let the second factor equal zero
...
Combine like terms on both sides
...
1
The solutions for the quadratic equation 4x2 = 81 are x = −32
1
and x = 32
...
Transform the equation so that all terms are
on one side and are equal to zero
...
Combine like terms on each side
...
The equation factors into
Applying the zero product property (if (a)(b) = 0,
then a = 0 or b = 0 or both = 0), the first factor
or the second factor or both must equal zero
...
Combine like terms on each side
...
5r
5
Simplify
...
Add 12 to both sides of the equation
...
r = −25
(5r − 12) = 0
5r − 12 + 12 = 0 + 12
5r = 12
Divide both sides by 5
...
r = 25
25r2 − 144 = 144 − 144
25r2 − 144 = 0
(5r + 12)(5r − 12) = 0
...
408
...
Using the zero product property, subtract 6 from
both sides
...
(2n + 6)(n + 7) = 0
(2n + 6) = 0
2n + 6 − 6 = 0 − 6
2n = −6
2n
−6
= 2
Divide both sides by 2
...
n = −3
Let the second factor equal zero
...
n+7−7=0−7
Combine like terms on each side
...
233
501 Algebra Questions
3(c2 − 11c − 26) = 0
409
...
Factor the trinomial expression using the
trinomial factor form
...
Using the zero product property, add 13 to both sides
...
c = 13
Let the second factor equal zero
...
c+2−2=0−2
Combine like terms on each side
...
410
...
Subtract 144 from both sides
...
The expression is the difference of two
perfect squares
...
Subtract 12 from both sides of the equation
...
(10r + 12) = 0
10r + 12 − 12 = 0 − 12
10r = −12
Divide both sides by 10
...
Let the second factor equal zero
...
Combine like terms on each side
...
10r
10
100r2 − 144 = 144 − 144
100r2 − 144 = 0
(10r + 12)(10r − 12) = 0
...
r = 15
1
2
The solution for the quadratic equation 100r = 144 is r = ±15
...
Use the greatest common factor method
...
3(x − 6)(x − 6) = 0
Ignore the factor 3 in the expression
...
(x − 6) = 0
x−6+6=0+6
Combine like terms on each side
...
234
501 Algebra Questions
7(a2 − 3a − 4) = 0
412
...
Factor the trinomial expression using the trinomial
factor form
...
Using the zero product property, add 4 to both sides
...
a=4
Let the second factor equal zero
...
a+1−1=0−1
Combine like terms on each side
...
8( y2 + 7y + 12) = 0
413
...
Factor the trinomial expression using the trinomial
factor form
...
Using the zero product property, subtract 4 from
both sides
...
y = −4
Let the second factor equal zero
...
y+3−3=0−3
Simplify
...
414
...
Add 10 from both sides of the equation
...
Factor the trinomial expression using the trinomial
factor form
...
2x2 + 9x + 10 = −10 + 10
2x2 + 9x + 10 = 0
(2x + 5)(x + 2) = 0
Combine like terms on each side
...
2x
2
−5
= 2
1
Simplify terms
...
x+2=0
Subtract 2 to both sides
...
x = −2
1
2
The solutions for the quadratic equation 2x + x = 10 are x = −22
and x = −2
...
Transform the equation into the familiar
trinomial equation form
...
Combine like terms on each side
...
Using the zero product property, subtract 5
from both sides
...
(2x − 3)(2x + 5) = 0
(2x + 5) = 0
2x + 5 − 5 = 0 − 5
2x = −5
Divide both sides by 2
...
Let the second factor equal zero
...
Simplify
...
2x
2
4x2 + 4x − 15 = 15 − 15
4x2 + 4x − 15 = 0
−5
= 2
1
3
= 2
1
Simplify terms
...
416
...
Add 4 to both sides of the equation
...
Factor the trinomial expression using the trinomial
factor form
...
9x2 + 12x + 4 = −4 + 4
9x2 + 12x + 4 = 0
(3x + 2)(3x + 2) = 0
Simplify
...
3x
3
−2
= 3
−2
Simplify terms
...
236
501 Algebra Questions
417
...
Subtract 19x from both sides
...
Add 20 to both sides
...
Factor the trinomial expression using the
trinomial factor form
...
Simplify
...
3x
3
3x2 − 19x = 19x − 19x − 20
3x2 − 19x = −20
3x2 − 19x + 20 = −20 + 20
3x2 − 19x + 20 = 0
4
= 3
1
Simplify terms
...
x−5=0
Add 5 to both sides
...
x=5
1
2
The solutions for the quadratic equation 3x = 19x − 20 are x = 13
and x = 5
...
Transform the equation into the familiar
trinomial equation form
...
Simplify
...
Factor the trinomial expression using the
trinomial factor form
...
Using the zero product property, add 7 to
both sides
...
(4b − 7) = 0
4b − 7 + 7 = 0 + 7
4b = 7
Divide both sides by 4
...
b = 14
Now let the second term equal zero
...
b+3−3=0−3
Simplify
...
237
501 Algebra Questions
419
...
Subtract 7n from both sides of the equation
...
Simplify the equation
...
Factor the trinomial expression using the
trinomial factor form
...
Using the zero product property, add 3 to
both sides
...
(2n − 3) = 0
2n − 3 + 3 = 0 + 3
2n = 3
Divide both sides by 2
...
n = 12
Now set the second equal to zero
...
n+1−1=0−1
Simplify
...
420
...
Add 9b to both sides of the equation
...
Simplify
...
Using the zero product property, subtract 5
from both sides
...
(6b + 5) = 0
6b + 5 − 5 = 0 − 5
6b = −5
Divide both sides by 6
...
b = 6
Now set the second factor equal to zero
...
b+4−4=0−4
Simplify
...
238
501 Algebra Questions
421
...
Now factor the trinomial expression using the
trinomial factor form
...
5(3x + 4)(x − 6) = 0
Simplify
...
3x
3
−4
= 3
−4
Simplify terms
...
x−6+6=0+6
Simplify
...
–4
3
422
...
Subtract 52x from both sides of the equation
...
Simplify
...
Using the zero product property, add 3
to both sides
...
(7x − 3) = 0
7x − 3 + 3 = 0 + 3
7x = 3
Divide both sides by 7
...
x = 7
Now set the second factor equal to zero
...
x−7+7=0+7
Simplify
...
239
501 Algebra Questions
423
...
Add 36 to both sides of the equation
...
Factor out the greatest common factor from
each term
...
Ignore the numerical factor and set the first
factor equal to zero
...
Simplify terms
...
2z
2
Simplify terms
...
Subtract 2 from both sides
...
z = −12
(3z + 2) = 0
3z + 2 − 2 = 0 − 2
3z = −2
Divide both sides by 3
...
z = 3
36z2 + 78z + 36 = −36 + 36
36z2 + 78z + 36 = 0
6(6z2 + 13z + 6) = 0
6(2z + 3)(3z + 2) = 0
−3
= 2
1
−2
= 3
−2
1
The solutions for the quadratic equation 36z2 + 78z = −36 are z = −12
−2
and z = 3
...
Transform the equation into the
familiar trinomial equation form
...
12r2 + 40r − 192 = 192 − 40r + 40r − 192
Combine like terms
...
4(3r2 + 10r − 48) = 0
Now factor the trinomial
expression
...
(r + 6) = 0
Subtract 6 from both sides
...
r = −6
Now set the second factor
equal to zero
...
3r − 8 + 8 = 0 + 8
Simplify
...
3r
3
8
= 3
2
Simplify terms
...
240
501 Algebra Questions
425
...
Simplify terms
...
Combine like terms
...
Factor the trinomial expression
...
3(43x − 15)
24x2
3
= 3
2
8x = 43x − 15
8x2 + 15 − 43x = 43x − 15 + 15 − 43x
8x2 + 15 − 43x = 0
8x2 − 43x + 15 = 0
(8x − 3)(x − 5) = 0
8x − 3 = 0
8x
8
3
= 8
3
Simplify terms
...
x − 5 = 0
Add 5 to both sides
...
241
18
Simplifying
Radicals
This chapter will give you practice in operating with radicals
...
Nor do all trinomials with whole numbers have whole
numbers for solutions
...
The radical sign tells you to find the root of a number
...
Generally, a number has two
roots, one positive and one negative
...
The symbol − tells you
to find the negative root
...
Tips for Simplifying Radicals
Simplify radicals by completely factoring the radicand and taking out the
square root
...
Then you look for square roots that can be factored
out of the radicand
...
g
...
Then you can simplify their roots out of the radical sign
...
If you get rid of the denominator within the radical sign, you will no longer have a fractional radicand
...
2
x =
e
...
,
5
2
x 5 =
10
x = 1 · 10x = 10x
5 · 5
5
5
5
1
2
2
When there is a radical in the denominator, you can rationalize the expression as
follows:
3
√6
3
√6
3√6
1
= √6 · √6 = 6 = 26
If the radicands are the same, radicals can be added and subtracted as if the radicals were variables
...
g
...
e
...
, 43
· 75
= 4 · 73
· 5 = 2815
Quotient property of radicals
When dividing radicals, first divide the terms in front of the radicals, and then
divide the radicands
...
g
...
426
...
500
428
...
24x
5
243
501 Algebra Questions
34x
2
430
...
50a
432
...
5
12xy
434
...
33
65
25
436
...
(9a
)(3ab
)
438
...
· 32
9
440
...
√
2
√150
442
...
5
64
8
444
...
√72
446
...
3 · 3
4
· 27
10
√15 · √105
448
...
354
6
6√126
450
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
Underlined expressions show simplified result
...
First, factor the radicand
...
2
3
· 2 · = 23
500
= 5
·
· 10
10
427
...
5
·
· 10
10 = 105
Now take out the square root
...
First, factor the radicand
...
429
...
Now take out the
square root
...
Although it looks complex, you can still begin by factoring the terms in the
radical sign
...
2 · 2 = 23
3
Factoring out the squares leaves
This result can be written a few
different ways
...
First, factor the radicand and look for
squares
...
245
501 Algebra Questions
432
...
433
...
Use the identity property of
multiplication and multiply the expression
by 1 in a form useful for your purposes
...
3
5
=
3 5
5
5
=
35
5
434
...
Then see if it can be simplified
any further
...
Use the product property of radicals to
combine the radicands in the numerator
...
√12xy
√x
√12xy
√x
√12xy
√x
Factoring out the square roots results in
The x in the numerator and the denominator
divides out, leaving
√12xyx
· √x = x
√12xyx
x
Now factor the radicand
...
23y
...
In this expression, add the “like terms’’ as if
the similar radicals were similar
variables
...
436
...
Now simplify the radicand
...
437
...
2b
2b · b = 27aa
2b2 = 27a · ab = 27a2b
(9a)(3ab
) = 9 · 3aa
438
...
246
25
· 315
= 2 · 35
· 15
501 Algebra Questions
Now look for a perfect square in the radicand
...
65
· 15 = 675
= 625
· 3 = 6 · 53
= 303
Or if you just factor the
radicand, you will see the
perfect square as 5 times 5
...
You can start by using the product property
· 32
9 · 32 =
9
16
to simplify the expression
...
You should recognize the perfect squares 9
and 16
...
√16
√9
16
√16
= √
· 32
9 · 32
9
16
4
16√2
4
· 32
= 316
· 2 = 3 · 42
= 3
440
...
Now simplify the radicand by factoring so
that any perfect squares will come out of
the radical sign
...
441
...
√160
√2
√4 · 4 · 2 · 5
4√2 · 5
√2
Use the product property in the numerator
...
4√2 · 5
= √2 = √
2
4√2 · √5
= √2
4√2 · √5
√2
= 45
442
...
Rationalize the denominator
...
Factor the radicand seeking perfect
squares
...
Divide out the common factor in the
numerator and denominator and you’re done
...
You use the quotient property to begin
8
√8
5√4 · 2
=
5
8
√64
5√4 · 2
8
Simplify the numerator
...
444
...
The 5 becomes part of the numerator
...
2128
2
–2
5√2
= 4
· 128
·2
· =
denominator
...
2
· 2
2
Divide out the common factor in the numerator
and denominator
...
2
· 2 · 2
·2·
· 2 = –2 · 2 · 2 = –8
2
445
...
You can simplify the whole numbers in the
numerator and denominator by a factor of 3
...
−4
248
√3 · √2
√2
−43
−43
in one radical sign
...
Simplify the perfect squares in the numerator
and denominator
...
Use the commutative property of
multiplication
...
Using the product property, put all terms
3√3
3√3
6√2
Now rationalize the denominator
...
446
...
√9 · 3
=
=
√36 · 2
6√2
· 33
= −4 · 3 · 3
· 3
· 3 · 3
· 3
= −12 · 3 = −36
=
3 ·
3
3·3
4
10
4 · 10
√4 · 10
3 · 3 = √9
4 · 10
√4 · 10
√9
2√10
= 3
2
310
...
Begin by factoring the radicand of the
second radical
...
Why? Because
you will then have the product of two
identical radicals
...
= −57
449
...
Start this one by using
the product property to combine the
radicands
...
354
· 6 = 39
6
· 6 · = 33
6
·3·
·6
Simplify the radical
...
Then 3 times 18 equals 54
...
Begin by rationalizing the denominator
...
18
18
Simplify the whole numbers in the
numerator and denominator
...
18
3
3
Simplify the numerator and divide
out common factors in the
√9 · 9 · 2 · 2 · 7
9 · 2√7
18√7
= = = 67
numerator and denominator
...
However, the focus here is to use radicals to solve equations
...
When you
use a radical to solve an equation, you must be aware of the positive and
negative roots
...
When one of the solutions does not
work, it is called an extraneous solution
...
Tips for Solving Radical Equations
■
■
■
Squaring both sides of an equation is a valuable tool when
solving radical equations
...
Isolate the radical on one side of an equation before using the
squaring property
...
g
...
501 Algebra Questions
■
For second-degree equations, use the radical sign on both sides of the
equation to find a solution for the variable
...
Solve the following radical equations
...
451
...
x2 = 135
453
...
2a
= 24
455
...
4x
+6=8
457
...
5x
− 4 + 3 = 12
459
...
5x
− 6 + 3 = 11
461
...
33x
+ 1 = 15
463
...
3 = 10 − 100x
−
1
465
...
−7
= 10 − 25x
39
+
467
...
√5x + 1
469
...
x = 3x
+4
471
...
x = 7x
0
− 1
473
...
475
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
The solution is underlined
...
Use the radical sign on both sides of the equation
...
Check the first solution in the original equation
...
x2 = 49
x = ±7
(7)2 = 49
49 = 49
(−7)2 = 49
49 = 49
Both solutions, x = ±7, check out
...
Use the radical sign on both sides of the
equation
...
Check the first solution in the original
equation
...
Both solutions, x = ±315
, check out
...
First, square both sides of the equation
...
Check by substituting in the original equation
...
So when you substitute 121 into the original
equation, only the positive root n
= 11 is to be
considered
...
Although this may
seem trivial at this point, as the radical equations
become more complex, this will become important
...
Isolate the radical on one side of the equation
...
Simplify terms
...
Simplify terms
...
The solution a = 144 checks out
...
Begin by adding 4 to both sides to isolate the radical
...
Square both sides of the equation
...
Divide both sides by 2
...
The solution x = 32 checks out
...
(4x
)
+ 6 2 = 82
4x + 6 = 64
4x = 58
Divide both sides by 4 and simplify
...
5
456
...
Check the solution in the original equation
...
The solution x = 14
...
457
...
Square both sides of the equation
...
Subtract 4 from both sides and divide by 3
...
Simplify terms
...
253
)
+6=8
4(14
...
Subtract 3 from both sides of the equation
isolating the radical
...
Simplify terms on both sides
...
x = 5 = 17
Check your solution in the original equation
...
Find the positive square root of 81
...
The solution x = 17 checks out
...
Square both sides of the equation
...
Subtract 9 from both sides and then divide by 4
...
Simplify the expression under the radical sign
...
The solution does not check out
...
h
460
...
Square both sides of the equation
...
Add 6 to both sides and divide the result by 5
...
Simplify the expression under the radical
...
The solution x = 14 checks out
...
Subtract 14 from both sides to isolate the radical
...
Subtract 9 from both sides
...
Check the solution in the original equation
...
The square root of 121 is 11
...
254
85
(4x
)
+ 9 2 = (−13)2
4x + 9 = 169
x = 40
4(40
+
9) = −13
169
= −13
13 ≠ −13
5x
−6=8
(5x
− 6 2 = 82
)
5x − 6 = 64
x = 14
5(14)
−
6 + 3 = 11
64
+ 3 = 11
8 + 3 = 11
11 = 11
9
− x = 11
9 − x = 121
−x = 112
x = −112
9
− (−1 + 14 = 25
12)
121
+ 14 = 25
25 = 25
501 Algebra Questions
462
...
Square both sides of the equation
...
Subtract 1 from both sides of the equation and
divide by 3
...
Simplify the expression under the radical sign
...
The solution x = 8 checks out
...
Subtract 7 from both sides of the equation
...
Square both sides of the equation
...
Multiply both sides by negative 1
...
Simplify terms under the radical
...
The solution x = −25 checks out
...
Add 100x
−
1 to both
sides of the equation
...
3 + 100x
−
1 = 10
Now subtract 3 from both sides
...
100x − 1 = 49
Add 1 to both sides of the equation and
divide by 100
...
5
Check the solution in the original equation
...
5)
−1
Simplify the expression under the radical sign
...
3 = 10 − 7
The solution x = 0
...
3=3
465
...
Square both sides of the equation
...
Check the solution in the original equation
...
The solution x = 50 checks out
...
Add 25x
39
+ to both
sides of the equation
...
25x
39
+ − 7 = 10
Add 7 to both sides of the equation
...
25x + 39 = 289
Subtract 39 from both sides and divide
the result by 25
...
7 = 10 − 25(10)
+ 39
−
Simplify the expression under the radical sign
...
−7 = −7
The solution x = 10 checks out
...
To isolate the radical on one side of the equation,
add 4 to both sides and divide the result by 3
...
Subtract 43 from both sides and divide by 13
...
Simplify the expression under the radical sign
...
The solution x = 6 checks out
...
To isolate the radical on one side of the equation,
multiply both sides by 5x
+1
...
Square both sides of the equation
...
28 = 75x
+1
4 = 5x
+1
16 = 5x + 1
3=x
Check the solution in the original equation
...
28
√16
Divide the numerator by the positive square root of 16
...
7=7
469
...
Square
both sides
...
x2 + 2x − 8 = 0
The result is a quadratic equation
...
(Refer to
Chapter 16 for practice and tips for
256
=7
=7
=7
501 Algebra Questions
factoring quadratic equations
...
Let the first factor equal zero and solve
for x
...
Check the solution in the original equation
...
The radical sign calls for a positive root
...
x = −4 is an
example of an extraneous root
...
Subtract 2 from both sides
...
Evaluate the expression under the
radical sign
...
Therefore, the only solution for the
equation is x = 2
...
With the radical alone on one side of the
equation, square both sides
...
Put it into standard
form and factor the equation using the
trinomial factor form to find the
solutions
...
Letting each factor equal zero and solving
for x results in two possible solutions,
x = 4 and/or −1
...
The solution checks out
...
Therefore, x ≠ −1
...
The only solution for the original equation
is x = 4
...
Square both sides of the equation
...
x2 − x − 12 = 0
The resulting quadratic equation may have
up to two solutions
...
x2 − x − 12 = (x − 4)(x + 3) = 0
Let the first factor equal zero and
solve for x
...
x + 3 = 0, so x = −3
Check each solution in the original
equation to rule out an extraneous solution
...
4 = 16
The solution x = 4 checks out
...
(−3) = (12
You could simplify the expression under the
radical sign to get the square root of 9
...
So, x = −3 is
not a solution
...
Square both sides of the equation
...
Factor the quadratic equation to find the
solutions, and check each in the original
equation to rule out any extraneous
solution
...
The second factor will give the
solution x = 2
...
Simplify the expression under the
radical sign
...
Now check the second solution to the
quadratic equation in the original
...
There are two solutions to the original
equation,
258
x2 = 7x − 10
x2 − 7x + 10 = 0
x2 − 7x + 10 = (x − 5)(x − 2) = 0
(5) = 7(5)
−
10
5 = 25
or 5 = 5
(2) = 7(2)
−
10
2 = 4
or 2 = 2
x = 2 and x = 5
501 Algebra Questions
473
...
Transform the equation into a
quadratic equation
...
Let the first factor equal zero and
solve for x
...
When you substitute −0
...
That cannot be true for the
original equation, so x ≠ −0
...
Substitute 1
...
Simplify the terms on each side of
the equal sign
...
5
...
Square both sides of the equation
...
Multiply both sides of the equation
by 2 to simplify the fraction
...
Let the first factor equal zero and
solve for x
...
Check the first possible solution in
the original equation
...
4x + 3 = 4x2
4x2 − 4x − 3 = 0
4x2 − 4x − 3 = (2x + 1)(2x − 3) = 0
2x + 1 = 0, so x = −0
...
5
4(1
...
5)
9
= 3 or 3 = 3
7
2 − 2x = x2
7
0 = x2 + 2x − 2
0 = 2x2 + 7x − 4
2x2 + 7x − 4 = (2x − 1)(x + 4) = 0
1
2x − 1 = 0, so x = 2
x + 4 = 0, so x = −4
− 72(1
) = (2)
2
2
1
− 74 = 2
− 134 =
1 = 2
4
2
1
1
2
1
2
So x = is a solution
...
1
= 2
− 72( 4)
2
= ( 4)
−
Simplify the expression
...
There is one solution
1
for the original equation, x = 2
...
Square both sides of the equation
...
Multiply the equation by 2 to
eliminate the fraction
...
Letting each factor of the trinomial
factors equal zero results in two
possible solutions for the original
1
equation, x = 4 and/or x = −22
...
3
x2 = 2x + 10
3
x2 − 2x − 10 = 0
2x2 − 3x − 20 = 0
2x2 − 3x − 20 = (x − 4)(2x + 5) = 0
4=
3(4) + 10
2
Simplify the radical expression
...
Check the second possible solution
−21 =
3(−21) + 10
in the original equation
...
1
Therefore, x = −22 is not a solution
to the original equation
...
260
20
Solving Equations
with the
Quadratic Formula
In this chapter, you will have the opportunity to practice solving equations using the quadratic formula
...
The quadratic formula will allow you
to find solutions for any quadratic equation that can be put in the form
ax2 + bx + c = 0, where a, b, and c are numbers
...
The solutions found using the
a
quadratic formula are also called the roots of the equation
...
Some will be in the form
of a radical
...
The ± in the quadratic equation tells you that there will be
two solutions, one when you add the radical and one when you subtract it
...
Use the values
for a, b, and c in the quadratic equation to determine the solution for
the original equation
...
When you are asked to find the solution to the nearest hundredth, you
can use a calculator to find the value of the radical
...
Reduce answers to their
simplest form or to the simplest radical form
...
476
...
2x2 − 7x − 30 = 0
478
...
18x2 + 9x + 1 = 0
480
...
14x2 = 12x + 32
482
...
5x2 = 27
484
...
3x2 + 11x − 7 = 0
486
...
x2 = 2
488
...
x2 = 20x − 19
490
...
x2 + 10x + 11 = 0
492
...
7x2 = 4(3x + 1)
1
3
494
...
5x2 − 12x + 1 = 0
262
501 Algebra Questions
Find the solution to the following equations to the nearest hundredth
...
11r2 − 4r − 7 = 0
497
...
4y2 = 16y − 5
499
...
4c2 − 11c + 2 = 0
501
...
The operations performed within these symbols are intended to show how to evaluate the
various terms that make up the entire expression
...
Once a single number appears within these parentheses,
the parentheses are no longer needed and need not be used the next time the entire expression is
written
...
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions
...
The solutions are underlined
...
The equation is in the proper form
...
Substitute the values into the quadratic
equation
...
x=
–2 ± 4
– –32
2
Evaluate the square root of 36
...
First add the terms in the
numerator, and then subtract them
...
477
...
First, list the values for a, b, and c
...
Simplify the expression
...
The two solutions for the variable x
are x = 6 and x = -2
...
264
a=2
x=
b = −7 c = −30
−(−7) ± √(−7)2 − 4(2)(−30)
2(2)
x=
7 ± √49 − (−240)
4
7 + 17
24
7 − 17
−10
=
7 ± √289
4
x = 4 = 4 = 6 and
x = 4 = 4 = −2
...
The equation is in the proper form
...
Substitute the values into the
quadratic equation
...
Find the two solutions for x by
adding and then subtracting
in the numerator
...
479
...
First, list the values for a, b, and c
...
a = 18
Simplify the expression
...
b=9
c=1
x=
−(9) ± √(9)2 − 4(18)(1)
2(18)
x=
−9 ± √81 − 72
36
−9
=
−6
+3
−9 ± √9
36
=
−9 ± 3
36
−1
x = 36 = 36 = 6 and
−12
−1
x = 36 = 3
The two solutions for the variable x
−1
−1
are x = 6 and x = 3
...
First transform the equation into the
proper form
...
Combine like terms on both
sides
...
Substitute the values into the
quadratic equation
...
Find the two solutions for x by
adding and then subtracting
in the numerator
...
265
=
−17 ± 31
12
501 Algebra Questions
481
...
Add (−12x − 32)
to both sides of the equation
...
14x2 − 12x − 32 = 0
Now list the values for a, b,
and c
...
x =
2(14)
Simplify the expression
...
x=
12 ± √144 + 1,792
28
12 + 44
56
12 − 44
−32
12 ± 1,936
28
=
=
12 ± 44
28
x = 2
= 28 = 2 and
8
1
x = 28 = 28 = −17
1
The two solutions for the variable x are x = 2 and x = −17
...
However,
the solution would have been the same
...
482
...
But you
could write it as 4x2 + 5x + 0 = 0, and then
your values would be a = 4, b = 5, and c = 0
...
Simplify the expression
...
x=
−(5) ± √(5)2 − 4(4)(0)
2(4)
5 ± √25 − 0
x = −
=
8
−5 ± 5
8
−
5+5
0
x=
8 = 8 = 0 and
−
−
5−5
10
1
−
x=
8 = 8 = 14
The two solutions for the variable x are x = 0 and x = −114
...
Subtract 27 from both sides of the
equation
...
So you could write the equation in
the proper form like this:
5x2 + 0x − 27 = 0
Now list the values for a, b, and c
...
x =
2(5)
Simplify the expression
...
1
= ±126
The two solutions for the variable x
1
are x = ±126
...
Transform the equation into the
desired form
...
Combine like terms on both sides
...
Substitute the values into the
quadratic equation
...
Since there is no rational number
equal to the square root of a
negative number, there are
no solutions for this equation
...
List the values of a, b,
b = 11
=
18 ± √−16
10
c = −7
and c
...
a=3
x=
−(11) ± √(11)2 − 4(3)(−7)
2(3)
Simplify the expression
...
486
...
a = 5 b = 52 c = 20
Substitute the values into the
quadratic equation
...
x=
−(52) ± √(52)2 − 4(5)(20)
2(5)
−52
± √2,704 − 400
10
=
−52
± √2,304
10
Find the two solutions for x by
adding and then subtracting
−52 + 48
−4
in the numerator
...
4 and
52 − 48
−100
x = −10 = 10 = −10
The two solutions for the
variable x are x = −0
...
267
=
−52 ± 48
10
501 Algebra Questions
487
...
Then add 5x + 2 to both
sides of the equation
...
Substitute the values into
the quadratic equation
...
2x2 = −5x − 2
2x2 + 5x + 2 = 0
a=2 b=5 c=2
x=
−(5) ± √(5)2 − 4(2)(2)
2(2)
−5
−2
+3
−5 ± √25 − 16
2(2)
=
−1
−5
=
−5 ± 3
4
−8
−3
x = 4 = 4 = 2 and x = 4 = 4 = −2
1
The two solutions for the variable x are x = −2 and x = −2
...
Transform the equation by
subtracting 5 from both sides
...
Substitute the values into the
quadratic equation
...
x=
−8 ± 64
0
+ 2
2
x=
−8 ± 4
· 21
2
=
=
−8 ± 84
2
−8 ± 221
2
= −4 ± 21
The solution for the variable x is x = −4 ± 21
...
Transform the equation by subtracting
20x and adding 19 to both sides
of the equation
...
Substitute the values into the
quadratic equation
...
x = 2 = 2 = 2
Find the two solutions for x by adding
and then subtracting in the
20 + 18
38
numerator
...
490
...
Use the distributive property of multiplication on the right side of the equation
...
List the values of a, b, and c
...
Simplify the expression
...
491
...
a=1
Substitute the values into the
quadratic equation
...
−
−
c = 11
−(10) ± √(102 − 4(1)(11)
2(1)
x=
−
b = 10
−
10 ± √100 − 44
10 ± √56
10 ± √4 · 14
10 ± 2√14
= = = = −5 ± 14
x=
2
2
2
2
The two solutions for the variable x are x = −5 + 14
and x = −5 − 14
...
List the values of a, b, and c
...
Simplify the expression
...
c = −6
b = 18
x=
−(18) ± √(18)2 − 4(24)(−6)
2(24)
x=
−18 ± √324 + 576
48
−18
+ 30
12
−18
− 30
−48
−18 ± √900
48
=
=
−18 ± 30
48
1
x = 48 = 48 = 4 and
x = 48 = 48 = −1
1
The two solutions for the variable x are x = 4 and x = −1
...
Transform the equation into the
proper form
...
Then subtract (12x + 4) from
both sides
...
Substitute the values into the
quadratic equation
...
x=
12 ± √144 + 112
14
Find the two solutions for x by
adding and then subtracting in
the numerator
...
269
12 ± √256
14
=
12 ± 16
14
501 Algebra Questions
494
...
Using the distributive property,
you get
Simplify the terms
...
Substitute the values into the
quadratic equation
...
4x2 + 9x − 36 = 0
a = 4 b = 9 c = −36
−(9) ± √(9)2 − 4(4)(−36)
2(4)
x=
−9
−9
± √81 + 576
± √657
x = 8 = 8
Simplify the expression
...
a = 5 b = −12 c = 1
495
...
Substitute the values into the
−(−12) ± √(−12)2 − 4(5)(1)
quadratic equation
...
12 ± √144 − 20
12 ± √124
12 ± √4 · 31
12 ± 2√31
6 ± √31
= = = =
x=
2(5)
2(5)
2(5)
2(5)
5
6 + √31
6 − √31
The two solutions for the variable x are x = 5 and x = 5
...
List the values of a, b, and c
...
x=
−(−4) ± √(−4)2 − 4(11)(−7)
2(11)
4 ± √16 + 308
4 ± √324
4 ± 18
Simplify the expression
...
r = 22 = 22 = 1 and
4 − 18
−14
r = 22 = 22 = −0
...
64
...
List the values of a, b, and c
...
m=
Simplify the expression
...
17
...
c = −8
−(21) ± √(21)2 − 4(3)(−8)
2(3)
−21
± √441 + 96
−21
−21
+ 23
...
17
−21
− 23
...
17
± √537
m = 6 = 6
m = 6 = 6 = 0
...
36
The two solutions for the variable m are m = 0
...
36
...
Transform the equation by subtracting
4y2 − 16y + 5 = 16y − 5 − 16y + 5
4y2 − 16y + 5 = 0
a = 4 b = −16 c = 5
16y from and adding 5 to both sides
...
List the values of a, b, and c
...
Simplify the expression
...
32
...
32)
29
...
y = 8 = 8 = 3
...
32)
2
...
34
The two solutions for the variable y are y = 3
...
34
...
List the values of a, b, and c
...
s=
b = 12
c = −1
−(12) ± √(12)2 − 4(5)(−1)
2(5)
−12
± √144 + 20
−12
± √164
Simplify the expression
...
81
...
81
0
...
s = 10 = 1
0 = 0
...
81
−24
...
48
The two solutions for the variable s are s = 0
...
48
...
List the values of a, b, and c
...
c=
Simplify the expression
...
43
20
...
43
1
...
43
c = 8 = 8 = 2
...
20
The two solutions for the variable c are c = 2
...
20
...
List the values of a, b, and c
...
k=
−(−32) ± √(−32)2 − 4(11)(10)
2(11)
Simplify the expression
...
17
=
32 ± √584
22
56
...
55 and
32 − 24
...
8322 = 0
...
55 and k = 0
...
271
32 ± 24
Title: Algebra Question 501
Description: Algebra Question From Learning Express Newyork. The Learning Express Skill Builder in Focus Writing Team is comprised of experts in test preparation, as well as educators and teachers who specialize in language arts and math.
Description: Algebra Question From Learning Express Newyork. The Learning Express Skill Builder in Focus Writing Team is comprised of experts in test preparation, as well as educators and teachers who specialize in language arts and math.