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COLLEGE ALGEBRA ǀ MATH 114 ǀ NOTRE DAME UNIVERSITY
PROBLEMS WITH SOLUTIONS TO COLLEGE ALGEBRA TEXTBOOK BY PAUL R
...
A
Solve for x if possible:
1
...
) 2x + 19 = 35
SOLUTION:

STEPS:

2x + 19 = 35
2x = 35 – 19

transpose constant term and change its sign

2x = 16

perform arithmetic operations
1

( 2x = 16 ) (2)

multiply both side by constant’s reciprocal of x

x=8

and evaluate for the value of x as a Solution if any

Since x has a value of 8, then it is the solution that we will see on the graph using Geogebra App
3
...
) 6x + 13 = 11x – 10
SOLUTION:

STEPS:

6x + 13 = 11x – 10
6x – 11x = – 10 – 13

transpose constant term and collect like terms

-5x = – 23

perform arithmetic operations for like terms
1
−5

( -5x = -23) ( )

x=

𝟐𝟑

𝟑

or 𝟒 𝟓

𝟓

multiply both side by constant’s reciprocal of x
and evaluate for the value of x as a Solution if any

Since x has a value of 23/5 , then it is the solution that we will see on the graph using Geogebra App
5
...
) x2 + 4x + 5 = (x+2)(x+7)
SOLUTION:

STEPS:

x2 + 4x + 5 = (x+2)(x+7)
x2 + 4x + 5 = 7x + 2x + 14

simplify the expression

x2 - x2 + 4x - 7x - 2x = 14 – 5

transpose constant term and collect like terms

-5x = 9
1
−5

multiply both side by constant’s reciprocal of x

( -5x = 9) ( )
𝟗

𝟒

x = − 𝟓 or −𝟏 𝟓

and evaluate for the value of x as a Solution if any

Since x has a value of -9/5 , then it is the solution that we will see on the graph using Geogebra App

7
...
) (2x + 3)(3x+2) = (x - 5)(6x + 1)
SOLUTION:

STEPS:

(2x + 3)(3x+2) = (x - 5)(6x + 1)
6x2 + 4x + 9x + 6 = 6x2 + x - 30x – 5

simplify the expression

6x2 - 6x2 + 4x + 9x - x + 30x = -5 – 6

transpose constant term and collect like terms

42x = -11
1

multiply both side by constant’s reciprocal of x

( 42x = -11) (42)
𝟏𝟏

x = - 𝟒𝟐

and evaluate for the value of x as a Solution if any

Since x has a value of - 11/42 , then it is the solution that we will see on the graph using Geogebra App
9
...
) (x + a)(x + b) = (x + c)(x + d)
SOLUTION:

STEPS:

(x + a)(x + b) = (x + c)(x + d)
x2 + bx + ax + ab = x2 + dx + cx + cd

simplify the expression

x2 - x2 + bx + ax - dx - cx = cd - ab

transpose constant term and collect like terms

(b + a - d - c) x = cd - ab

factor out x from the expression

[ (b + a - d - c) x = cd - ab ] (

x=

1
)
b+a−d−c

𝐜𝐝−𝐚𝐛

multiply both side by b + a – d –c reciprocal
and evaluate for the value of x as a Solution if any

𝐛+𝐚−𝐝−𝐜

11
...
) (2x + 5)2 –x(4x – 5) = 100
SOLUTION:

STEPS:

(2x + 5)2 –x(4x – 5) = 100
4x2 + 20x + 25 –x(4x – 5) = 100

expand the binomial expression and apply

4x2 + 20x + 25 - 4x2 + 5x = 100

distribution property

4x2 - 4x2 + 20 x + 5x = 100 - 25

transpose constant term and collect like terms

1

( 25x = 75) (25)

multiply both side by constant’s reciprocal of x

x=3

and evaluate for the value of x as a Solution if any

13
...
) (2x – a)2 – 4x2 = x + 4a
SOLUTION:

STEPS:

(2x – a)2 – 4x2 = x + 4a

expand the binomial expression

4x2 - 4ax + a2 - 4x2 = x + 4a

transpose constant term and collect like terms

- 4ax - x = 4a - a2
(- 4a – 1) x = 4a - a2

factor out x from the expression

[ (- 4a – 1) x = 4a - a2 ] (

x=

1
)
−4a−1

𝟒𝐚 − 𝐚𝟐

multiply both side by 1 / (-4a-1) assuming -4a-1 ≠ 0
and evaluate for the value of x as a Solution if any

− 𝟒𝐚 – 𝟏

15
...
) x2 + a(a – 2x) = a2 + x(x – 2a)
SOLUTION:

STEPS:

x2 + a(a – 2x) = a2 + x(x – 2a)

apply distribution property

x2 + a2 – 2ax = a2 + x2 – 2ax
x2 – x2 – 2ax + 2ax = a2 - a2

transpose constant term and collect like terms

0+0=0

and evaluate for the value of x as a Solution if any

0=0
It means that this equation has No Solution
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) (x – 5)(5 – x) = 5(2x – 3) – x2
SOLUTION:

STEPS:

(x – 5)(5 – x) = 5(2x – 3) – x2

apply distribution property

5x – x2 – 25 + 5x = 10x – 15 – x2
-x2 + x2 + 5x + 5x – 10x = -15 + 25

transpose constant term and collect like terms

0 + 10x – 10x = 10

and evaluate for the value of x as a Solution if any

0 + 0 = 10
0 = 10
This statement is false
...


18
...
Therefore, this equation has No Solution
...
) Solve for r : P(1 + rn) = A
SOLUTION:

STEPS:

P(1 + rn) = A
[ P(1 + rn) = A ] (

A

A
P

–1](

1
n

1
n

)

–1)(

P

A− P
P

r=(

transpose constant term “1”

-1

A

r =(

r=

P

P

[ rn =

multiply both side by 1/P assuming P ≠ 0

)

A

1 + rn =
rn =

1
P

) (

1
n

multiply both side by 1/n assuming n ≠ 0

)

)

evaluate and simplify terms on the right side if any

𝐀− 𝐏
𝐏𝐧

20
...
) Solve for ꭍ: S =

𝑛
2

(𝑎+ ꭍ )

SOLUTION:
S=

𝑛
2

STEPS:

(𝑎+ ꭍ )
𝑛

2
n

[S=2 (𝑎+ ꭍ )](
2𝑆

)

multiply both side by 2/n

= 𝑎+ ꭍ

n
2𝑆

transpose variable “a”

–a = ꭍ

n

𝟐𝑺

∫=

- a=

𝐧

𝟐𝑺 −𝐧𝐚

evaluate and simplify terms on the right side if any

𝐧
1

22
...
) Solve for d: p + ghd + 2 v 2 d = c
SOLUTION:

STEPS:

1

p + ghd + 2 v 2 d = c

transpose variable “p”

1

ghd + 2 v 2 d = c – p

factor out d from the expression

1
2

(gh + v 2 ) d = c – p
1

[ (gh + 2 v 2 ) d = c – p ] (

1
1
2

gh + v2

d=

c−p
1
gh + v2
2

𝟐(𝐜−𝐩)

d = 𝟐𝐠𝐡 + 𝐯𝟐

=

c−p
1
gh + v2
2

2

)

2(c−p)

* 2 = 2gh + v2

multiply both side by 1 / (gh +

1 2
v )
2

evaluate and simplify terms on the right side if any

24
...
(a) Solve this equation for F

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