# Name:_________________________ Period:________ Date:_____________ Algebra 1

```Name:_________________________
Algebra 1
Chapter 3 Test Review Solution Guide
Period:________
Date:_____________
Solve each equation – No Decimals!
1. 3( x  5)  2 x  15
3x  15  2x  15
5x  15  15
5x  30
x6
2.
3.
distribute the 3
combine like terms(3x + 2x)
divide both sides by 5
1
(28 x  24)  13 x  3
3
1

3 (28 x  24)   13 x  33
3

28x  24  39x  9
 24  11x  9
 33  11x
 33
x
11
3  x
5
x  7  33
8
5

8 x  7   338
8

5x  56  264
5x  320
x  64
multiply both sides by 3 to get rid of the fraction
distribute the 3 on both sides
subtract 9x on both sides
subtract 9 on both sides
divide both sides by 11
simplified the fraction
multiply both sides by 8 to get rid of fraction
distribute 8
divide by 5 on both sides
4. 4x  7  2x  5x  3  8x
2x  7  3x  3
5x  7  3
5x  10
x  2
combine like terms on both sides
subtract 7 on both sides
divide by 5 on both sides
5. 3x  2(1  x)  x  2
3x  2  2x  x  2
x2  x2
2  2
No Solution
distribute the 2 into each term in the parenthesis
combine like terms (3x – 2x)
subtract x on both sides
the numbers are not equal, so there is no solution for x
6.
5x  2 4 x  4

3
2
5
x

2

  4x  4 
6

6
 3   2 
2(5 x  2)  (4 x  4)3
10x  4  12x  12
 4  2x  12
 16  2x
8  x
7. 5 x  ( x  3)  5
5x  x  3  5
4x  3  5
4x  8
x2
8. 2(4 x  7)  (2  3x)  9
8x  14  2  3x  9
11x  16  9
11x  25
3
x2
11
multiply both sides by the common denominator 6
simplify with the denominators
distribute on both sides
subtract 10x on both sides
subtract 12 on both sides
divide both sides by 2
distribute the negative into the parenthesis
combine like terms (5x – x)
divide both sides by 4
distribute the 2 and the negative
combine like terms (8x + 3x) and (-14 – 2)
divide both sides by 11
9. 52( x  3)  ( x  2)  5x  20
52x  6  x  2  5x  20
5( x  4)  5 x  20
5x  20  5x  20
20  20
Identity
10. 5x  23(1  x)  2(1  x)  22
5x  23  3x  2  2x  22
5 x  2(5 x  1)  22
5x  10x  2  22
 5x  2  22
 5x  20
x  4
distribute the 2 and the negative in the brackets
combine like terms in brackets and change the
brackets to parenthesis
distribute the 5 in the parenthesis
subtract 5x on both sides
Whenever to constants are remaining and they
are equal, then it is an identity
distribute the 3 and the -2 inside the brackets
combine like terms in the brackets
distribute the 2 into the parenthesis
combine like terms (5x – 10x)
subtract 2 on both sides
divide both sides by -5
11. 2(5 x  4)  3( x  5)  8(2 x  7)
10x  8  3x  15  16x  56
7 x  7  16x  56
7  9x  56
63  9x
7x
12.
13.
x  1 3x  6 3


10
5
2
 x  1 3x  6   3 
10

   10
5  2
 10
( x  1)  2(3x  6)  (3)5
x  1  6x  12  15
 5x  13  15
 5x  2
2
x
5
3x  1 4 x  2 3


3
5
5
 3x  1 4 x  2   3 
15

   15
5  5
 3
5(3x  1)  3(4 x  2)  (3)3
15x  5  12x  6  9
3x  1  9
3x  10
1
x3
3
distribute the 2, -3, and 8 into the parenthesis
combine like terms (10x – 3x) and (-8 + 15)
subtract 7x on both sides
divide by 9 on both sides
multiply both sides by 10 to get rid of fractions
distribute the 10 and simplify
distribute the -2 in the parenthesis
Combine like terms (x – 6x) and (1 – 12)
divide both sides by -13
Multiply both sides by the common denominator
Simplify denominators and 15
Distribute the 5 and -3
Combine like terms (15x – 12x) and (3 – 6)
Divide by 3 on both sides
14. 23( x  2)  ( x)  5( x  3)  2(1  x)
23x  6  x  5x  15  2  2x Distributed the 3, negative, 5, and -2
2(4 x  6)  7 x  13
Combine like terms
Distributed the 2 into the parenthesis
8x  12  7 x  13
Subtracted 7x on both sides
x  12  13
Subtract 12 on both sides
x 1
15. The sum of three consecutive odd integers is 81. Find the integers!
(n) + (n+2) + (n+4) = 81
3n + 6 = 81
3n = 75
n = 25
So, 25, 27, and 29 are the three consecutive integers that add up to 81!!!
16. One train leaves the station at noon traveling 60 mph. At 2:00 pm a second train
leaves the station on a parallel track traveling 90 mph. How long will it take the
faster train to overtake the slower one?
Rate
60 mph
90 mph
Slower Train
Faster Train
60t = 90(t – 2)
60t = 90t – 180
-30t = -180
t = 6 hours
Time
t
t–2
Distance
60t
90(t – 2)
(This is the time for SLOWER train!!)
t–2
=6–2
= 4 hours for the faster train to catch up to the slower train!)
17. The length of a rectangle is 5 more than twice the width. The perimeter is 46.
Find the length and width of the rectangle.
L = 5 + 2W
P = 2L + 2W
46 = 2(5 + 2W) + 2W
46 = 10 + 4W + 2W
46 = 10 + 6W
36 = 6W
6=W
L = 5 + 2W
L = 5 + 2(6)
L = 5 + 12
L = 17
```