Download Solving Equations Chapter Test Review Directions: Complete this

Unit 1: Solving Equations
Solving Equations Chapter Test Review
Directions: Complete this Review on your own paper. Go back and review the lesson for any
answers that are incorrect.
Part 1: Two Step Equations and Equations with the Distributive Property (Less 1-7)
1. -3x + 7 = -21
2. 5(x+4) = -10
3. -5(y+6) + 3y = 12
4. -15.5 = 2(-3.1x + 4) – 8
Part 2: Solving Equations with Fractions (Less 8)
1. ¼(3x – 1) – ¾ = -4
2.
+ 9 =
Part 3: Solving Literal Equations and Using Formulas (Less 9)
1. Solve for F.
2. Solve for W.
C = 5/9(F – 32)
P = 2(L+ W)
3. The formula for converting temperatures from the Kelvin scale to the Celsius scale is: C = K-273.
•
•
Solve this formula for K.
Suppose you wanted to change 280° C to the Kelvin scale. What would the equivalent
temperature be in degrees Kelvin?
4. If the length of a rectangle measures 4x – 6 units and the width measures 2x+1, units find the
measurement of the length if the perimeter is 98 units.
Part 4: Solving Equations with Variables on Both Sides (Less 10)
1. 8x + 3 = 2(-2x – 5) + 4
2. 4 − = + 6
Part 5: Word Problems (Less 11)
1. A technician charges $16.50 an hour plus parts to repair televisions. The parts for John’s TV cost
$155. The total bill was $196.25. How many hours did the technician work on John’s TV?
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Unit 1: Solving Equations
2. Jenny has $25 and is saving at the rate of $9 per week. Brian has $100 and is spending $6 per
week. After how many weeks will Jenny and Brian have the same amount of money? How much will
they have?
3. A rectangle has a width of 4x. The length of the rectangle is 3 less than 2 times the width. The
perimeter of the rectangle is 114 units. What is the length of the longest side of the rectangle?
Part 6: Absolute Value Equations (Lesson 12)
1. Solve for x:
|x – 3| = 7
2. Solve for y:
|y + 9| - 4 = 11
3. Solve for s: |s – 5| + 10 = 4
4. Solve for x: 4|x – 7| = 36
5. Solve for y: 6|y+2| -10 = -4
6. 3|2x-4| = 9x+27
7. Which set of numbers is the solution to the following equation: |3x – 8| +4 = 15
A. {1, -19/3}
B.{-1, 19/3}
C. {-11/3, 19/3}
Copyright © 2011 Karin Hutchinson – Algebra-class.com
D. {11/3, -19/3}
Unit 1: Solving Equations
Solving Equations Chapter Test Review – Answer Key
Directions: Complete this Review on your own paper. Go back and review the lesson for any
answers that are incorrect.
Part 1: Two Step Equations and Equations with the Distributive Property (Less 1-7)
1. -3x + 7 = -21
2. 5(x+4) = -10
-3x + 7 -7 = -21 – 7
Subtract 7
5x +20 = -10
Distribute 5
-3x = -28
Simplify: (-21-7=-28)
5x+20-20 = -10 -20
Subtract 20
-3x/-3 = -28/-3
Divide by -3
5x = -30
Simplify: (-10-20 = -30)
5x/5 = -30/5
Divide by 5
x = 28/3 or 9.3
x = -6
3. -5(y+6) + 3y = 12
4. -15.5 = 2(-3.1x + 4) – 8
-5y – 30 + 3y = 12
Distribute -5
-15.5 = -6.2x +8 – 8
Distribute 2
-2y – 30 = 12
Combine -5y+3y = -2y
-15.5 = -6.2x
Simplify: 8-8 = 0
-2y – 30+30 = 12+30
Add 30
-15.5/-6.2 = -6.2x/-6.2
Divide by -6.2
-2y = 42
Simplify: 12+30=42
2.5 = x
-2y/-2 = 42/-2
Divide by -2
Y = -21
Part 2: Solving Equations with Fractions (Less 8)
1. ¼(3x – 1) – ¾ = -4
4[1/4(3x-1)- 1 – ¾] = -4(4)
2.
Multiply by 4
3x-1 – 3 = -16
Simplify
3x – 4 = -16
Combine: -1 – 3 = -4
3x-4+4 = -16+4
Add 4
3x = -12
Simplify: -16+4=-12
3x/3 = -12/3
Divide by 3
X = -4
Simplify
Copyright © 2011 Karin Hutchinson – Algebra-class.com
+ 9 =
6[ + 9] = (6)
Multiply by 6
2p + 54 = 15
Simplify
2p +54-54 = 15-54
Subtract 54
2p = -39
Simplify: 15-54 = -39
2p/2 = -39/2
Divide by 2
P = -39/2 or -19.5
Unit 1: Solving Equations
Part 3: Solving Literal Equations and Using Formulas (Less 9)
1. Solve for F.
2. Solve for W.
C = 5/9(F – 32)
P = 2(L+ W)
c/(5/9) = 5/9(F-32)/5/9
9/5c = f – 32
9/5c + 32 = f – 32+32
9/5c +32 = f
Divide by 5/9
Simplify
Add 32
Or if you get rid of the fraction first:
9(C) = 9[5/9(F-32)]
Multiply by 9
9C = 5(F-32)
Simplify
9c = 5f – 160
Distribute 5
9c +160= 5f – 160+160
Add 160
9c +160 = 5f
Simplify
(9c +160)/5 = 5f/5
Divide by 5
(9c+160)/5 = f
Simplifies to: 9/5c + 32 = f
P/2 = 2(L+W)/2
P/2 = L+ W
P/2 – L = L-L +W
P/2 – L = W
Divide by 2
Simplify
Subtract L
Or if you distribute first:
P = 2L + 2W
Distribute 2
P – 2L = 2L – 2L +2W
Subtract 2L
P-2L = 2W
Simplify
(P-2L)/2 = 2W/2
Divide by 2
(P-2L)/2 = W
Simplify
Simplifies to: P/2 – L = W
3. The formula for converting temperatures from the Kelvin scale to the Celsius scale is: C = K-273.
•
Solve this formula for K.
C = K – 273
Original Problem
C +273 = K -273+273
Add 273 to both sides
C + 273 = K
•
Suppose you wanted to change 280° C to the Kelvin scale. What would the equivalent
temperature be in degrees Kelvin?
K = C +273
where C = 280
K = 280 + 273
K = 553 degrees
The equivalent temperature would be 553 degrees Kelvin.
4. If the length of a rectangle measures 4x – 6 units and the width measures 2x+1, units find the
measurement of the length if the perimeter is 98 units.
L = 4x-6 W = 2x+1
P = 98
P = 2L + 2W
Perimeter Formula
98 = 2(4x-6) + 2(2x+1)
Substitute for P, L and W
98 = 8x – 12 + 4x +2
Distribute
98 = 12x – 10
Combine like terms
98 +10 = 12x – 10 +10
Add 10 to both sides
108 = 12x
Simplify
108/12 = 12x/12
Divide by 12
9 = x Since x = 9, the length equals: 4(9) – 6 which is: 30 units. The length is 30 units.
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Unit 1: Solving Equations
Part 4: Solving Equations with Variables on Both Sides (Less 10)
1. 8x + 3 = 2(-2x – 5) + 4
2. 4 − = + 6
8x +3 = -4x -10 + 4
Distribute 2
8x + 3 = -4x-6
Combine: -10+4 = -6
8x +3-3 = -4x -6-3
Subtract 3
8x = -4x -9
Simplify: -6-3 = -9
8x +4x = -4x+4x – 9
Add 4x
12x = -9
Simplify: 8x+4x = 12x
12x/12 = -9/12
Divide by 12
X = -9/12 or -3/4
3[4 − ] = 3[ + 6]
Multiply by 3
12 – y = 2y + 18
Simplify
12-18 – y = 2y +18-18
Subtract 18
-6 – y = 2y
Simplify: 12-18 = -6
-6 – y + y = 2y + y
Add y
-6 = 3y
Simplify: 2y+y = 3y
-6/3 = 3y/3
Divide by 3
-2 = y
Part 5: Word Problems (Less 11)
1. A technician charges $16.50 an hour plus parts to repair televisions. The parts for John’s TV cost
$155. The total bill was $196.25. How many hours did the technician work on John’s TV?
First write an equation: rate(# hours) + parts = total
16.50 x + 155 = 196.25
Let x = # of hours
Equation for this problem – Now solve for x.
16.50x + 155 – 155 = 196.25-155
Subtract 155 from both sides
16.50x = 41.25
Simplify: 196.25-155 = 41.25
16.50x/16.50 = 41.25/16.50
Divide by 16.50
X = 2.5
It took the technician 2.5 hours to work on John’s tv.
2. Jenny has $25 and is saving at the rate of $9 per week. Brian has $100 and is spending $6 per
week. After how many weeks will Jenny and Brian have the same amount of money? How much will
they have?
Equation for Jenny: 9x+25
Equation for Brian: 100 – 6x
Let x = number of weeks
Since we want to know when they will have the same amount of money, we will set them equal to each
other.
9x+25 = 100-6x
Now solve for x.
9x +6x +25 = 100 – 6x +6x
Add 6x to both sides
15x +25 = 100
Simplify
15x + 25 – 25 = 100 – 25
Subtract 25
15x = 75
Simplify: 100-25 = 75
15x/15 = 75 / 15
Divide by 14
X=5
After 5 weeks they will have the amount of money. They will have $70.
9(5)+25 = 70 or 100-6(5) = 70
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Unit 1: Solving Equations
3. A rectangle has a width of 4x. The length of the rectangle is 3 less than 2 times the width. The
perimeter of the rectangle is 114 units. What is the length of the longest side of the rectangle?
W = 4x
L = 2(4x) – 3 or 8x-3 (3 less than 2 times the width)
P = 2L + 2W
Perimeter formula
114 = 2(8x-3) + 2(4x)
Substitute for P, L and W
114 = 16x -6 + 8x
Distribute 2
114 = 24x -6
Combine: 16x+8x = 24x
114 +6 = 24x – 6+6
Add 6 to both sides
120 = 24x
Simplify: 114+6 = 120
120/24 = 24x/24
Divide by 24
P = 114
5=x
The problem asks for the length of the longest side. Let’s substitute:
W = 4(5)
L = 8(5) - 3
W = 20
L = 37
The length of the longest side is 37 units.
Part 6: Absolute Value Equations
1. Solve for x:
|x – 3| = 7
The absolute value is already isolated on the left hand side, so we can write two linear equations:
X–3=7
or
x – 3 = -7
x – 3+3 = 7+3
x – 3 +3 = -7 + 3
x = 10
x = -4
{-4, 10}
Copyright © 2011 Karin Hutchinson – Algebra-class.com
Unit 1: Solving Equations
2. Solve for y:
|y + 9| - 4 = 11
We must first isolate the absolute value on the left hand side by adding 4 to both sides.
|y +9 | - 4 + 4 = 11 + 4
Add 4 to both sides
|y+9| = 15
Simplify
y+9 = 15
or
y+9 = -15
2 linear equations
y+9 -9 = 15- 9
y +9 -9 = -15 – 9
Subtract 9
y=6
y = -24
3. Solve for s: |s – 5| + 10 = 4
We must first isolate the absolute value on the left hand side by subtracting 10 from both sides.
|s - 5 | +10 – 10 = 4-10
Subtract 10 from both sides
|s-5| = - 6
We can stop here. An absolute value expression cannot be equal to a negative number. The solution will
be the empty set.
{Ø}
4. Solve for x: 4|x – 7| = 36
We must first isolate the absolute value on the left hand side by dividing by 4 on both sides.
4|x-7| = 36
4
4
Divide by 4 on both sides
|x-7| = 9
X–7=9
X – 7+7 = 9+7
X = 16
Result of dividing by 4
or
x – 7 = -9
x – 7 +7 = -9 +7
x = -2
{-2, 16}
Copyright © 2011 Karin Hutchinson – Algebra-class.com
2 linear equations
Add 7 to both sides
Unit 1: Solving Equations
5. Solve for y: 6|y+2| -10 = -4
6|y+2| -10 +10 = -4 +10
Add 10 to begin isolating the absolute value on the left side
6|y +2| = 6
Simplify
6|y+2| = 6
6
6
Divide by 6 to isolate the absolute value
|y+2| = 1
y +2 = 1
y+2 -2 = 1-2
y = -1
Simplify
or
y+2 = -1
y +2 -2 = -1 -2
y = -3
2 linear equations
Subtract 2
Simplify
{-3, -1}
6. 3|2x-4| = 9x+27
3|2x – 4| = 9x +27
3
3
Divide by 3 on both sides in order to isolate the absolute value
|2x – 4| = 3x +9
Result of dividing by 3
2x – 4 = 3x +9
2x – 3x – 4 = 3x – 3x +9
-x – 4 = 9
-x – 4+4 = 9+4
-x = 13
-x/-1 = 13/-1
x = -13
or
(Subtract 3x)
(Simplify)
(Add 4)
(Divide by -1)
2x – 4 = -(3x + 9)
2x – 4 = -3x -9
2x +3x – 4 = -3x +3x – 9
5x – 4 = -9
5x – 4+4 = -9 +4
5x = -5
5x/5 = -5/5
x = -1
2 linear equations
Distribute -1
Add 3x
Simplify
Add 4
Simplify
Divide by 5
**When you check the answers, -13 does not work. Therefore, the only solution is: {-1}
7. Which set of numbers is the solution to the following equation: |3x – 8| +4 = 15
A. {1, -19/3}
B.{-1, 19/3}
C. {-11/3, 19/3}
|3x – 8| = 4 – 4 = 15 – 4
D. {11/3, -19/3}
Subtract 4 from both sides
|3x-8| = 11
3x – 8 = 11
3x – 8 +8 = 11+8
3x = 19
3x/3 = 19/3
x = 19/3
or
{-1, 19/3}
Copyright © 2011 Karin Hutchinson – Algebra-class.com
3x – 8 = -11
3x -8 +8 = -11+8
3x = -3
3x/3 = -3/3
x = -1
2 linear equations
Add 8 to both sides
Simplify
Divide by 3