Algebra 1
Unit 1: Solving Equations
Lesson 12: Solving and Graphing Absolute Value Equations
Definition of Absolute Value
The absolute value of a number is _______________________________________________________
The absolute value of a number is always __________________________________________________
Representation of Absolute Value
Example 1: Evaluate an Expression with
Absolute Value
REMEMBER:
|x| = x
and
|-x| = x
**Absolute value bars act as ______________
_______________________. Perform all
operations inside the absolute value bars first.
Example 2: Solve the equation for x. Then
check your solutions.
|x | = 9
Evaluate: 2|3a – 2| + 3
if a = -5
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1
Unit 1: Solving Equations
Example 3: Solve the equation for x. Then
check your solutions.
Example 4: Solve the equation for x. Then
check your solutions.
|y + 6| = 9
|½y -3| - 5 = 9
Check Your Answers:
Check Your Answers:
Example 5
Example 6
|2x +5| +9 = 0
A chicken that you are cooking must reach an
internal temperature of 165°. A typical meat
thermometer is accurate within plus or minus 2.5°.
Write an equation to determine the least and
greatest temperatures of the meat.
•
•
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Solve the equation.
To what temperature should you bake the
chicken to ensure that the minimal
temperature is reached?
Algebra 1
Unit 1: Solving Equations
Lesson 12: Solving and Graphing Absolute Value Equations – Practice Problems
Part 1: Decide whether the given number is a solution to the equation. (Hint: substitute the
number into the equation. If the equation represents a true statement, then the number is a
solution.)
1. |2x-8| = 2
where x = -5
3. |2/3x – 4| = 4/3
5.
where x = 8
½ |3-4y| + 7 = 25
2. |6 – 4y| = 30
where y =-7
4. 3|8x +4| - 6 = 30
where x = -2
where y = -9
Part 2: Rewrite the absolute value equation as two linear equations. (This step will get you ready
for solving the equations)
1. |4x – 7| = 9
2. |-3x + 4| = 16
3. |2/3y + 8|- 10 = 20
4. 6 + |1/4x – 8| = 30
Part 3: Solve the following equations and check your answers.
1. |3x – 10| = 34
2. |1/2s – 12| = 5
3. |-4x + 3| = 57
4. |.45x + 4| = 13
5. |8r – 5| - 12 = 17
6. |5x – 3| + 10 = 3
7. |3/4y – 9| - 8 = -7/8
8. |.5x + 4| + 4 = 8.25
9. 18 + |.25x – 10| = 24
10. |4x – 2.5| - 12 = 7.5
Part 4: Write an absolute value equation for each problem. Then solve the equation.
1. In order to brew coffee, you must have a brewing temperature of 195° F, plus or minus 5
degrees.
• Write an equation that could be used to find the minimum and maximum temperatures of a
cup of coffee freshly brewed.
• Find the maximum and minimum temperatures for brewing coffee.
2. The typical length of a high school gymnasium is 90 feet, give or take 10 feet.
• Write an equation that could be used to find the minimum and maximum lengths of a high
school gymnasium. Then find the minimum length of a gymnasium.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1
Unit 1: Solving Equations
3. A box of cereal weighing 16 ounces, may not exactly weigh 16 ounces. The weight of a box of
cereal can vary plus, or minus .4 ounces.
• Write an equation that could be used to determine the minimum and maximum weights of a
box of cereal.
• If you bought 3 boxes of cereal, what is the total maximum weight of all three boxes
together?
1. Decide whether the two answers given are a solution to the
equation. Explain your answer. (3 points)
|9 – 3x| + 5 = 38
Solution set: {-8, -14}
2. Solve the following equations. Check your answers. (3 points each)
1. |2/3x – 9| = 21
2. 12 + |9x + 5.5| = 29
3. Normal body temperature is 98.6°F. Your body temperature can change by as much a 1°F
throughout the day. (3 points)
• Write an equation that can be used to determine the minimum and maximum normal body
temperature of a person on any given day.
• Find the maximum normal body temperature of a person.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1
Unit 1: Solving Equations
Lesson 12: Solving and Graphing Absolute Value Equations – Practice Problems
Answer Key
Part 1: Decide whether the given number is a solution to the equation. (Hint: substitute the
number into the equation. If the equation represents a true statement, then the number is a
solution.)
1. |2x-8| = 2
where x = -5
|2(-5)-8| = 2
|-18| = 2
Substitute -5 for x.
Evaluate inside of |abs|
18 = 2
2. |6 – 4y| = 30
|6-4(-7)| = 30
|34| = 30
34 ≠ 30
|-18| = 18
where y =-7
Substitute -7 for y.
Evaluate inside of |abs|
|34| = 34
18 ≠ 2
Since the two sides are not equal, we can
conclude that -5 is not a solution to this
equation.
3. |2/3x – 4| = 4/3
|2/3(8) - 4| = 4/3
|4/3| = 4/3
where x = 8
Substitute 8 for x.
Evaluate inside of |abs|
4/3 = 4/3
|4/3| = 4/3
Since the two sides are equal, we can
conclude that 8 is a solution to this equation.
5.
Since the two sides are not equal, we can
conclude that -7 is not a solution to this
equation.
½ |3-4y| + 7 = 25
4. 3|8x +4| - 6 = 30
3|8(-2)+4| - 6=30
3|-12| -6 = 30
Substitute -9 for y
½|39| +7 = 25
Evaluate inside of |abs|
½(39) + 7 = 25
|39| = 39
26.5 = 25
½(39) + 7 = 26.5
Substitute -2 for x.
Evaluate inside of |abs|
3(12) – 6 = 30
|-12| = 12
30 = 30
3(12) -6 = 30
Since the two sides are equal, we can
conclude that -2 is a solution to this equation
where y = -9
½|3 – 4(-9)| + 7 = 25
where x = -2
26.5 ≠ 25
Since the two sides are not equal, we can
conclude that -9 is not a solution to this
equation.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1
Unit 1: Solving Equations
Part 2: Rewrite the absolute value equation as two linear equations. (This step will get you ready
for solving the equations)
1. |4x – 7| = 9
2. |-3x + 4| = 16
The answer inside the absolute value sign can
be positive 9 or negative 9. So, let one equation
equal 9 and the other equation equal -9.
4x-7 = 9
4x – 7 = -9
The answer inside the absolute value sign can
be positive 16 or negative 16. So, let one
equation equal 16 and the other equation equal
-16.
-3x +4 =16
-3x +4 = -16
Only difference is one answer is 9 and the other
-9.
3. |2/3y + 8|- 10 = 20
4. 6 + |1/4x – 8| = 30
Step 1: Isolate the absolute value on the left
side of the equal sign by adding 10 to both
sides.
Step 1: Isolate the absolute value on the left
side of the equal sign by subtracting 6 from both
sides.
|2/3y+8|-10 +10 = 20+10
Add 10
|2/3y+8| = 30
20+10 = 30
6 - 6 + |1/4x – 8| = 30 – 6
|1/4x – 8| = 24
Subtract 6
30-6 = 24
The answer inside the absolute value sign can
be positive 30 or negative 30. So, let one
equation equal 30 and the other equation equal
-30.
The answer inside the absolute value sign can
be positive 24 or negative 24. So, let one
equation equal 24 and the other equation equal
-24.
2/3y+8 = 30
1/4x-8 = 24
2/3y +8 = -30
1/4x-8 = -24
Part 3: Solve the following equations and check your answers (using a calculator if possible).
1. |3x – 10| = 34
2. |1/2s – 12| = 5
Step 1: Since the absolute value is isolated on
the left side, we can write two equations.
Step 1: Since the absolute value is isolated on
the left side, we can write two equations.
3x-10 = 34
1/2s - 12 = 5
3x – 10 = -34
3x-10 = 34
3x – 10 + 10 = 34 + 10
Add 10
3x = 44
Simplify
3x/3 = 44/3
Divide by 3
x = 44/3
______________________________________
3x – 10 = -34
3x – 10 + 10 = -34 + 10
Add 10
3x = -24
Simplify
3x/3 = -24/3
Divide by 3
x = -8
Solution Set: {44/3, -8}
1/2s - 12 = -5
1/2s - 12 = 5
1/2s – 12 + 12 = 5+12
Add 12
1/2s = 17
Simplify
(2)1/2s = 17(2)
Multiply by 2
s = 34
______________________________________
1/2s - 12 = -5
1/2s – 12 + 12 = -5+12
Add 12
1/2s = 7
Simplify
(2)1/2s = 7(2)
Multiply by 2
s = 14
Solution Set: {14, 34 }
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1
Unit 1: Solving Equations
3. |-4x + 3| = 57
4. |.45x + 4| = 13
Step 1: Since the absolute value is isolated on
the left side, we can write two equations.
Step 1: Since the absolute value is isolated on
the left side, we can write two equations.
-4x +3 = 57
.45x +4 = 13
-4x+3 = -57
-4x+3 = 57
-4x +3 – 3 = 57 – 3
Subtract 3
-4x = 54
Simplify
-4x/-4 = 54/-4
Divide by -4
x = -27/2 or -13.5
Simplify
______________________________________
-4x+3 = -57
-4x +3 – 3 = -57 – 3
Subtract 3
-4x = -60
Simplify
-4x/-4 = -60/-4
Divide by -4
x = 15
Simplify
Solution Set: {-13.5, 15}
.45x +4 = -13
.45x +4 = 13
.45x +4 – 4 = 13 – 4
Subtract 4
.45x = 9
Simplify
.45x/.45 = 9/.45
Divide by .45
x = 20
Simplify
______________________________________
.45x +4 = - 13
.45x +4 – 4 = -13 – 4
Subtract 4
.45x = -17
Simplify
.45x/.45 = -17/.45
Divide by .45
x = 37.78
Simplify
Solution Set: {20, 37.78}
5. |8r – 5| - 12 = 17
6. |5x – 3| + 10 = 3
Step 1: Isolate the absolute value on the left
side by adding 12 to both sides.
Step 1: Isolate the absolute value on the left
side by subtracting 10 from both sides.
|8r – 5| - 12 + 12 = 17+ 12
|5x-3| +10 -10 = 3-10
|8r – 5| = 29
|5x-3| = -7
New Equation
Step 2: Write 2 linear equations.
8r – 5 = 29
8r – 5 = -29
8r – 5 = 29
8r – 5 + 5 = 29 + 5
Add 5
8r = 34
Simplify
8r/8 = 34/8
Divide by 8
r = 17/4 or 4.25
Simplify
______________________________________
8r – 5 = -29
8r – 5 + 5 = -29 + 5
Add 5
8r = -24
Simplify
8r/8 = -24/8
Divide by 8
r =-3
Simplify
Solution Set: {-3, 17/4}
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
New Equation
Stop Here!!! And think……
Can any expression inside of an absolute value
sign equal a negative number?
NO, the absolute value of any expression must
be positive; therefore, we don’t need to go any
further. The solution set is: the empty set:
{Ø}.
Algebra 1
Unit 1: Solving Equations
7. |3/4y – 9| - 8 = -7/8
8. |.5x + 4| + 4 = 8.25
Step 1: Isolate the absolute value on the left
side by adding 8 to both sides.
Step 1: Isolate the absolute value on the left
side by subtracting 4 from both sides.
|3/4y – 9| - 8 +8 = -7/8 +8
|3/4y – 9| = 57/8
|.5x + 4| + 4 -4 = 8.25 - 4
|.5x + 4|= 4.25
New Equation
New Equation
Step 2: Write 2 linear equations.
3/4y – 9 = 57/8
3/4y – 9 = -57/8
Step 2: Write 2 linear equations.
.5x + 4= 4.25
.5x + 4= - 4.25
3/4y – 9 = 57/8
3/4y -9 + 9 = 57/8 + 9
Add 9
3/4y = 129/8
Simplify
(4/3)3/4y = 129/8(4/3)
Multiply by 4/3
y = 43/2 or 21.5
Simplify
______________________________________
3/4y – 9 = -57/8
3/4y -9 + 9 = -57/8 + 9
Add 9
3/4y = 15/8
Simplify
(4/3)3/4y = 15/8(4/3)
Multiply by 4/3
y =5/2 or 2.5
Simplify
.5x + 4 = 4.25
.5x + 4 – 4 = 4.25 – 4
Subtract 4
.5x = .25
Simplify
.5x/.5 = .25/.5
Divide by .5
x= .5 or 1/2
______________________________________
.5x + 4 = - 4.25
.5x + 4 – 4 = -4.25 – 4
Subtract 4
.5x = -8.25
Simplify
.5x/.5 = -8.25/.5
Divide by .5
x= -16.5
Solution Set: {-16.5, 0.5}
Solution Set: {5/2, 43/2}
9. 18 + |.25x – 10| = 24
10. |4x – 2.5| - 12 = 7.5
Step 1: Isolate the absolute value on the left
side by subtracting 18 from both sides.
Step 1: Isolate the absolute value on the left
side by adding 12 to both sides.
18 - 18 + |.25x – 10| = 24 - 18
|.25x – 10| = 6
New Equation
|4x – 2.5| - 12 + 12 = 7.5+12
|4x – 2.5| = 19.5
New Equation
Step 2: Write 2 linear equations.
.25x – 10 = 6
.25x – 10 = -6
Step 2: Write 2 linear equations.
4x – 2.5 = 19.5
4x – 2.5 = -19.5
.25x – 10 = 6
.25x – 10 + 10 = 6 + 10
Add 10
.25x = 16
Simplify
.25x/.25 = 16/.25
Divide by .25
x = 64
Simplify
______________________________________
.25x – 10 = -6
.25x – 10 + 10 = -6 + 10
Add 10
.25x = 4
Simplify
.25x/.25 = 4/.25
Divide by .25
x = 16
Simplify
4x – 2.5 = 19.5
4x – 2.5 + 2.5= 19.5 +2.5
Add 2.5
4x = 22
Simplify
4x/4 = 22/4
Divide by 4
x = 11/2 or 5.5
Simplify
______________________________________
4x – 2.5 = -19.5
4x – 2.5 + 2.5= -19.5 +2.5
Add 2.5
4x = -17
Simplify
4x/4 = -17/4
Divide by 4
x = -17/4 or -4.25
Simplify
Solution Set: {16, 64}
Solution Set: {-17/4, 11/2} or {-4.25, 5.5}
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1
Unit 1: Solving Equations
Part 4: Write an absolute value equation for each problem. Then solve the equation.
1. In order to brew coffee, you must have a brewing temperature of 195° F, plus or minus 5
degrees.
• Write an equation that could be used to find the minimum and maximum temperatures of a
cup of coffee freshly brewed.
In this problem we know that the difference between the actual temperature and 195 must be with plus or
minus 5 degrees. Therefore, we know our variable – 195 must be within the absolute value sign, and that
value must equal 5 degrees.
Let t = the brewing temperature of the coffee.
|t – 195| = 5
•
is the equation we can use to find the minimum and maximum temperatures.
Find the maximum and minimum temperatures for brewing coffee.
We can solve this equation to find the minimum and maximum temperatures of the coffee. Let’s solve for t.
|t – 195| = 5
t – 195 = 5
t – 195 = -5
Write two linear equations.
t – 195 +195 = 5 + 195
t – 195+195 = -5 + 195
Add 195 to both equations
t = 200
t = 190
Simplify
The minimum temperature for brewing coffee is 190 degrees and the maximum temperature for
brewing coffee is 200 degrees.
2. The typical length of a high school gymnasium is 90 feet, give or take 10 feet.
• Write an equation that could be used to find the minimum and maximum lengths of a high
school gymnasium. Then find the minimum length of a gymnasium.
In this problem, we know that difference between the length of the gymnasium and 90 feet must be plus or
minus 10 feet.
Let x = the actual length of a gymnasium.
|x – 90| = 10 is the equation we can use to find the minimum or maximum length of the gymnasium.
Now, we can solve to find minimum length of the gymnasium.
|x-90| = 10
x – 90 = 10
x – 90 = -10
Write two linear equations.
x – 90 + 90 = 10 + 90
x – 90 + 90 = -10 + 90
Add 90 to both sides
x = 100
x = 80
Simplify
The minimum length of a gymnasium is 80 feet.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1
Unit 1: Solving Equations
3. A box of cereal weighing 16 ounces, may not exactly weigh 16 ounces. The weight of a box of
cereal can vary plus, or minus .4 ounces.
• Write an equation that could be used to determine the minimum and maximum weights of a
box of cereal.
In this problem, we know that the difference between the actual weight of a box of cereal and 16 is plus or
minus .4 ounces.
Let x = the actual weight of the cereal.
|x – 16| = .4
cereal.
•
is the equation that we can use to determine the minimum and maximum weights of the
If you bought 3 boxes of cereal, what is the total maximum weight of all three boxes
together?
Using the equation from above: | x- 16| = .4
x – 16 = .4
x – 16 = -.4
Write two linear equations
x – 16 + 16 = .4 + 16
x – 16 + 16 = -.4 + 16
Add 16
x = 16.4
x = 15.6
Simplify
The maximum weight of one box is 16.4 ounces. Therefore, the maximum weight of 3 boxes of cereal
would be 49.2 ounces (16.4 * 3 = 49.2)
1. Decide whether the two answers given are a solution to the
equation. Explain your answer. (3 points)
|9 – 3x| + 5 = 38
Solution set: {-8, -14}
In order to decide whether the given answers are solution to the equation, we must substitute each solution
into the original equation. If the expression is true, then the solution set is correct.
|9 – 3x| + 5 = 38
|9 – 3(-8)| + 5 = 38
|33| +5 = 38
33+5 = 38
38 = 38
Solution set: {-8, -14}
Substitute -8 for x.
This is a true statement
|9 – 3x| + 5 = 38
|9 – 3(-14)| + 5 = 38
|51| +5 = 38
51+5 = 38
56 ≠ 38
Substitute -14
Since 56 does not equal 38, -14 is not a
solution. Therefore, the solution set is
incorrect. -8 is a solution, but -14 is not a
solution.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com
Algebra 1
Unit 1: Solving Equations
2. Solve the following equations. Check your answers. (3 points each)
1. |2/3x – 9| = 21
2. 12 + |9x + 5.5| = 29
Step 1: Since the absolute value is isolated on
the left side, we can write two equations.
Step 1: Isolate the absolute value on the left
side by subtracting 12 from both sides.
2/3x – 9 = 21
12 - 12+ |9x + 5.5| = 29 - 12
|9x + 5.5| = 17
New Equation
2/3x – 9 = -21
2/3x – 9 = 21
2/3x – 9 +9 = 21 +9
Add 9
2/3x = 30
Simplify
(3/2)2/3x = 30(3/2)
Multiply by 3/2
x = 45
______________________________________
2/3x – 9 = -21
2/3x – 9 +9 = -21 +9
Add 9
2/3x =-12
Simplify
Multiply by 3/2
(3/2)2/3x = -12(3/2)
x = -18
Step 2: Write 2 linear equations.
9x + 5.5 = 17
9x + 5.5 = -17
9x + 5.5 = 17
9x + 5.5 – 5.5 = 17 – 5.5
Subtract 5.5
9x = 11.5
Simplify
9x / 9 = 11.5 / 9
Divide by 9
x = 23/18 or 1.28
______________________________________
9x + 5.5 = -17
9x + 5.5 – 5.5 = -17-5.5
Subtract 5.5
9x = -22.5
Simplify
9x / 9 = -22.5 / 9
Divide by 9
x = -2.5
Solution Set: {-18, 45}
Solution Set: {-2.5, 1.28}
3. Normal body temperature is 98.6°F. Your body temperature can change by as much a 1°F
throughout the day. (3 points)
• Write an equation that can be used to determine the minimum and maximum normal body
temperature of a person on any given day.
We know that the difference in actual body temperature and 98.6 degrees is plus or minus 1 degree.
Let x = actual body temperature.
|x – 98.6| = 1 is the equation used to determine the minimum and maximum body temperature.
•
Find the maximum normal body temperature of a person.
|x – 98.6| = 1
Given equation
x – 98.6 = 1
x – 98.6 = -1
Write two linear equations
x – 98.6 + 98.6 = 1 + 98.6
x- 98.6 + 98.6 = -1 + 98.6
Add 98.6
x = 99.6
x = 97.6
Simplify.
The minimum normal body temperature of a person is 97.6°F and the maximum normal body
temperature is 99.6°F.
Copyright © 2009-2012 Karin Hutchinson – Algebra-class.com