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Algebra 1
Inequalities
Lesson 7: Solving Absolute Value Inequalities
Absolute Value Inequalities with Less Than (<)
|x| < 5
Absolute Value Inequalities with Greater Than (>)
|x| > 5
For all real numbers, a and b, if b>0, then the following statements are true.
1. If |a| < b, then:
-b < a < b
2. If |a| > b, then: a>b or a< -b
EXAMPLES
If | x+3| <2, then: -2 < x+3 < 2
If |x+3| > 2, then:
x+3 > 2
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
or x+3 < -2
Algebra 1
Inequalities
Example 1
Solve and graph the following absolute value inequality:
|2x – 5| < 10
Example 2
Solve and graph the following absolute value inequality:
|2x – 6| > 8
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
Example 3
**Isolate the absolute value FIRST***
|2x – 2| + 3 > 9
Example 4
Example 5
Josie is going on vacation to Mexico in January. The
average temperature is 73°F, but the highs and lows
each day can differ from the average by 9°F.
|x +2| > -3
•
•
Write an absolute value inequality to describe
the weather in Mexico on any given day in
January.
Solve the inequality to find the range of
temperatures for Mexico in January.
|x + 2| < -3
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
Lesson 7: Solving Absolute Value Inequalities – Practice Problems
Part 1: Write an absolute value inequality for each of the following:
1. all numbers between -7 and 7.
2. all numbers greater than 4 or less than -4
3. all numbers less than 20.5 and greater than -20.5
4. all numbers less than -12 or greater than 12
5.
6.
7.
8.
Part 2: Solve each inequality and graph the solution on a number line.
1. |2 + 5x| < 13
2. |4n – 12| > 18
3. |6-4n| > 22
4. |-5y-8| < 6
5. |-3y + 20| > -5
6. |-8y +2| +1 < 9
7. |3y -1| < -3
8. 5 + | -5y +2| < 10
9. |9y +6| - 4 > -2
10. |3(3y – 9) +1| -1 > 13
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
11. |2(4y – 8)| < 6
12. |1/2(4y – 6)| +2 > 1
Part 3: Write an absolute value inequality to match each problem. Then solve.
1. John is applying for a job that has an average starting salary of $56,000, but the actual starting
salary may vary from the average by as much as $3500. Write an absolute value inequality that
represents this situation. Find the range of John’s possible starting salary.
2. A typical box of cereal that is advertised as weighing 24 ounces, may often vary by .5 ounces.
(Manufacturers have a “tolerance” of about .5 ounces per box of cereal.) Write and solve an
absolute value inequality that describes the acceptable weights for a 24 ounce box of cereal.
3. A juice bottler has a tolerance of .02 liters for every two liter bottle manufactured. Write an
absolute value inequality describing the capacity of bottles that are outside of the acceptable
range.
Part 1: Write an absolute value inequality to describe each
graph. (2 points each)
1.
2.
Part 2: Solve and graph each absolute value inequality. (3 points each)
1. |-3x – 5| > 4
2. 3 + |2x – 3| < 10
3. The recommend weight range for a regulation volley ball is between 260 and 280 grams. Write
an absolute value inequality for the weight range of a volleyball.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
Lesson 7: Solving Absolute Value Inequalities – Practice Problems
Answer Key
Part 1: Write an absolute value inequality for each of the following:
1. all numbers between -7 and 7.
Since this says all number “between”, I know that I can use the less than (<) symbol.
|x| < 7
(translates to: -7< x< 7 which means “all numbers between -7and 7”)
2. all numbers greater than 4 or less than -4
This statement uses the word “ or” which indicates that we need to write 2 different inequalities. Therefore,
we must use the greater than (>) symbol.
|x| > 4
(translates to: x >4 or
x<-4)
3. all numbers less than 20.5 and greater than -20.5
The word “and” tells me that this is an intersection of sets. Therefore, I must use the less than (<) symbol.
|x| < 20.5
(translates to: -20.5 < x < 20.5
or x < 20.5 and
x > -20.5)
4. all numbers less than -12 or greater than 12
This statement uses the word “ or” which indicates that we need to write 2 different inequalities. Therefore,
we must use the greater than (>) symbol.
|x| > 12
(which translates to: x >12 or x < -12)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
5.
This graph shows a union of two sets, which means that we are using the word “or”. This also means that
we must use the greater than (>) symbol.
|x| > 12
(translates to: x> 12 or x < -12)
6.
This graph shows an intersection of sets which means that we must use the less than (<) symbol.
|x| < 18
(translates to: -18 < x < 18)
7.
This graph shows an intersection of sets which means that we must use the less than (<) symbol.
|x| < 2
(translates to: -2 < x < 2)
8.
This graph shows a union of two sets, which means that we are using the word “or”. This also means that
we must use the greater than (>) symbol.
|x| > 10
(translates to: x> 10 or x < -10)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
Part 2: Solve each inequality and graph the solution on a number line.
1. |2 + 5x| < 13
This absolute value inequality has the less than (<) symbol. Therefore, we will solve it using an intersection of
sets:
-13 < 2+5x < 13
-13 -2 < 2-2 +5x < 13-2
Subtract 2 from all three sides.
-15 < 5x < 11
Simplify
-15/5 <x < 11/5
Divide by 5 on all three sides
-3 < x < 2.2
Simplify
2. |4n – 12| > 18
This absolute value inequality has the greater than (>) symbol. Therefore, we will solve it using a union of sets
and the word “or”.
4n – 12 > 18
OR
4n – 12 < -18
4n – 12 +12 > 18+12
Add 12
4n – 12 +12 < -18 +12
Add 12
4n > 30
Simplify
4n < -6
Simplify
4n/4 > 30/4
Divide by 4
4n/4 < -6/4
Divide 4
n > 7.5
Or
n< -1.5
n > 7.5 Or n< -1.5
3. |6-4n| > 22
This absolute value inequality has the greater than (>) symbol. Therefore, we will solve it using a union of sets
and the word “or”.
6-4n > 22
OR
6-4n < -22
6 -6 – 4n > 22 – 6
Subtract 6
6-6-4n < -22 – 6
Subtract 6
-4n > 16
Simplify
-4n < -28
Simplify
-4n/-4 > 16/-4
Divide by -4
-4n/-4 < -28/-4
Divide by -4
n < -4
OR
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
n>7
Algebra 1
Inequalities
4. |-5y-8| < 6
This absolute value inequality has the less than (<) symbol. Therefore, we will solve it using an intersection of
sets:
-6 < -5y – 8 < 6
-6 +8 < -5y – 8 +8 < 6+8
Add 8 to all three sides
2 < -5y < 14
Simplify
2/-5 < -5y/-5 < 14/-5
Divide by -5 on all three sides
-.4 > y > -2.8
Reverse the sign when you divide by a negative number.
This can be rewritten as: -2.8 < y < -.4 (when you flip sides, you must reverse the
sign)
5. |-3y + 20| > -5
This absolute value inequality has the greater than (>) symbol. Therefore, we will solve it using a union of sets
and the word “or”.
-3y +20 > -5
OR
-3y +20 < -5
BUT: We can STOP here because we know that all absolute values are positive or zero. Therefore, |-3y+20|
must be greater than -5 because it will be a positive number (or 0).
The solution is: All real numbers
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
6. |-8y +2| +1 < 9
First you must isolate the absolute value by subtracting 1 from both sides:
|-8y +2| +1 – 1 < 9-1
Subtract 1 from both sides
|-8y +2| < 8
This absolute value inequality has the less than (<) symbol. Therefore, we will solve it using an intersection of
sets:
-8< -8y +2 < 8
-8 – 2 < -8y +2-2 < 8-2
Subtract 2 from all three sides
-10 < -8y < 6
Simplify
-10/-8 < -8y/-8 < 6/-8
Divide by -8 on all three sides
1.25 > y > -.75
Reverse the inequality sign when you divide by a negative number
This can also be rewritten as: -.75 < y < 1.25
7. |3y -1| < -3
STOP: We don’t need to solve this absolute value inequality. Since we know that all absolute values must be
positive or zero, we know that |3y -1| ≠ a negative number. This statement says that the absolute value will be
less than -3, which means it would have to be negative.
The solution set is the empty set. There are no real solutions. {Ø}
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
8. 5 + | -5y +2| < 10
First you must isolate the absolute value by subtracting 5 from both sides:
5 -5 + |-5y +2| < 10-5
Subtract 5 from both sides
|-5y +2| < 5
This absolute value inequality has the less than (<) symbol. Therefore, we will solve it using an intersection of
sets:
-5 < -5y+2 < 5
- 5 -2 < -5y +2-2 < 5-2
Subtract 2 from all three sides
-7 < -5y < 3
Simplify
-7/-5 < -5y/-5 < 3/-5
Divide by -5 on all three sides
1.4 > y > -.6
Reverse the inequality sign when you divide by a negative number
This can also be rewritten as: -.6 < y < 1.4
9. |9y +6| - 4 > -2
First we must isolate the absolute value on the left side by adding 4 to both sides.
|9y + 6 | -4+4 > -2 +4
Add 4 to both sides
|9y +6| > 2
This absolute value inequality has the greater than (>) symbol. Therefore, we will solve it using a union of sets
and the word “or”.
9y+6 > 2
OR
9y +6 < -2
9y +6 – 6 > 2 – 6
Subtract 6
9y + 6 – 6 < -2 - 6
9y > -4
Simplify
9y < -8
9y/9 > -4/9
Divide by 9
9y/9 < -8/9
y >- 4/9 (-.44)
y < -8/9 (.89)
y > -4/9 (.44) OR
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
y < -8/9 (-.89)
Algebra 1
Inequalities
10. |3(3y – 9) +1| -1 > 13
First we must isolate the absolute value on the left side by adding 1 to both sides.
|3(3y – 9) +1| -1 +1 > 13 +1
Add 1 to both sides
|3(3y – 9) +1| > 14
This absolute value inequality has the greater than (>) symbol. Therefore, we will solve it using a union of sets
and the word “or”.
3(3y – 9)+1 > 14
OR
3(3y – 9)+1 < -14
9y – 27 +1 > 14
Distribute the 3
9y – 27 +1 < -14
Distribute 3
9y -26 > 14
Combine like terms
9y – 26 < -14
like terms
9y – 26 +26 > 14+26
Add 26
9y – 26+26 < -14+26
9y > 40
Simplify
9y < 12
9y/9 > 40/9
Divide by 9
9y/9 < 12/9
y > 40/9 or 4.44
Add 26
Divide by 9
y < 12/9 or 1.3
y > 40/9 (4.44) OR
y < 12/9 (1.33)
11. |2(4y – 8)| < 6
This absolute value inequality has the less than (<) symbol. Therefore, we will solve it using an intersection of
sets:
-6 < 2(4y – 8) < 6
-6 < 8y – 16 < 6
Distribute the 2
-6 +16 < 8y -16+16 < 6+16
Add 16 to both sides
10 < 8y < 22
Simplify
10/8 < 8y/8 < 22/8
Divide by 8 on both sides
1.25 < y < 2.75
Simplify
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
12. |1/2(4y – 6)| +2 > 1
First we must isolate the absolute value on the left side by subtracting 2 from both sides.
|1/2(4y – 6)| +2 - 2 > 1 -2
Subtract 2 from both sides
|1/2(4y – 6)| > -1
** We can stop here because the absolute value expression is greater than -1. We know that an
absolute value expression cannot be negative, but it will always be greater than -1. So the solution
is: all real numbers.
Part 3: Write an absolute value inequality to match each problem. Then solve.
1. John is applying for a job that has an average starting salary of $56,000, but the actual starting
salary may vary from the average by as much as $3500. Write an absolute value inequality that
represents this situation. Find the range of John’s possible starting salary.
Let x= John’s starting salary
|Average salary – John’s starting salary |
by as much as
|56,000 – x|
≤
$3500
3500
The absolute value equation is: |56,000 – x| ≤ 3500
-3500 ≤ 56000 – x ≤ 3500
-3500 -56000 ≤ 56000 -56000- x ≤ 3500 – 56000
Subtract 56000 from all sides
-59500 ≤ -x ≤ -52500
Simplify
-59500/-1 ≤ -x/-1 ≤ -52500/-1
Divide by -1
59500 ≥ x≥ 52500
52500 ≤ x ≤ 59500
John’s starting salary range is between $52500 and $59500.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Reverse sign since you divided by -1
Flip sides and Flip sign
Algebra 1
Inequalities
2. A typical box of cereal that is advertised as weighing 24 ounces, may often vary by .5 ounces.
(Manufacturers have a “tolerance” of about .5 ounces per box of cereal.) Write and solve an
absolute value inequality that describes the acceptable weights for a 24 ounce box of cereal.
Let x= John’s starting salary
|Advertised weight – actual weight |
|24 – x|
by as much as
.5
.5
≤
The absolute value equation is: |24 – x| ≤ .5
-.5 ≤ 24 – x ≤ .5
-.5 -24 ≤ 24-24- x ≤ .5-24
Subtract 24 from all sides
--24.5 ≤ -x ≤ -23.5
Simplify
--24.5/-1 ≤ -x/-1 ≤ -23.5/-1
Divide by -1
24.5 ≥ x≥ 23.5
Reverse sign since you divided by -1
Flip sides and Flip sign
23.5 ≤ x ≤ 24.5
The acceptable weights for a 24 ounce box of cereal are between 23.5 ounces and 24.5 ounces.
3. A juice bottler has a tolerance of .02 liters for every two liter bottle manufactured. Write an
absolute value inequality describing the capacity of bottles that are outside of the acceptable
range.
Let x = actual capacity of 2 liter bottles
|2 – actual capacity| > .02
We are using greater than because we want to
know what capacities are outside of the
acceptable range.
|2 – x| > .02
Split into 2 inequalities.
2-x > .02
Or
2-x < -.02
2-2 –x > .02 – 2
Subtract 2
2-2 –x < -.02 – 2
Subtract 2
-x > -1.98
Simplify
-x < -2.02
Simplify
-x/-1 > -1.98 / -1
Divide by -1
-x/-1 < -2.02/-1
Divide by -1
x < 1.98
x > 2.02
Capacities outside of the acceptable range are capacities less than 1.98 ounces or greater than 2.02
ounces.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
Part 1: Write an absolute value inequality to describe each
graph. (2 points each)
1.
-35 ≤ x ≤ 35 is the compound inequality
(Use less than since it is an intersection of sets.)
|x| ≤ 35
2.
X ≤ -10 OR
|x| ≥ 10
x ≥ 10 is the compound inequality
(use greater than since it’s a union of sets and uses the word “or:”)
Part 2: Solve and graph each absolute value inequality. (3 points each)
1. |-3x – 5| > 4
This absolute value inequality has the greater than (>) symbol. Therefore, we will solve it using a union of sets
and the word “or”.
-3x - 5 > 4
OR
-3x – 5 < -4
-3x – 5+5 > 4+5
Add 5
-3x – 5+5 < -4+5
Add 5
-3x > 9
Simplify
-3x < 1
Simplify
-3x/-3 > 9/-3
Divide by -3
-3x/-3 < 1/-3
Divide by -3
x < -3
Reverse sign
x > -1/3
Reverse sign
x< -3 or x > -1/3
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Algebra 1
Inequalities
2. 3 + |2x – 3| < 10
We must first isolate the absolute value on the left side by subtracting three from both sides.
3-3 +| 2x – 3|< 10 -3
Subtract 3 from both sides
|2x -3| < 7
Simplify
-7 < 2x -3 < 7
Write as compound inequality since it’s a less than symbol
-7 +3 < 2x – 3 + 3 < 7+3
Add 3 to all three sides
-4 < 2x < 10
Simplify
-4/2 < 2x/2 < 10/2
Divide by 2 on all three sides
-2 < x < 5
Simplify
3. The recommend weight range for a regulation volley ball is between 260 and 280 grams. Write
an absolute value inequality for the weight range of a volleyball.
In this problem, you are basically given the answer. The weight range is between 260 and 270 grams. So we
need to figure out the number in the “middle” so that we can write the absolute value inequality.
The number in the middle is 270 with a gain or loss of 10 grams either way. Therefore, the absolute value
inequality is:
| 270 – x| < 10
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