SOLVING ABSOLUTE VALUE INEQUALITIES

The general form of an absolute value inequality are

• |ax + b| ≤ k
• |ax + b| ≥ k
• |ax + b| > k
• |ax + b| < k

For example :

If |ax + b| ≤ k

Rewrite the inequalities, remove the absolute value symbol, and divide it into two branches.

ax + b ≤ k (or) ax + b ≥ -k

Solve each inequality and write the solution set if the variable is an element of the set of integers.

Problem 1 :

│x│ > 9

Solution :

x > 9 (or) x < -9

Hence the solution is {….-12, -11, -10, 10, 11, 12…}.

Problem 2 :

│y + 2│ > 7

Solution :

y + 2 > 7 (or) y + 2 < -7

y > 5 (or) y < -9

Hence the solution is {….-12, -11, -10, 6, 7, 8….}.

Problem 3 :

│b + 6│ ≤ 5

Solution :

b + 6 ≤ 5 (or) b + 6 ≥ -5

b ≤ -1 (or) b ≥ -11

Hence the solution is -11 ≤ b ≤ -1.

{…-11, -10, -9,…-3, -2, -1}

Problem 4 :

│x - 3│ < 4

Solution :

x - 3 < 4 (or) x - 3 > -4

x < 7 (or) x > -1

Hence the solution is -1 < x < 7.

{0, 1, 2, 3, 4, 5, 6}

Problem 5 :

│y + 6│ > 13

Solution :

y + 6 > 13 (or) y + 6 < -13

y > 7 (or) y < -19

Hence the solution is {…-22, -21, -20, 8, 9, 10…}.

Problem 6 :

│2b - 7│ ≥ 9

Solution :

2b - 7 ≥ 9 (or) 2b - 7 ≤ -9

2b ≥ 16 (or) 2b ≤ -2

b ≥ 8 (or) b ≤ -1

Hence the solution is {…-3, -2, -1, 8, 9, 10…}.

Problem 7 :

│6 - 3x│ < 15

Solution :

6 - 3x < 15 (or) 6 - 3x > -15

-3x < 9 (or) -3x > -21

-x < 3 (or) x > 7

x > -3 (or) x < 7

Hence the solution is -3 < x < 7.

{-2, -1, 0, 1, 2, 3, 4, 5, 6}

Problem 8 :

│8 + 4b│ ≥ 0

Solution :

8 + 4b ≥ 0 (or) 8 + 4b ≤ -0

4b ≥ -8 (or) 4b ≤ -8

b ≥ -2 (or) b ≤ -2

Hence the solution is set of all integers.

Problem 9 :

│5 - b│ + 4 < 9

Solution :

│5 - b│ + 4 < 9

Subtract 4 from each side.

│5 - b│ < 5

5 - b < 5 (or) 5 - b > -5

-b < 0 (or) -b > -10

b > 0 (or) b < 10

Hence the solution is 0 < b < 10.

{1, 2, 3, 4, 5, 6, 7, 8, 9}

Problem 10 :

│11 - 2b│ - 6 > 11

Solution :

│11 - 2b│ - 6 > 11

Add 6 from each side.

│11 - 2b│ > 17

11 - 2b > 17 (or) 11 - 2b < -17

-2b > 6 (or) -2b < -28

-b > 3 (or) b < 14

b < -3 (or) b > 14

Hence the solution is 14 < b < -3.

{…-6, -5, -4, 15, 16, 17…}

Problem 11 :

│6 - 3b│ + 4 < 3

Solution :

│6 - 3b│ + 4 < 3

Subtract 4 from each side.

│6 - 3b│ < -1

Absolute value cannot be less than 0. So there is no solution.

Problem 12 :

│7 - x│ + 2 ≤ 12

Solution :

│7 - x│ + 2 ≤ 12

Subtract 2 from each side.

│7 - x│ ≤ 10

7 - x ≤ 10 (or) 7 - x ≥ - 10

-x ≤ 3 (or) -x ≥ -17

x ≥ -3 (or) x ≤ 17

Hence the solution is -3 ≤ x ≤ 17.

{-3, -2, -1,….15, 16, 17}

Problem 13 :

You are buying a new computer. The table shows the prices of computers in a store advertisement. You are willing to pay the mean price with an absolute deviation of at most $100. How many of the computer prices meet your condition?

Solution :

Total prices = 10

Total amount = (890 + 650 + 660 + 450 + 725 + 750 + 370 + 670 + 650 + 825)

= 6640

mean = 6640/10

= 664

Mean devation = |x - mean|

= |x - 664|

Mean deviation must be atleast 100

|x - 664| ≤ 100

564 ≤ x ≤ 764

The prices you will consider must be at least $564 and at most $764. Six prices meet your condition: $750, $650, $660, $670, $650, and $725.

Problem 14 :

Write an absolute value inequality that represents the situation. Then solve the inequality.

The difference between the areas of the figures is less than 2.

Solution :

Area of triangle = 1/2 x base x height

= (1/2) ⋅ 4 ⋅ (x + 6)

= 2(x + 6)

Area of rectangle = length ⋅ width

= 2 ⋅ 6 

= 12

Since the area should be positive

|2(x + 6) - 12| < 2

|2x + 12 - 12| < 2

|2x | < 2

2x < 2 or 2x > -2

x < 1 or x > -1