Solving Absolute Value Equations

Solving Absolute Value Equations of the Type | x | = k .

Absolute value equations are useful in determining distance and error measurements.

The examples that we will consider are:

| x | = 3

| x – 6 | = 4

| 2 x – 3 | = 9

| x + 7 | = 2

| x+ 8 | = | 3x – 4 |

Example 1 : Solve for x : | x | = 3

Solution.

This equation is asking us to find all numbers, x , that are 3 units from zero on the number line.

We must consider numbers both to the right and to the left of zero on the number line.

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Notice that both 3 and -3 are three units from zero.

The solution is: x = 3 or x = −3 .

Example 1 suggests a rule that we can use when solving absolute value equations.

If c is a positive number, then | x | = c is equivalent to x = c or x = c.

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Example 2 : Solve for x : | x – 6 | = 4

Solution.

Step 1. Break the equation up into two equivalent equations using the rule: If | x | = c then x = c or x = - c .

| x – 6 | = 4 is equivalent to x – 6 = 4 or x – 6 = – 4

Step 2. Solve each equation .

x – 6 + 6 = 4 + 6

x = 10

x – 6 + 6 = – 4 + 6


x = 2

Step 3 . Check the solutions.

| 10 – 6 | = | 4 | = 4

| 2 – 6 | = | 4 | = 4

The solutions are x = 10 and x = 2 .

Example 3 : Solve for x : | 2 x – 3 | = 9

Solution.

Step 1.

Break the equation up into two equivalent equations using the rule: If | x | = c then x = c or x = - c .

| 2 x – 3 | = 9 is equivalent to 2 x – 3 = 9 or 2 x – 3 = -9

Step 2. Solve each equation .

2 x – 3 = 9 or 2 x – 3 = -9

2 x – 3 + 3 = 9 + 3 or 2 x – 3 + 3 = -9 + 3

2 x = 12 or 2 x = -6

2 x ÷ 2 = 12 ÷ 2 or 2 x ÷ 2 = -6 ÷ 2

x = 6 or x = -3

Step 3 . Check the solutions.

x = 6: | 2(6) – 3 | = | 12 – 3 | = | 9 | = 9

x = -3: | 2(-3) – 3 | = | -6 – 3 | = | -9 | = 9

The solutions are x = 6 and x = -3 .

Example 4 : Solve for x : | x + 7 | = 2

Solution.

The absolute value of a number is never negative. This equation has no solution .

Solving Absolute Value Equations of the Type | x | = | y |.

If the absolute values of two expressions are equal, then either the two expressions are equal, or they are opposites.

If x and y represent algebraic expressions, | x | = | y | is equivalent to x = y or x = y.

Example 5 : Solve for x : | x + 8 | = | 3 x – 4 |

Solution.

Step 1. Break the equation up into two equivalent equations .

| x + 8 | = | 3 x – 4 | is equivalent to x + 8 = 3 x – 4 or x + 8 = (3 x – 4)

Step 2. Solve each equation.

x + 8 = 3 x – 4 or x + 8 = (3 x – 4)

x + 8 = 3 x – 4 or x + 8 = 3 x + 4

x + 8 – x = 3 x – 4 – x or x + 8 + 3 x = -3 x + 4 + 3 x

8 = 2 x – 4 or 4 x + 8 = 4

8 + 4 = 2 x – 4 + 4 or 4 x + 8 – 8 = 4 – 8

12 = 2 x or 4 x = – 4

12 ÷ 2 = 2 x ÷ 2 or 4 x ÷ 4 = – 4 ÷ 4

6 = x or x = – 1

Step 3 . Check the solutions.

x = 6: | 6 + 8 | = | 3(6) – 4 |

| 14 | = | 18 – 4 |

| 14 | = | 14 |

14 = 14

x = 1: | 1 + 8 | = | 3( 1) – 4 |

| 7 | = | 3 – 4 |

| 7 | = | 7 |

7 = 7

The solutions are x = 6 and x = – 1 .

|12t - 3| = -8
|3x + 1| = |2x + 4|
Solve |2x + 3| = 19