# Solve Equations with Absolute Value

This is a tutorial on solving equations with absolute value. Detailed solutions and explanations are included.

Example 1:Solve the equation

|x + 6 | = 7
Solution to Example 1:
• If |x + 6 | = 7, then
a) x + 6 = 7
or
b) x + 6 = -7
• Solve equation a)
x + 6 = 7
x = 1
• Solve equation b)
x + 6 = -7
x = -13
Check solutions:
• solution x = 1
Left Side of Equation for x = 1.
|1 + 6 |
= | 7 |
= 7
Right Side of Equation for x = 1.
7
• x = -13
Left Side of Equation for x = 1.
|-13 + 6 |
= | -7 |
= 7
Right Side of Equation for x = 1.
7

The solutions to the given equation are x = 1 and x = -13

Matched Exercise 1: Solve the equation
|-x - 8 | = 10

Example 2: Solve the equation

-2 |x / 2 + 3 | - 4 = -10
Solution to Example 2:
• Given
-2 |x / 2 + 3 | - 4 = -10
• We first write the equation in the form | A | = B. Add 4 to both sides and group like terms
-2|x / 2 + 3 | = -6
• Divide both sides by -2
|x / 2 + 3 | = 3
• We now proceed as in example 1 above, the equation
|x / 2 + 3 | = 3 gives two equations.

a) x / 2 + 3 = 3
or
b) x / 2 + 3 = -3
• Solve equation a)
x / 2 + 3 = 3
• to obtain
x = 0
• Solve equation b)
x / 2 + 3 = -3
• to obtain
x = -12
Check solutions:
• x = 0
Left Side of Equation for x = 0.
-2 |x / 2 + 3 | - 4
= -2| 3 | - 4
= -10
Right Side of Equation for x = 1.
-10
• x = -12
Left Side of Equation for x = -12.
-2 |x / 2 + 3 | - 4
= -2 |-12 / 2 + 3 | - 4
= -2 |-6 + 3 | - 4
= -2(3) - 4
= -10
Right Side of Equation for x = -12.
-10
The solutions to the given equation are x = 0 and x = -12 Matched Exercise 2: Solve the equation
4 |x + 2| - 30 = -10

Example 3: Solve the equation

|2 x - 2 | = x + 1
Solution to Example 3:
• If 2 x - 2 ≥ 0 which is equivalent to x ≥ 1, then |2 x - 2 | = 2 x - 2 and the given equation becomes
2 x - 2 = x + 1
• Add 2 - x to both sides
x = 3
• Since x = 3 satisfies the condition x ≥ 1, it is a solution.
• If 2x - 2 < 0 which is equivalent to x < 1, then |2 x - 2 | = -(2 x - 2) and the given equation becomes
-(2 x - 2) = x + 1
• Solve for x to obtain
x = 1 / 3
• Since x = 1 / 3 satisfies the condition x < 1, it is a solution.
Check solutions
• x = 3
Left Side of Equation for x = 3.
|2 x - 2 |
= |2*3 - 2 |
= 4
Right Side of Equation for x = 3.
x + 1
= 3 + 1
= 4
• x = 1/3
Left Side of Equation for x = 1 / 3.
|2 x - 2 |
= |2*(1/3) - 2 |
= 4 / 3
Right Side of Equation for x = 1 / 3.
x + 1
= 4 / 3
The solutions to the given equation are x = 3 and x = 1 / 3 Matched Exercise 3: Solve the equation
- 4|x + 2 | = x - 8

Example 4: Solve the equation

|x2 - 4| = x + 2
Solution to Example 3:
• If x2 - 4 ≥ 0 ,or x2 ≥ 4, then | x2 - 4 | = x2 - 4 and the given equation becomes
x2 - 4 = x + 2
• Add - (x + 2) to both sides
x2 - 4 -( x + 2) = 0
• Factor the left term
(x - 2)(x + 2) -( x + 2) = 0
(x + 2)(x - 2 -1) = 0
(x + 2)(x - 3) = 0
• Using the factor theorem, we can write two simpler equations
x + 2 = 0
or
x - 3 = 0
• Solve the above equations for x to find two values of x that make the left side of the equation equal to zero.
x = -2 and x = 3.
• Both values satisfy the condition x2 ≥ 4 and are solutions to the given equation.
x = -2 and x = 3.
• If x2 - 4 < 0 ,or x2 < 4, then | x2 - 4 | = -(x2 - 4) and the given equation becomes.
-(x2 - 4) = x + 2
-(x2 - 4) - ( x + 2) = 0
• Factor the left term.
-(x - 2)(x + 2) - ( x + 2) = 0
(x - 2)(x + 2) + ( x + 2) = 0
(x - 2)(x + 2) + ( x + 2) = 0
(x + 2)(x - 2 + 1) = 0
(x + 2)(x - 1) = 0
• Two values make the left side of the above equation equal to zero
x = -2 and x = 1.
• Only x = 1 satisfies the condition x2 < 4
Check solutions:
• x = -2
Right Side of Equation = | x2 - 4 |
= | (-2)2 - 4 | = 0
Left Side of Equation = x + 2 = -2 + 2 = 0
• x = 3 Left Side of Equation = | x2 - 4 |
= | 32 - 4 |
= | 5 |
= 5 Right Side of Equation = x + 2 = 3 + 2 = 5
• x = 1
Left Side of Equation = | x2 - 4 |
= | 12 - 4 | = | - 3 | = 3 Right Side of Equation = x + 2 = 1 + 2 = 3
Conclusion
The solutions to the given equation are x = -2, x = 1 and x = 3.
Matched Exercise 4: Solve the equation
|x2 - 16 | = x - 4

Solve the following absolute value equations
a) | x - 4 | = 9
b) | x 2 + 4 | = 5
c) | x 2 - 9 | = x + 3
d) | x + 1 | = x - 3
e) | -x | = 2

a) -5 , 13
b) -1 , 1
c) -3 , 2 , 4
d) no real solutions
e) -2 , 2

More references and links on how to Solve Equations, Systems of Equations and Inequalities and Step by Step Solver for Equation With Absolute Value.