Solve Equations with Absolute Value
This is a tutorial on solving equations with absolute value. Detailed solutions and explanations are included.
Examples with Solutions
Example 1
Solve the equation
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
 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
Example 2
Solve the equation
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
2x / 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
 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
Matched Exercise 2: Solve the equation
Example 3
Solve the equation
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.
 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
Matched Exercise 3: Solve the equation
Example 4
Solve the equation
If x^{2}  4 ≥ 0 ,or x^{2} ≥ 4, then  x^{2}  4  = x^{2}  4 and the given equation becomes
x^{2}  4 = x + 2

Add  (x + 2) to both sides
x^{2}  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 x^{2} ≥ 4 and are solutions to the given equation.
x = 2 and x = 3.

If x^{2}  4 < 0 ,or x^{2} < 4, then  x^{2}  4  = (x^{2}  4) and the given equation becomes.
(x^{2}  4) = x + 2
(x^{2}  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 x^{2} < 4

x = 2
Right Side of Equation =  x^{2}  4 
=  (2)^{2}  4  = 0
Left Side of Equation = x + 2 = 2 + 2 = 0

x = 3
Left Side of Equation =  x^{2}  4 
=  3^{2}  4 
=  5 
= 5 Right Side of Equation = x + 2 = 3 + 2 = 5 
x = 1
Left Side of Equation =  x^{2}  4 
=  1^{2}  4  =   3  = 3 Right Side of Equation = x + 2 = 1 + 2 = 3
The solutions to the given equation are x = 2, x = 1 and x = 3.
Matched Exercise 4: Solve the equation
Exercises
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
Answers to Above Exercises
a) 5 , 13
b) 1 , 1
c) 3 , 2 , 4
d) no real solutions
e) 2 , 2
More Refernces and Links
Solve Equations, Systems of Equations and InequalitiesStep by Step Solver for Equation With Absolute Value .