Solve Equations with Absolute Value
Solve equations with absolute value; including examples and questions with detailed solutions and explanations.
Review of Absolute Value
The rules you need to know in order to be able to solve the question in this tutorial.1)  x  = 0 if x = 0
2)  x  = x if x > 0
3)  x  =  x if x < 0
4) The equation  x  = k with k < 0 has no real solutions.
5) The equation  x  = k , k ≥ 0 is equivalent to x = k or x =  k
Examples with Solutions
Example 1Solve the equation and check the answers found.

If x + 6  = 7, then (see rule 5 above)
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 and check the answers found.

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 and check the answers found.

If 2 x  2 ≥ 0 which is equivalent to x ≥ 1, then 2 x  2  = 2 x  2 (see rule 2 above) 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) (see rule 3 above) 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 and check the answers found.

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
Solutions to the Above Matched Exercises
Matched Exercise 1
Solve the equationThe above equation has two solutions
x = 2
x = 18
Matched Exercise 2
Solve the equationThe above equation has two solutions
x = 3
x = 7
Matched Exercise 3
Solve the equationThe above equation has two solutions
x = 0
x = 16/3
Matched Exercise 4
Solve the equationThe above equation has one solution
x = 4
More Exercises with Answers
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 References and Links
Absolute Value Equations And InequalitiesSolve Equations, Systems of Equations and Inequalities
Step by Step Solver for Equation With Absolute Value .