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 1
Solve the equation and check the answers found.
|x + 6 | = 7
Solution to Example 1:
  • 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
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

Solution to Matched Exercise


Example 2
Solve the equation and check the answers found.
-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

Solution to Matched Exercise


Example 3
Solve the equation and check the answers found.
|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 (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.
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

Solution to Matched Exercise


Example 4
Solve the equation and check the answers found.
|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

Solution to Matched Exercise


Solutions to the Above Matched Exercises

Matched Exercise 1

Solve the equation
|-x - 8 | = 10
Answer to Matched Exercise 1:
The above equation has two solutions
x = 2
x = -18

Matched Exercise 2

Solve the equation
4 |x + 2 | - 30 = -10
Answer to Matched Exercise 2:
The above equation has two solutions
x = 3
x = -7

Matched Exercise 3

Solve the equation
- 4 | x + 2 | = x - 8
Answer to Matched Exercise 3:
The above equation has two solutions
x = 0
x = -16/3

Matched Exercise 4

Solve the equation
|x2 - 16 | = x - 4
Answer to Matched Exercise 4:
The 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 Inequalities
Solve Equations, Systems of Equations and Inequalities
Step by Step Solver for Equation With Absolute Value .

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