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.