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Solution - Absolute value equations

Exact form: x=72,-83
x=\frac{7}{2} , -\frac{8}{3}
Mixed number form: x=312,-223
x=3\frac{1}{2} , -2\frac{2}{3}
Decimal form: x=3.5,2.667
x=3.5 , -2.667

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

Use the rules:
|x|=|y|x=±y and |x|=|y|±x=y
to write all four options of the equation
|5x+1|=|x+15|
without the absolute value bars:

|x|=|y||5x+1|=|x+15|
x=+y(5x+1)=(x+15)
x=y(5x+1)=(x+15)
+x=y(5x+1)=(x+15)
x=y(5x+1)=(x+15)

When simplified, equations x=+y and +x=y are the same and equations x=y and x=y are the same, so we end up with only 2 equations:

|x|=|y||5x+1|=|x+15|
x=+y , +x=y(5x+1)=(x+15)
x=y , x=y(5x+1)=(x+15)

2. Solve the two equations for x

11 additional steps

(5x+1)=(x+15)

Subtract from both sides:

(5x+1)-x=(x+15)-x

Group like terms:

(5x-x)+1=(x+15)-x

Simplify the arithmetic:

4x+1=(x+15)-x

Group like terms:

4x+1=(x-x)+15

Simplify the arithmetic:

4x+1=15

Subtract from both sides:

(4x+1)-1=15-1

Simplify the arithmetic:

4x=151

Simplify the arithmetic:

4x=14

Divide both sides by :

(4x)4=144

Simplify the fraction:

x=144

Find the greatest common factor of the numerator and denominator:

x=(7·2)(2·2)

Factor out and cancel the greatest common factor:

x=72

12 additional steps

(5x+1)=-(x+15)

Expand the parentheses:

(5x+1)=-x-15

Add to both sides:

(5x+1)+x=(-x-15)+x

Group like terms:

(5x+x)+1=(-x-15)+x

Simplify the arithmetic:

6x+1=(-x-15)+x

Group like terms:

6x+1=(-x+x)-15

Simplify the arithmetic:

6x+1=15

Subtract from both sides:

(6x+1)-1=-15-1

Simplify the arithmetic:

6x=151

Simplify the arithmetic:

6x=16

Divide both sides by :

(6x)6=-166

Simplify the fraction:

x=-166

Find the greatest common factor of the numerator and denominator:

x=(-8·2)(3·2)

Factor out and cancel the greatest common factor:

x=-83

3. List the solutions

x=72,-83
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x+1|
y=|x+15|
The equation is true where the two lines cross.

Why learn this

We encounter absolute values almost daily. For example: If you walk 3 miles to school, do you also walk minus 3 miles when you go back home? The answer is no because distances use absolute value. The absolute value of the distance between home and school is 3 miles, there or back.
In short, absolute values help us deal with concepts like distance, ranges of possible values, and deviation from a set value.