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Solution - Absolute value equations

Exact form: x=12,-23
x=\frac{1}{2} , -\frac{2}{3}
Decimal form: x=0.5,0.667
x=0.5 , -0.667

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
|5x1|=|x3|
without the absolute value bars:

|x|=|y||5x1|=|x3|
x=+y(5x1)=(x3)
x=y(5x1)=(x3)
+x=y(5x1)=(x3)
x=y(5x1)=(x3)

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||5x1|=|x3|
x=+y , +x=y(5x1)=(x3)
x=y , x=y(5x1)=(x3)

2. Solve the two equations for x

13 additional steps

(-5x-1)=(-x-3)

Add to both sides:

(-5x-1)+x=(-x-3)+x

Group like terms:

(-5x+x)-1=(-x-3)+x

Simplify the arithmetic:

-4x-1=(-x-3)+x

Group like terms:

-4x-1=(-x+x)-3

Simplify the arithmetic:

4x1=3

Add to both sides:

(-4x-1)+1=-3+1

Simplify the arithmetic:

4x=3+1

Simplify the arithmetic:

4x=2

Divide both sides by :

(-4x)-4=-2-4

Cancel out the negatives:

4x4=-2-4

Simplify the fraction:

x=-2-4

Cancel out the negatives:

x=24

Find the greatest common factor of the numerator and denominator:

x=(1·2)(2·2)

Factor out and cancel the greatest common factor:

x=12

14 additional steps

(-5x-1)=-(-x-3)

Expand the parentheses:

(-5x-1)=x+3

Subtract from both sides:

(-5x-1)-x=(x+3)-x

Group like terms:

(-5x-x)-1=(x+3)-x

Simplify the arithmetic:

-6x-1=(x+3)-x

Group like terms:

-6x-1=(x-x)+3

Simplify the arithmetic:

6x1=3

Add to both sides:

(-6x-1)+1=3+1

Simplify the arithmetic:

6x=3+1

Simplify the arithmetic:

6x=4

Divide both sides by :

(-6x)-6=4-6

Cancel out the negatives:

6x6=4-6

Simplify the fraction:

x=4-6

Move the negative sign from the denominator to the numerator:

x=-46

Find the greatest common factor of the numerator and denominator:

x=(-2·2)(3·2)

Factor out and cancel the greatest common factor:

x=-23

3. List the solutions

x=12,-23
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x1|
y=|x3|
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.