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

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

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

2. Solve the two equations for x

13 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

-4x-3=(5x-1)-5x

Group like terms:

-4x-3=(5x-5x)-1

Simplify the arithmetic:

4x3=1

Add to both sides:

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

Simplify the arithmetic:

4x=1+3

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

Move the negative sign from the denominator to the numerator:

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

12 additional steps

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

Expand the parentheses:

(x-3)=-5x+1

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

6x-3=(-5x+1)+5x

Group like terms:

6x-3=(-5x+5x)+1

Simplify the arithmetic:

6x3=1

Add to both sides:

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

Simplify the arithmetic:

6x=1+3

Simplify the arithmetic:

6x=4

Divide both sides by :

(6x)6=46

Simplify the fraction:

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=|x3|
y=|5x1|
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.