Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: x=4,12
x=4 , \frac{1}{2}
Decimal form: x=4,0.5
x=4 , 0.5

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
|x+3|=|3x5|
without the absolute value bars:

|x|=|y||x+3|=|3x5|
x=+y(x+3)=(3x5)
x=y(x+3)=(3x5)
+x=y(x+3)=(3x5)
x=y(x+3)=(3x5)

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||x+3|=|3x5|
x=+y , +x=y(x+3)=(3x5)
x=y , x=y(x+3)=(3x5)

2. Solve the two equations for x

13 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

-2x+3=(3x-5)-3x

Group like terms:

-2x+3=(3x-3x)-5

Simplify the arithmetic:

2x+3=5

Subtract from both sides:

(-2x+3)-3=-5-3

Simplify the arithmetic:

2x=53

Simplify the arithmetic:

2x=8

Divide both sides by :

(-2x)-2=-8-2

Cancel out the negatives:

2x2=-8-2

Simplify the fraction:

x=-8-2

Cancel out the negatives:

x=82

Find the greatest common factor of the numerator and denominator:

x=(4·2)(1·2)

Factor out and cancel the greatest common factor:

x=4

12 additional steps

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

Expand the parentheses:

(x+3)=-3x+5

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

4x+3=5

Subtract from both sides:

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

Simplify the arithmetic:

4x=53

Simplify the arithmetic:

4x=2

Divide both sides by :

(4x)4=24

Simplify the fraction:

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

3. List the solutions

x=4,12
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|x+3|
y=|3x5|
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.