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Solution - Absolute value equations

Exact form: x=6,32
x=6 , \frac{3}{2}
Mixed number form: x=6,112
x=6 , 1\frac{1}{2}
Decimal form: x=6,1.5
x=6 , 1.5

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
|3x|=|5x12|
without the absolute value bars:

|x|=|y||3x|=|5x12|
x=+y(3x)=(5x12)
x=y(3x)=(5x12)
+x=y(3x)=(5x12)
x=y(3x)=(5x12)

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||3x|=|5x12|
x=+y , +x=y(3x)=(5x12)
x=y , x=y(3x)=(5x12)

2. Solve the two equations for x

9 additional steps

3x=(5x-12)

Subtract from both sides:

(3x)-5x=(5x-12)-5x

Simplify the arithmetic:

-2x=(5x-12)-5x

Group like terms:

-2x=(5x-5x)-12

Simplify the arithmetic:

2x=12

Divide both sides by :

(-2x)-2=-12-2

Cancel out the negatives:

2x2=-12-2

Simplify the fraction:

x=-12-2

Cancel out the negatives:

x=122

Find the greatest common factor of the numerator and denominator:

x=(6·2)(1·2)

Factor out and cancel the greatest common factor:

x=6

8 additional steps

3x=-(5x-12)

Expand the parentheses:

3x=5x+12

Add to both sides:

(3x)+5x=(-5x+12)+5x

Simplify the arithmetic:

8x=(-5x+12)+5x

Group like terms:

8x=(-5x+5x)+12

Simplify the arithmetic:

8x=12

Divide both sides by :

(8x)8=128

Simplify the fraction:

x=128

Find the greatest common factor of the numerator and denominator:

x=(3·4)(2·4)

Factor out and cancel the greatest common factor:

x=32

3. List the solutions

x=6,32
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3x|
y=|5x12|
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.