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Solution - Absolute value equations

Exact form: x=-2,-15
x=-2 , -\frac{1}{5}
Decimal form: x=2,0.2
x=-2 , -0.2

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
|8x2|=|12x+6|
without the absolute value bars:

|x|=|y||8x2|=|12x+6|
x=+y(8x2)=(12x+6)
x=y(8x2)=(12x+6)
+x=y(8x2)=(12x+6)
x=y(8x2)=(12x+6)

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||8x2|=|12x+6|
x=+y , +x=y(8x2)=(12x+6)
x=y , x=y(8x2)=(12x+6)

2. Solve the two equations for x

13 additional steps

(8x-2)=(12x+6)

Subtract from both sides:

(8x-2)-12x=(12x+6)-12x

Group like terms:

(8x-12x)-2=(12x+6)-12x

Simplify the arithmetic:

-4x-2=(12x+6)-12x

Group like terms:

-4x-2=(12x-12x)+6

Simplify the arithmetic:

4x2=6

Add to both sides:

(-4x-2)+2=6+2

Simplify the arithmetic:

4x=6+2

Simplify the arithmetic:

4x=8

Divide both sides by :

(-4x)-4=8-4

Cancel out the negatives:

4x4=8-4

Simplify the fraction:

x=8-4

Move the negative sign from the denominator to the numerator:

x=-84

Find the greatest common factor of the numerator and denominator:

x=(-2·4)(1·4)

Factor out and cancel the greatest common factor:

x=2

12 additional steps

(8x-2)=-(12x+6)

Expand the parentheses:

(8x-2)=-12x-6

Add to both sides:

(8x-2)+12x=(-12x-6)+12x

Group like terms:

(8x+12x)-2=(-12x-6)+12x

Simplify the arithmetic:

20x-2=(-12x-6)+12x

Group like terms:

20x-2=(-12x+12x)-6

Simplify the arithmetic:

20x2=6

Add to both sides:

(20x-2)+2=-6+2

Simplify the arithmetic:

20x=6+2

Simplify the arithmetic:

20x=4

Divide both sides by :

(20x)20=-420

Simplify the fraction:

x=-420

Find the greatest common factor of the numerator and denominator:

x=(-1·4)(5·4)

Factor out and cancel the greatest common factor:

x=-15

3. List the solutions

x=-2,-15
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|8x2|
y=|12x+6|
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.