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Solution - Absolute value equations

Exact form: x=1
x=1

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|x8||x+6|=0

Add |x+6| to both sides of the equation:

|x8||x+6|+|x+6|=|x+6|

Simplify the arithmetic

|x8|=|x+6|

2. 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
|x8|=|x+6|
without the absolute value bars:

|x|=|y||x8|=|x+6|
x=+y(x8)=(x+6)
x=y(x8)=((x+6))
+x=y(x8)=(x+6)
x=y(x8)=(x+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||x8|=|x+6|
x=+y , +x=y(x8)=(x+6)
x=y , x=y(x8)=((x+6))

3. Solve the two equations for x

5 additional steps

(x-8)=(x+6)

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

-8=(x+6)-x

Group like terms:

-8=(x-x)+6

Simplify the arithmetic:

8=6

The statement is false:

8=6

The equation is false so it has no solution.

11 additional steps

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

Expand the parentheses:

(x-8)=-x-6

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2x8=6

Add to both sides:

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

Simplify the arithmetic:

2x=6+8

Simplify the arithmetic:

2x=2

Divide both sides by :

(2x)2=22

Simplify the fraction:

x=22

Simplify the fraction:

x=1

4. Graph

Each line represents the function of one side of the equation:
y=|x8|
y=|x+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.