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Solution - Absolute value equations

Exact form: x=23,29
x=\frac{2}{3} , \frac{2}{9}
Decimal form: x=0.667,0.222
x=0.667 , 0.222

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|=2|x-13|
without the absolute value bars:

|x|=|y||x|=2|x-13|
x=+y(x)=2(x-13)
x=-y(x)=2(-(x-13))
+x=y(x)=2(x-13)
-x=y-(x)=2(x-13)

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|=2|x-13|
x=+y , +x=y(x)=2(x-13)
x=-y , -x=y(x)=2(-(x-13))

2. Solve the two equations for x

8 additional steps

x=2·(x+-13)

Expand the parentheses:

x=x·2+(-1·2)3

Simplify the arithmetic:

x=2x+-23

Subtract from both sides:

x-2x=(2x+-23)-2x

Simplify the arithmetic:

-x=(2x+-23)-2x

Group like terms:

-x=(2x-2x)+-23

Simplify the arithmetic:

-x=-23

Multiply both sides by :

-x·-1=(-23)·-1

Remove the one(s):

x=(-23)·-1

Remove the one(s):

x=23

12 additional steps

x=2·(-(x+-13))

Expand the parentheses:

x=2·(-x+13)

x=-x·2+(1·2)3

Group like terms:

x=(-1·2)x+(1·2)3

Multiply the coefficients:

x=-2x+(1·2)3

Simplify the arithmetic:

x=-2x+23

Add to both sides:

x+2x=(-2x+23)+2x

Simplify the arithmetic:

3x=(-2x+23)+2x

Group like terms:

3x=(-2x+2x)+23

Simplify the arithmetic:

3x=23

Divide both sides by :

(3x)3=(23)3

Simplify the fraction:

x=(23)3

Simplify the arithmetic:

x=2(3·3)

x=29

3. List the solutions

x=23,29
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|x|
y=2|x-13|
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.