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Solution - Absolute value equations

Exact form: x=4,12
x=4 , 12

Other Ways to Solve

Absolute value equations

Step-by-step explanation

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

|x|+2|x6|=0

Add 2|x6| to both sides of the equation:

|x|+2|x6|2|x6|=2|x6|

Simplify the arithmetic

|x|=2|x6|

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
|x|=2|x6|
without the absolute value bars:

|x|=|y||x|=2|x6|
x=+y(x)=2(x6)
x=y(x)=2((x6))
+x=y(x)=2(x6)
x=y(x)=2(x6)

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|x6|
x=+y , +x=y(x)=2(x6)
x=y , x=y(x)=2((x6))

3. Solve the two equations for x

9 additional steps

x=-2·(x-6)

Expand the parentheses:

x=-2x-2·-6

Simplify the arithmetic:

x=2x+12

Add to both sides:

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

Simplify the arithmetic:

3x=(-2x+12)+2x

Group like terms:

3x=(-2x+2x)+12

Simplify the arithmetic:

3x=12

Divide both sides by :

(3x)3=123

Simplify the fraction:

x=123

Find the greatest common factor of the numerator and denominator:

x=(4·3)(1·3)

Factor out and cancel the greatest common factor:

x=4

11 additional steps

x=-2·(-(x-6))

Expand the parentheses:

x=-2·(-x+6)

x=-2·-x-2·6

Group like terms:

x=(-2·-1)x-2·6

Multiply the coefficients:

x=2x-2·6

Simplify the arithmetic:

x=2x12

Subtract from both sides:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x=12

Multiply both sides by :

-x·-1=-12·-1

Remove the one(s):

x=-12·-1

Simplify the arithmetic:

x=12

4. List the solutions

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

5. Graph

Each line represents the function of one side of the equation:
y=|x|
y=2|x6|
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.