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Solution - Absolute value equations

Exact form: x=4,47
x=4 , \frac{4}{7}
Decimal form: x=4,0.571
x=4 , 0.571

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
|6x12|=|x+8|
without the absolute value bars:

|x|=|y||6x12|=|x+8|
x=+y(6x12)=(x+8)
x=y(6x12)=(x+8)
+x=y(6x12)=(x+8)
x=y(6x12)=(x+8)

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||6x12|=|x+8|
x=+y , +x=y(6x12)=(x+8)
x=y , x=y(6x12)=(x+8)

2. Solve the two equations for x

11 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

5x12=8

Add to both sides:

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

Simplify the arithmetic:

5x=8+12

Simplify the arithmetic:

5x=20

Divide both sides by :

(5x)5=205

Simplify the fraction:

x=205

Find the greatest common factor of the numerator and denominator:

x=(4·5)(1·5)

Factor out and cancel the greatest common factor:

x=4

10 additional steps

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

Expand the parentheses:

(6x-12)=-x-8

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

7x-12=(-x-8)+x

Group like terms:

7x-12=(-x+x)-8

Simplify the arithmetic:

7x12=8

Add to both sides:

(7x-12)+12=-8+12

Simplify the arithmetic:

7x=8+12

Simplify the arithmetic:

7x=4

Divide both sides by :

(7x)7=47

Simplify the fraction:

x=47

3. List the solutions

x=4,47
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|6x12|
y=|x+8|
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.