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Solution - Absolute value equations

Exact form: x=-12,34
x=-\frac{1}{2} , \frac{3}{4}
Decimal form: x=0.5,0.75
x=-0.5 , 0.75

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
|2x4|=|6x2|
without the absolute value bars:

|x|=|y||2x4|=|6x2|
x=+y(2x4)=(6x2)
x=y(2x4)=(6x2)
+x=y(2x4)=(6x2)
x=y(2x4)=(6x2)

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||2x4|=|6x2|
x=+y , +x=y(2x4)=(6x2)
x=y , x=y(2x4)=(6x2)

2. Solve the two equations for x

13 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

4x4=2

Add to both sides:

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

Simplify the arithmetic:

4x=2+4

Simplify the arithmetic:

4x=2

Divide both sides by :

(-4x)-4=2-4

Cancel out the negatives:

4x4=2-4

Simplify the fraction:

x=2-4

Move the negative sign from the denominator to the numerator:

x=-24

Find the greatest common factor of the numerator and denominator:

x=(-1·2)(2·2)

Factor out and cancel the greatest common factor:

x=-12

12 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

8x-4=(-6x+2)+6x

Group like terms:

8x-4=(-6x+6x)+2

Simplify the arithmetic:

8x4=2

Add to both sides:

(8x-4)+4=2+4

Simplify the arithmetic:

8x=2+4

Simplify the arithmetic:

8x=6

Divide both sides by :

(8x)8=68

Simplify the fraction:

x=68

Find the greatest common factor of the numerator and denominator:

x=(3·2)(4·2)

Factor out and cancel the greatest common factor:

x=34

3. List the solutions

x=-12,34
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x4|
y=|6x2|
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.