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Solution - Absolute value equations

Exact form: x=14,58
x=\frac{1}{4} , \frac{5}{8}
Decimal form: x=0.25,0.625
x=0.25 , 0.625

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
|2x2|=|6x3|
without the absolute value bars:

|x|=|y||2x2|=|6x3|
x=+y(2x2)=(6x3)
x=y(2x2)=(6x3)
+x=y(2x2)=(6x3)
x=y(2x2)=(6x3)

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||2x2|=|6x3|
x=+y , +x=y(2x2)=(6x3)
x=y , x=y(2x2)=(6x3)

2. Solve the two equations for x

11 additional steps

(2x-2)=(6x-3)

Subtract from both sides:

(2x-2)-6x=(6x-3)-6x

Group like terms:

(2x-6x)-2=(6x-3)-6x

Simplify the arithmetic:

-4x-2=(6x-3)-6x

Group like terms:

-4x-2=(6x-6x)-3

Simplify the arithmetic:

4x2=3

Add to both sides:

(-4x-2)+2=-3+2

Simplify the arithmetic:

4x=3+2

Simplify the arithmetic:

4x=1

Divide both sides by :

(-4x)-4=-1-4

Cancel out the negatives:

4x4=-1-4

Simplify the fraction:

x=-1-4

Cancel out the negatives:

x=14

10 additional steps

(2x-2)=-(6x-3)

Expand the parentheses:

(2x-2)=-6x+3

Add to both sides:

(2x-2)+6x=(-6x+3)+6x

Group like terms:

(2x+6x)-2=(-6x+3)+6x

Simplify the arithmetic:

8x-2=(-6x+3)+6x

Group like terms:

8x-2=(-6x+6x)+3

Simplify the arithmetic:

8x2=3

Add to both sides:

(8x-2)+2=3+2

Simplify the arithmetic:

8x=3+2

Simplify the arithmetic:

8x=5

Divide both sides by :

(8x)8=58

Simplify the fraction:

x=58

3. List the solutions

x=14,58
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x2|
y=|6x3|
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.