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Solution - Absolute value equations

Exact form: =133,103
=\frac{13}{3} , \frac{10}{3}
Mixed number form: =413,313
=4\frac{1}{3} , 3\frac{1}{3}
Decimal form: =4.333,3.333
=4.333 , 3.333

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
|+3|=|6x23|
without the absolute value bars:

|x|=|y||+3|=|6x23|
x=+y(+3)=(6x23)
x=y(+3)=(6x23)
+x=y(+3)=(6x23)
x=y(+3)=(6x23)

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||+3|=|6x23|
x=+y , +x=y(+3)=(6x23)
x=y , x=y(+3)=(6x23)

2. Solve the two equations for

7 additional steps

(3)=(6x-23)

Swap sides:

(6x-23)=(3)

Add to both sides:

(6x-23)+23=(3)+23

Simplify the arithmetic:

6x=(3)+23

Simplify the arithmetic:

6x=26

Divide both sides by :

(6x)6=266

Simplify the fraction:

x=266

Find the greatest common factor of the numerator and denominator:

x=(13·2)(3·2)

Factor out and cancel the greatest common factor:

x=133

10 additional steps

(3)=-(6x-23)

Expand the parentheses:

(3)=-6x+23

Swap sides:

-6x+23=(3)

Subtract from both sides:

(-6x+23)-23=(3)-23

Simplify the arithmetic:

-6x=(3)-23

Simplify the arithmetic:

6x=20

Divide both sides by :

(-6x)-6=-20-6

Cancel out the negatives:

6x6=-20-6

Simplify the fraction:

x=-20-6

Cancel out the negatives:

x=206

Find the greatest common factor of the numerator and denominator:

x=(10·2)(3·2)

Factor out and cancel the greatest common factor:

x=103

3. List the solutions

=133,103
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|+3|
y=|6x23|
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.