Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: v=32
v=\frac{3}{2}
Mixed number form: v=112
v=1\frac{1}{2}
Decimal form: v=1.5
v=1.5

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
|3v6|=|3v3|
without the absolute value bars:

|x|=|y||3v6|=|3v3|
x=+y(3v6)=(3v3)
x=y(3v6)=(3v3)
+x=y(3v6)=(3v3)
x=y(3v6)=(3v3)

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||3v6|=|3v3|
x=+y , +x=y(3v6)=(3v3)
x=y , x=y(3v6)=(3v3)

2. Solve the two equations for v

5 additional steps

(3v-6)=(3v-3)

Subtract from both sides:

(3v-6)-3v=(3v-3)-3v

Group like terms:

(3v-3v)-6=(3v-3)-3v

Simplify the arithmetic:

-6=(3v-3)-3v

Group like terms:

-6=(3v-3v)-3

Simplify the arithmetic:

6=3

The statement is false:

6=3

The equation is false so it has no solution.

12 additional steps

(3v-6)=-(3v-3)

Expand the parentheses:

(3v-6)=-3v+3

Add to both sides:

(3v-6)+3v=(-3v+3)+3v

Group like terms:

(3v+3v)-6=(-3v+3)+3v

Simplify the arithmetic:

6v-6=(-3v+3)+3v

Group like terms:

6v-6=(-3v+3v)+3

Simplify the arithmetic:

6v6=3

Add to both sides:

(6v-6)+6=3+6

Simplify the arithmetic:

6v=3+6

Simplify the arithmetic:

6v=9

Divide both sides by :

(6v)6=96

Simplify the fraction:

v=96

Find the greatest common factor of the numerator and denominator:

v=(3·3)(2·3)

Factor out and cancel the greatest common factor:

v=32

3. Graph

Each line represents the function of one side of the equation:
y=|3v6|
y=|3v3|
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.