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

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

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

2. Solve the two equations for v

5 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

3=6

The statement is false:

3=6

The equation is false so it has no solution.

12 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

6v3=6

Add to both sides:

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

Simplify the arithmetic:

6v=6+3

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=|3v3|
y=|3v6|
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.