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Solution - Absolute value equations

Exact form: v=4
v=4

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
|2v10|=|2v6|
without the absolute value bars:

|x|=|y||2v10|=|2v6|
x=+y(2v10)=(2v6)
x=y(2v10)=(2v6)
+x=y(2v10)=(2v6)
x=y(2v10)=(2v6)

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||2v10|=|2v6|
x=+y , +x=y(2v10)=(2v6)
x=y , x=y(2v10)=(2v6)

2. Solve the two equations for v

5 additional steps

(2v-10)=(2v-6)

Subtract from both sides:

(2v-10)-2v=(2v-6)-2v

Group like terms:

(2v-2v)-10=(2v-6)-2v

Simplify the arithmetic:

-10=(2v-6)-2v

Group like terms:

-10=(2v-2v)-6

Simplify the arithmetic:

10=6

The statement is false:

10=6

The equation is false so it has no solution.

12 additional steps

(2v-10)=-(2v-6)

Expand the parentheses:

(2v-10)=-2v+6

Add to both sides:

(2v-10)+2v=(-2v+6)+2v

Group like terms:

(2v+2v)-10=(-2v+6)+2v

Simplify the arithmetic:

4v-10=(-2v+6)+2v

Group like terms:

4v-10=(-2v+2v)+6

Simplify the arithmetic:

4v10=6

Add to both sides:

(4v-10)+10=6+10

Simplify the arithmetic:

4v=6+10

Simplify the arithmetic:

4v=16

Divide both sides by :

(4v)4=164

Simplify the fraction:

v=164

Find the greatest common factor of the numerator and denominator:

v=(4·4)(1·4)

Factor out and cancel the greatest common factor:

v=4

3. Graph

Each line represents the function of one side of the equation:
y=|2v10|
y=|2v6|
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.