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Solution - Absolute value equations

Exact form: v=12,125
v=12 , \frac{12}{5}
Mixed number form: v=12,225
v=12 , 2\frac{2}{5}
Decimal form: v=12,2.4
v=12 , 2.4

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
3|v4|=|2v|
without the absolute value bars:

|x|=|y|3|v4|=|2v|
x=+y3(v4)=(2v)
x=y3(v4)=(2v)
+x=y3(v4)=(2v)
x=y3((v4))=(2v)

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|v4|=|2v|
x=+y , +x=y3(v4)=(2v)
x=y , x=y3(v4)=(2v)

2. Solve the two equations for v

8 additional steps

3·(v-4)=2v

Expand the parentheses:

3v+3·-4=2v

Simplify the arithmetic:

3v12=2v

Subtract from both sides:

(3v-12)-2v=(2v)-2v

Group like terms:

(3v-2v)-12=(2v)-2v

Simplify the arithmetic:

v-12=(2v)-2v

Simplify the arithmetic:

v12=0

Add to both sides:

(v-12)+12=0+12

Simplify the arithmetic:

v=0+12

Simplify the arithmetic:

v=12

10 additional steps

3·(v-4)=-(2v)

Expand the parentheses:

3v+3·-4=-(2v)

Simplify the arithmetic:

3v-12=-(2v)

Add to both sides:

(3v-12)+2v=(-2v)+2v

Group like terms:

(3v+2v)-12=(-2v)+2v

Simplify the arithmetic:

5v-12=(-2v)+2v

Simplify the arithmetic:

5v12=0

Add to both sides:

(5v-12)+12=0+12

Simplify the arithmetic:

5v=0+12

Simplify the arithmetic:

5v=12

Divide both sides by :

(5v)5=125

Simplify the fraction:

v=125

3. List the solutions

v=12,125
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=3|v4|
y=|2v|
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.