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Solution - Absolute value equations

Exact form: y=3,9
y=3 , 9

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
|2y3|=|3y+12|
without the absolute value bars:

|x|=|y||2y3|=|3y+12|
x=+y(2y3)=(3y+12)
x=y(2y3)=(3y+12)
+x=y(2y3)=(3y+12)
x=y(2y3)=(3y+12)

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||2y3|=|3y+12|
x=+y , +x=y(2y3)=(3y+12)
x=y , x=y(2y3)=(3y+12)

2. Solve the two equations for y

11 additional steps

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

5y-3=(-3y+12)+3y

Group like terms:

5y-3=(-3y+3y)+12

Simplify the arithmetic:

5y3=12

Add to both sides:

(5y-3)+3=12+3

Simplify the arithmetic:

5y=12+3

Simplify the arithmetic:

5y=15

Divide both sides by :

(5y)5=155

Simplify the fraction:

y=155

Find the greatest common factor of the numerator and denominator:

y=(3·5)(1·5)

Factor out and cancel the greatest common factor:

y=3

11 additional steps

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

Expand the parentheses:

(2y-3)=3y-12

Subtract from both sides:

(2y-3)-3y=(3y-12)-3y

Group like terms:

(2y-3y)-3=(3y-12)-3y

Simplify the arithmetic:

-y-3=(3y-12)-3y

Group like terms:

-y-3=(3y-3y)-12

Simplify the arithmetic:

y3=12

Add to both sides:

(-y-3)+3=-12+3

Simplify the arithmetic:

y=12+3

Simplify the arithmetic:

y=9

Multiply both sides by :

-y·-1=-9·-1

Remove the one(s):

y=-9·-1

Simplify the arithmetic:

y=9

3. List the solutions

y=3,9
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2y3|
y=|3y+12|
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.