Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: y=4,16
y=4 , 16

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
|2y2|=|3y+18|
without the absolute value bars:

|x|=|y||2y2|=|3y+18|
x=+y(2y2)=(3y+18)
x=y(2y2)=(3y+18)
+x=y(2y2)=(3y+18)
x=y(2y2)=(3y+18)

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||2y2|=|3y+18|
x=+y , +x=y(2y2)=(3y+18)
x=y , x=y(2y2)=(3y+18)

2. Solve the two equations for y

11 additional steps

(2y-2)=(-3y+18)

Add to both sides:

(2y-2)+3y=(-3y+18)+3y

Group like terms:

(2y+3y)-2=(-3y+18)+3y

Simplify the arithmetic:

5y-2=(-3y+18)+3y

Group like terms:

5y-2=(-3y+3y)+18

Simplify the arithmetic:

5y2=18

Add to both sides:

(5y-2)+2=18+2

Simplify the arithmetic:

5y=18+2

Simplify the arithmetic:

5y=20

Divide both sides by :

(5y)5=205

Simplify the fraction:

y=205

Find the greatest common factor of the numerator and denominator:

y=(4·5)(1·5)

Factor out and cancel the greatest common factor:

y=4

11 additional steps

(2y-2)=-(-3y+18)

Expand the parentheses:

(2y-2)=3y-18

Subtract from both sides:

(2y-2)-3y=(3y-18)-3y

Group like terms:

(2y-3y)-2=(3y-18)-3y

Simplify the arithmetic:

-y-2=(3y-18)-3y

Group like terms:

-y-2=(3y-3y)-18

Simplify the arithmetic:

y2=18

Add to both sides:

(-y-2)+2=-18+2

Simplify the arithmetic:

y=18+2

Simplify the arithmetic:

y=16

Multiply both sides by :

-y·-1=-16·-1

Remove the one(s):

y=-16·-1

Simplify the arithmetic:

y=16

3. List the solutions

y=4,16
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2y2|
y=|3y+18|
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.