Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: y=3,17
y=3 , 17

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
|3y2|=|4y+19|
without the absolute value bars:

|x|=|y||3y2|=|4y+19|
x=+y(3y2)=(4y+19)
x=y(3y2)=(4y+19)
+x=y(3y2)=(4y+19)
x=y(3y2)=(4y+19)

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||3y2|=|4y+19|
x=+y , +x=y(3y2)=(4y+19)
x=y , x=y(3y2)=(4y+19)

2. Solve the two equations for y

11 additional steps

(3y-2)=(-4y+19)

Add to both sides:

(3y-2)+4y=(-4y+19)+4y

Group like terms:

(3y+4y)-2=(-4y+19)+4y

Simplify the arithmetic:

7y-2=(-4y+19)+4y

Group like terms:

7y-2=(-4y+4y)+19

Simplify the arithmetic:

7y2=19

Add to both sides:

(7y-2)+2=19+2

Simplify the arithmetic:

7y=19+2

Simplify the arithmetic:

7y=21

Divide both sides by :

(7y)7=217

Simplify the fraction:

y=217

Find the greatest common factor of the numerator and denominator:

y=(3·7)(1·7)

Factor out and cancel the greatest common factor:

y=3

11 additional steps

(3y-2)=-(-4y+19)

Expand the parentheses:

(3y-2)=4y-19

Subtract from both sides:

(3y-2)-4y=(4y-19)-4y

Group like terms:

(3y-4y)-2=(4y-19)-4y

Simplify the arithmetic:

-y-2=(4y-19)-4y

Group like terms:

-y-2=(4y-4y)-19

Simplify the arithmetic:

y2=19

Add to both sides:

(-y-2)+2=-19+2

Simplify the arithmetic:

y=19+2

Simplify the arithmetic:

y=17

Multiply both sides by :

-y·-1=-17·-1

Remove the one(s):

y=-17·-1

Simplify the arithmetic:

y=17

3. List the solutions

y=3,17
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3y2|
y=|4y+19|
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.