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Solution - Absolute value equations

Exact form: y=-2,-45
y=-2 , -\frac{4}{5}
Decimal form: y=2,0.8
y=-2 , -0.8

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
|3y+3|=|2y+1|
without the absolute value bars:

|x|=|y||3y+3|=|2y+1|
x=+y(3y+3)=(2y+1)
x=y(3y+3)=(2y+1)
+x=y(3y+3)=(2y+1)
x=y(3y+3)=(2y+1)

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||3y+3|=|2y+1|
x=+y , +x=y(3y+3)=(2y+1)
x=y , x=y(3y+3)=(2y+1)

2. Solve the two equations for y

7 additional steps

(3y+3)=(2y+1)

Subtract from both sides:

(3y+3)-2y=(2y+1)-2y

Group like terms:

(3y-2y)+3=(2y+1)-2y

Simplify the arithmetic:

y+3=(2y+1)-2y

Group like terms:

y+3=(2y-2y)+1

Simplify the arithmetic:

y+3=1

Subtract from both sides:

(y+3)-3=1-3

Simplify the arithmetic:

y=13

Simplify the arithmetic:

y=2

10 additional steps

(3y+3)=-(2y+1)

Expand the parentheses:

(3y+3)=-2y-1

Add to both sides:

(3y+3)+2y=(-2y-1)+2y

Group like terms:

(3y+2y)+3=(-2y-1)+2y

Simplify the arithmetic:

5y+3=(-2y-1)+2y

Group like terms:

5y+3=(-2y+2y)-1

Simplify the arithmetic:

5y+3=1

Subtract from both sides:

(5y+3)-3=-1-3

Simplify the arithmetic:

5y=13

Simplify the arithmetic:

5y=4

Divide both sides by :

(5y)5=-45

Simplify the fraction:

y=-45

3. List the solutions

y=-2,-45
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3y+3|
y=|2y+1|
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.