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Solution - Absolute value equations

Exact form: y=65,-8
y=\frac{6}{5} , -8
Mixed number form: y=115,-8
y=1\frac{1}{5} , -8
Decimal form: y=1.2,8
y=1.2 , -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+1|=|2y+7|
without the absolute value bars:

|x|=|y||3y+1|=|2y+7|
x=+y(3y+1)=(2y+7)
x=y(3y+1)=(2y+7)
+x=y(3y+1)=(2y+7)
x=y(3y+1)=(2y+7)

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+1|=|2y+7|
x=+y , +x=y(3y+1)=(2y+7)
x=y , x=y(3y+1)=(2y+7)

2. Solve the two equations for y

9 additional steps

(3y+1)=(-2y+7)

Add to both sides:

(3y+1)+2y=(-2y+7)+2y

Group like terms:

(3y+2y)+1=(-2y+7)+2y

Simplify the arithmetic:

5y+1=(-2y+7)+2y

Group like terms:

5y+1=(-2y+2y)+7

Simplify the arithmetic:

5y+1=7

Subtract from both sides:

(5y+1)-1=7-1

Simplify the arithmetic:

5y=71

Simplify the arithmetic:

5y=6

Divide both sides by :

(5y)5=65

Simplify the fraction:

y=65

8 additional steps

(3y+1)=-(-2y+7)

Expand the parentheses:

(3y+1)=2y-7

Subtract from both sides:

(3y+1)-2y=(2y-7)-2y

Group like terms:

(3y-2y)+1=(2y-7)-2y

Simplify the arithmetic:

y+1=(2y-7)-2y

Group like terms:

y+1=(2y-2y)-7

Simplify the arithmetic:

y+1=7

Subtract from both sides:

(y+1)-1=-7-1

Simplify the arithmetic:

y=71

Simplify the arithmetic:

y=8

3. List the solutions

y=65,-8
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3y+1|
y=|2y+7|
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.