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Solution - Absolute value equations

Exact form: y=-4,34
y=-4 , \frac{3}{4}
Decimal form: y=4,0.75
y=-4 , 0.75

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
|5y+1|=|3y7|
without the absolute value bars:

|x|=|y||5y+1|=|3y7|
x=+y(5y+1)=(3y7)
x=y(5y+1)=(3y7)
+x=y(5y+1)=(3y7)
x=y(5y+1)=(3y7)

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||5y+1|=|3y7|
x=+y , +x=y(5y+1)=(3y7)
x=y , x=y(5y+1)=(3y7)

2. Solve the two equations for y

11 additional steps

(5y+1)=(3y-7)

Subtract from both sides:

(5y+1)-3y=(3y-7)-3y

Group like terms:

(5y-3y)+1=(3y-7)-3y

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2y+1=7

Subtract from both sides:

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

Simplify the arithmetic:

2y=71

Simplify the arithmetic:

2y=8

Divide both sides by :

(2y)2=-82

Simplify the fraction:

y=-82

Find the greatest common factor of the numerator and denominator:

y=(-4·2)(1·2)

Factor out and cancel the greatest common factor:

y=4

12 additional steps

(5y+1)=-(3y-7)

Expand the parentheses:

(5y+1)=-3y+7

Add to both sides:

(5y+1)+3y=(-3y+7)+3y

Group like terms:

(5y+3y)+1=(-3y+7)+3y

Simplify the arithmetic:

8y+1=(-3y+7)+3y

Group like terms:

8y+1=(-3y+3y)+7

Simplify the arithmetic:

8y+1=7

Subtract from both sides:

(8y+1)-1=7-1

Simplify the arithmetic:

8y=71

Simplify the arithmetic:

8y=6

Divide both sides by :

(8y)8=68

Simplify the fraction:

y=68

Find the greatest common factor of the numerator and denominator:

y=(3·2)(4·2)

Factor out and cancel the greatest common factor:

y=34

3. List the solutions

y=-4,34
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5y+1|
y=|3y7|
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.