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Solution - Absolute value equations

Exact form: y=-4,-13
y=-4 , -\frac{1}{3}
Decimal form: y=4,0.333
y=-4 , -0.333

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
|5y2|=|7y+6|
without the absolute value bars:

|x|=|y||5y2|=|7y+6|
x=+y(5y2)=(7y+6)
x=y(5y2)=(7y+6)
+x=y(5y2)=(7y+6)
x=y(5y2)=(7y+6)

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||5y2|=|7y+6|
x=+y , +x=y(5y2)=(7y+6)
x=y , x=y(5y2)=(7y+6)

2. Solve the two equations for y

13 additional steps

(5y-2)=(7y+6)

Subtract from both sides:

(5y-2)-7y=(7y+6)-7y

Group like terms:

(5y-7y)-2=(7y+6)-7y

Simplify the arithmetic:

-2y-2=(7y+6)-7y

Group like terms:

-2y-2=(7y-7y)+6

Simplify the arithmetic:

2y2=6

Add to both sides:

(-2y-2)+2=6+2

Simplify the arithmetic:

2y=6+2

Simplify the arithmetic:

2y=8

Divide both sides by :

(-2y)-2=8-2

Cancel out the negatives:

2y2=8-2

Simplify the fraction:

y=8-2

Move the negative sign from the denominator to the numerator:

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-2)=-(7y+6)

Expand the parentheses:

(5y-2)=-7y-6

Add to both sides:

(5y-2)+7y=(-7y-6)+7y

Group like terms:

(5y+7y)-2=(-7y-6)+7y

Simplify the arithmetic:

12y-2=(-7y-6)+7y

Group like terms:

12y-2=(-7y+7y)-6

Simplify the arithmetic:

12y2=6

Add to both sides:

(12y-2)+2=-6+2

Simplify the arithmetic:

12y=6+2

Simplify the arithmetic:

12y=4

Divide both sides by :

(12y)12=-412

Simplify the fraction:

y=-412

Find the greatest common factor of the numerator and denominator:

y=(-1·4)(3·4)

Factor out and cancel the greatest common factor:

y=-13

3. List the solutions

y=-4,-13
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5y2|
y=|7y+6|
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.