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Solution - Absolute value equations

Exact form: y=8,-43
y=8 , -\frac{4}{3}
Mixed number form: y=8,-113
y=8 , -1\frac{1}{3}
Decimal form: y=8,1.333
y=8 , -1.333

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
|2y2|=|y+6|
without the absolute value bars:

|x|=|y||2y2|=|y+6|
x=+y(2y2)=(y+6)
x=y(2y2)=(y+6)
+x=y(2y2)=(y+6)
x=y(2y2)=(y+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||2y2|=|y+6|
x=+y , +x=y(2y2)=(y+6)
x=y , x=y(2y2)=(y+6)

2. Solve the two equations for y

7 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

y-2=(y+6)-y

Group like terms:

y-2=(y-y)+6

Simplify the arithmetic:

y2=6

Add to both sides:

(y-2)+2=6+2

Simplify the arithmetic:

y=6+2

Simplify the arithmetic:

y=8

10 additional steps

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

Expand the parentheses:

(2y-2)=-y-6

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

3y-2=(-y-6)+y

Group like terms:

3y-2=(-y+y)-6

Simplify the arithmetic:

3y2=6

Add to both sides:

(3y-2)+2=-6+2

Simplify the arithmetic:

3y=6+2

Simplify the arithmetic:

3y=4

Divide both sides by :

(3y)3=-43

Simplify the fraction:

y=-43

3. List the solutions

y=8,-43
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2y2|
y=|y+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.