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Solution - Absolute value equations

Exact form: y=52
y=\frac{5}{2}
Mixed number form: y=212
y=2\frac{1}{2}
Decimal form: y=2.5
y=2.5

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
|2y+4|=|2y6|
without the absolute value bars:

|x|=|y||2y+4|=|2y6|
x=+y(2y+4)=(2y6)
x=y(2y+4)=(2y6)
+x=y(2y+4)=(2y6)
x=y(2y+4)=(2y6)

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||2y+4|=|2y6|
x=+y , +x=y(2y+4)=(2y6)
x=y , x=y(2y+4)=(2y6)

2. Solve the two equations for y

13 additional steps

(-2y+4)=(2y-6)

Subtract from both sides:

(-2y+4)-2y=(2y-6)-2y

Group like terms:

(-2y-2y)+4=(2y-6)-2y

Simplify the arithmetic:

-4y+4=(2y-6)-2y

Group like terms:

-4y+4=(2y-2y)-6

Simplify the arithmetic:

4y+4=6

Subtract from both sides:

(-4y+4)-4=-6-4

Simplify the arithmetic:

4y=64

Simplify the arithmetic:

4y=10

Divide both sides by :

(-4y)-4=-10-4

Cancel out the negatives:

4y4=-10-4

Simplify the fraction:

y=-10-4

Cancel out the negatives:

y=104

Find the greatest common factor of the numerator and denominator:

y=(5·2)(2·2)

Factor out and cancel the greatest common factor:

y=52

6 additional steps

(-2y+4)=-(2y-6)

Expand the parentheses:

(-2y+4)=-2y+6

Add to both sides:

(-2y+4)+2y=(-2y+6)+2y

Group like terms:

(-2y+2y)+4=(-2y+6)+2y

Simplify the arithmetic:

4=(-2y+6)+2y

Group like terms:

4=(-2y+2y)+6

Simplify the arithmetic:

4=6

The statement is false:

4=6

The equation is false so it has no solution.

3. List the solutions

y=52
(1 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2y+4|
y=|2y6|
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.