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Solution - Absolute value equations

Exact form: y=-13,15
y=-\frac{1}{3} , \frac{1}{5}
Decimal form: y=0.333,0.2
y=-0.333 , 0.2

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
|8y|=2|y1|
without the absolute value bars:

|x|=|y||8y|=2|y1|
x=+y(8y)=2(y1)
x=y(8y)=2((y1))
+x=y(8y)=2(y1)
x=y(8y)=2(y1)

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||8y|=2|y1|
x=+y , +x=y(8y)=2(y1)
x=y , x=y(8y)=2((y1))

2. Solve the two equations for y

9 additional steps

8y=2·(y-1)

Expand the parentheses:

8y=2y+2·-1

Simplify the arithmetic:

8y=2y2

Subtract from both sides:

(8y)-2y=(2y-2)-2y

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

6y=2

Divide both sides by :

(6y)6=-26

Simplify the fraction:

y=-26

Find the greatest common factor of the numerator and denominator:

y=(-1·2)(3·2)

Factor out and cancel the greatest common factor:

y=-13

12 additional steps

8y=2·(-(y-1))

Expand the parentheses:

8y=2·(-y+1)

8y=2·-y+2·1

Group like terms:

8y=(2·-1)y+2·1

Multiply the coefficients:

8y=-2y+2·1

Simplify the arithmetic:

8y=2y+2

Add to both sides:

(8y)+2y=(-2y+2)+2y

Simplify the arithmetic:

10y=(-2y+2)+2y

Group like terms:

10y=(-2y+2y)+2

Simplify the arithmetic:

10y=2

Divide both sides by :

(10y)10=210

Simplify the fraction:

y=210

Find the greatest common factor of the numerator and denominator:

y=(1·2)(5·2)

Factor out and cancel the greatest common factor:

y=15

3. List the solutions

y=-13,15
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|8y|
y=2|y1|
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.