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Solution - Absolute value equations

Exact form: y=-1,-15
y=-1 , -\frac{1}{5}
Decimal form: y=1,0.2
y=-1 , -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+4|=2|y1|
without the absolute value bars:

|x|=|y||8y+4|=2|y1|
x=+y(8y+4)=2(y1)
x=y(8y+4)=2((y1))
+x=y(8y+4)=2(y1)
x=y(8y+4)=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+4|=2|y1|
x=+y , +x=y(8y+4)=2(y1)
x=y , x=y(8y+4)=2((y1))

2. Solve the two equations for y

12 additional steps

(8y+4)=2·(y-1)

Expand the parentheses:

(8y+4)=2y+2·-1

Simplify the arithmetic:

(8y+4)=2y-2

Subtract from both sides:

(8y+4)-2y=(2y-2)-2y

Group like terms:

(8y-2y)+4=(2y-2)-2y

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

6y+4=2

Subtract from both sides:

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

Simplify the arithmetic:

6y=24

Simplify the arithmetic:

6y=6

Divide both sides by :

(6y)6=-66

Simplify the fraction:

y=-66

Simplify the fraction:

y=1

16 additional steps

(8y+4)=2·(-(y-1))

Expand the parentheses:

(8y+4)=2·(-y+1)

(8y+4)=2·-y+2·1

Group like terms:

(8y+4)=(2·-1)y+2·1

Multiply the coefficients:

(8y+4)=-2y+2·1

Simplify the arithmetic:

(8y+4)=-2y+2

Add to both sides:

(8y+4)+2y=(-2y+2)+2y

Group like terms:

(8y+2y)+4=(-2y+2)+2y

Simplify the arithmetic:

10y+4=(-2y+2)+2y

Group like terms:

10y+4=(-2y+2y)+2

Simplify the arithmetic:

10y+4=2

Subtract from both sides:

(10y+4)-4=2-4

Simplify the arithmetic:

10y=24

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=-1,-15
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|8y+4|
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.