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Solution - Absolute value equations

Exact form: y=2,1
y=-2 , 1

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|y4|3|y|=0

Add 3|y| to both sides of the equation:

|y4|3|y|+3|y|=3|y|

Simplify the arithmetic

|y4|=3|y|

2. 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
|y4|=3|y|
without the absolute value bars:

|x|=|y||y4|=3|y|
x=+y(y4)=3(y)
x=y(y4)=3((y))
+x=y(y4)=3(y)
x=y(y4)=3(y)

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||y4|=3|y|
x=+y , +x=y(y4)=3(y)
x=y , x=y(y4)=3((y))

3. Solve the two equations for y

12 additional steps

(y-4)=3y

Subtract from both sides:

(y-4)-3y=(3y)-3y

Group like terms:

(y-3y)-4=(3y)-3y

Simplify the arithmetic:

-2y-4=(3y)-3y

Simplify the arithmetic:

2y4=0

Add to both sides:

(-2y-4)+4=0+4

Simplify the arithmetic:

2y=0+4

Simplify the arithmetic:

2y=4

Divide both sides by :

(-2y)-2=4-2

Cancel out the negatives:

2y2=4-2

Simplify the fraction:

y=4-2

Move the negative sign from the denominator to the numerator:

y=-42

Find the greatest common factor of the numerator and denominator:

y=(-2·2)(1·2)

Factor out and cancel the greatest common factor:

y=2

11 additional steps

(y-4)=3·-y

Group like terms:

(y-4)=(3·-1)y

Multiply the coefficients:

(y-4)=-3y

Add to both sides:

(y-4)+3y=(-3y)+3y

Group like terms:

(y+3y)-4=(-3y)+3y

Simplify the arithmetic:

4y-4=(-3y)+3y

Simplify the arithmetic:

4y4=0

Add to both sides:

(4y-4)+4=0+4

Simplify the arithmetic:

4y=0+4

Simplify the arithmetic:

4y=4

Divide both sides by :

(4y)4=44

Simplify the fraction:

y=44

Simplify the fraction:

y=1

4. List the solutions

y=2,1
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|y4|
y=3|y|
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.