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Solution - Absolute value equations

Exact form: y=-43,-25
y=-\frac{4}{3} , -\frac{2}{5}
Mixed number form: y=-113,-25
y=-1\frac{1}{3} , -\frac{2}{5}
Decimal form: y=1.333,0.4
y=-1.333 , -0.4

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
|4y3|=|y+1|
without the absolute value bars:

|x|=|y||4y3|=|y+1|
x=+y(4y3)=(y+1)
x=y(4y3)=(y+1)
+x=y(4y3)=(y+1)
x=y(4y3)=(y+1)

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||4y3|=|y+1|
x=+y , +x=y(4y3)=(y+1)
x=y , x=y(4y3)=(y+1)

2. Solve the two equations for y

11 additional steps

(-4y-3)=(-y+1)

Add to both sides:

(-4y-3)+y=(-y+1)+y

Group like terms:

(-4y+y)-3=(-y+1)+y

Simplify the arithmetic:

-3y-3=(-y+1)+y

Group like terms:

-3y-3=(-y+y)+1

Simplify the arithmetic:

3y3=1

Add to both sides:

(-3y-3)+3=1+3

Simplify the arithmetic:

3y=1+3

Simplify the arithmetic:

3y=4

Divide both sides by :

(-3y)-3=4-3

Cancel out the negatives:

3y3=4-3

Simplify the fraction:

y=4-3

Move the negative sign from the denominator to the numerator:

y=-43

12 additional steps

(-4y-3)=-(-y+1)

Expand the parentheses:

(-4y-3)=y-1

Subtract from both sides:

(-4y-3)-y=(y-1)-y

Group like terms:

(-4y-y)-3=(y-1)-y

Simplify the arithmetic:

-5y-3=(y-1)-y

Group like terms:

-5y-3=(y-y)-1

Simplify the arithmetic:

5y3=1

Add to both sides:

(-5y-3)+3=-1+3

Simplify the arithmetic:

5y=1+3

Simplify the arithmetic:

5y=2

Divide both sides by :

(-5y)-5=2-5

Cancel out the negatives:

5y5=2-5

Simplify the fraction:

y=2-5

Move the negative sign from the denominator to the numerator:

y=-25

3. List the solutions

y=-43,-25
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4y3|
y=|y+1|
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.