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Solution - Absolute value equations

Exact form: y=12,-32
y=\frac{1}{2} , -\frac{3}{2}
Mixed number form: y=12,-112
y=\frac{1}{2} , -1\frac{1}{2}
Decimal form: y=0.5,1.5
y=0.5 , -1.5

Other Ways to Solve

Absolute value equations

Step-by-step explanation

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

|4y|+|2y3|=0

Add |2y3| to both sides of the equation:

|4y|+|2y3||2y3|=|2y3|

Simplify the arithmetic

|4y|=|2y3|

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
|4y|=|2y3|
without the absolute value bars:

|x|=|y||4y|=|2y3|
x=+y(4y)=(2y3)
x=y(4y)=(2y3)
+x=y(4y)=(2y3)
x=y(4y)=(2y3)

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||4y|=|2y3|
x=+y , +x=y(4y)=(2y3)
x=y , x=y(4y)=(2y3)

3. Solve the two equations for y

8 additional steps

4y=-(2y-3)

Expand the parentheses:

4y=2y+3

Add to both sides:

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

Simplify the arithmetic:

6y=(-2y+3)+2y

Group like terms:

6y=(-2y+2y)+3

Simplify the arithmetic:

6y=3

Divide both sides by :

(6y)6=36

Simplify the fraction:

y=36

Find the greatest common factor of the numerator and denominator:

y=(1·3)(2·3)

Factor out and cancel the greatest common factor:

y=12

6 additional steps

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

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

4y=2y3

Subtract from both sides:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2y=3

Divide both sides by :

(2y)2=-32

Simplify the fraction:

y=-32

4. List the solutions

y=12,-32
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|4y|
y=|2y3|
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.