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Solution - Absolute value equations

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

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
|3y4|=|3y5|
without the absolute value bars:

|x|=|y||3y4|=|3y5|
x=+y(3y4)=(3y5)
x=y(3y4)=(3y5)
+x=y(3y4)=(3y5)
x=y(3y4)=(3y5)

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||3y4|=|3y5|
x=+y , +x=y(3y4)=(3y5)
x=y , x=y(3y4)=(3y5)

2. Solve the two equations for y

5 additional steps

(3y-4)=(3y-5)

Subtract from both sides:

(3y-4)-3y=(3y-5)-3y

Group like terms:

(3y-3y)-4=(3y-5)-3y

Simplify the arithmetic:

-4=(3y-5)-3y

Group like terms:

-4=(3y-3y)-5

Simplify the arithmetic:

4=5

The statement is false:

4=5

The equation is false so it has no solution.

12 additional steps

(3y-4)=-(3y-5)

Expand the parentheses:

(3y-4)=-3y+5

Add to both sides:

(3y-4)+3y=(-3y+5)+3y

Group like terms:

(3y+3y)-4=(-3y+5)+3y

Simplify the arithmetic:

6y-4=(-3y+5)+3y

Group like terms:

6y-4=(-3y+3y)+5

Simplify the arithmetic:

6y4=5

Add to both sides:

(6y-4)+4=5+4

Simplify the arithmetic:

6y=5+4

Simplify the arithmetic:

6y=9

Divide both sides by :

(6y)6=96

Simplify the fraction:

y=96

Find the greatest common factor of the numerator and denominator:

y=(3·3)(2·3)

Factor out and cancel the greatest common factor:

y=32

3. Graph

Each line represents the function of one side of the equation:
y=|3y4|
y=|3y5|
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.