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Solution - Absolute value equations

Exact form: u=32
u=\frac{3}{2}
Mixed number form: u=112
u=1\frac{1}{2}
Decimal form: u=1.5
u=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
|3u5|=|3u4|
without the absolute value bars:

|x|=|y||3u5|=|3u4|
x=+y(3u5)=(3u4)
x=y(3u5)=(3u4)
+x=y(3u5)=(3u4)
x=y(3u5)=(3u4)

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||3u5|=|3u4|
x=+y , +x=y(3u5)=(3u4)
x=y , x=y(3u5)=(3u4)

2. Solve the two equations for u

5 additional steps

(3u-5)=(3u-4)

Subtract from both sides:

(3u-5)-3u=(3u-4)-3u

Group like terms:

(3u-3u)-5=(3u-4)-3u

Simplify the arithmetic:

-5=(3u-4)-3u

Group like terms:

-5=(3u-3u)-4

Simplify the arithmetic:

5=4

The statement is false:

5=4

The equation is false so it has no solution.

12 additional steps

(3u-5)=-(3u-4)

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

6u5=4

Add to both sides:

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

Simplify the arithmetic:

6u=4+5

Simplify the arithmetic:

6u=9

Divide both sides by :

(6u)6=96

Simplify the fraction:

u=96

Find the greatest common factor of the numerator and denominator:

u=(3·3)(2·3)

Factor out and cancel the greatest common factor:

u=32

3. Graph

Each line represents the function of one side of the equation:
y=|3u5|
y=|3u4|
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.