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Solution - Absolute value equations

Exact form: u=917,3
u=\frac{9}{17} , 3
Decimal form: u=0.529,3
u=0.529 , 3

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
|7u|=|10u+9|
without the absolute value bars:

|x|=|y||7u|=|10u+9|
x=+y(7u)=(10u+9)
x=y(7u)=(10u+9)
+x=y(7u)=(10u+9)
x=y(7u)=(10u+9)

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||7u|=|10u+9|
x=+y , +x=y(7u)=(10u+9)
x=y , x=y(7u)=(10u+9)

2. Solve the two equations for u

5 additional steps

7u=(-10u+9)

Add to both sides:

(7u)+10u=(-10u+9)+10u

Simplify the arithmetic:

17u=(-10u+9)+10u

Group like terms:

17u=(-10u+10u)+9

Simplify the arithmetic:

17u=9

Divide both sides by :

(17u)17=917

Simplify the fraction:

u=917

10 additional steps

7u=-(-10u+9)

Expand the parentheses:

7u=10u9

Subtract from both sides:

(7u)-10u=(10u-9)-10u

Simplify the arithmetic:

-3u=(10u-9)-10u

Group like terms:

-3u=(10u-10u)-9

Simplify the arithmetic:

3u=9

Divide both sides by :

(-3u)-3=-9-3

Cancel out the negatives:

3u3=-9-3

Simplify the fraction:

u=-9-3

Cancel out the negatives:

u=93

Find the greatest common factor of the numerator and denominator:

u=(3·3)(1·3)

Factor out and cancel the greatest common factor:

u=3

3. List the solutions

u=917,3
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|7u|
y=|10u+9|
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.