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Solution - Absolute value equations

Exact form: u=-8,23
u=-8 , \frac{2}{3}
Decimal form: u=8,0.667
u=-8 , 0.667

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
|4u7|=|5u+1|
without the absolute value bars:

|x|=|y||4u7|=|5u+1|
x=+y(4u7)=(5u+1)
x=y(4u7)=(5u+1)
+x=y(4u7)=(5u+1)
x=y(4u7)=(5u+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||4u7|=|5u+1|
x=+y , +x=y(4u7)=(5u+1)
x=y , x=y(4u7)=(5u+1)

2. Solve the two equations for u

10 additional steps

(4u-7)=(5u+1)

Subtract from both sides:

(4u-7)-5u=(5u+1)-5u

Group like terms:

(4u-5u)-7=(5u+1)-5u

Simplify the arithmetic:

-u-7=(5u+1)-5u

Group like terms:

-u-7=(5u-5u)+1

Simplify the arithmetic:

u7=1

Add to both sides:

(-u-7)+7=1+7

Simplify the arithmetic:

u=1+7

Simplify the arithmetic:

u=8

Multiply both sides by :

-u·-1=8·-1

Remove the one(s):

u=8·-1

Simplify the arithmetic:

u=8

12 additional steps

(4u-7)=-(5u+1)

Expand the parentheses:

(4u-7)=-5u-1

Add to both sides:

(4u-7)+5u=(-5u-1)+5u

Group like terms:

(4u+5u)-7=(-5u-1)+5u

Simplify the arithmetic:

9u-7=(-5u-1)+5u

Group like terms:

9u-7=(-5u+5u)-1

Simplify the arithmetic:

9u7=1

Add to both sides:

(9u-7)+7=-1+7

Simplify the arithmetic:

9u=1+7

Simplify the arithmetic:

9u=6

Divide both sides by :

(9u)9=69

Simplify the fraction:

u=69

Find the greatest common factor of the numerator and denominator:

u=(2·3)(3·3)

Factor out and cancel the greatest common factor:

u=23

3. List the solutions

u=-8,23
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4u7|
y=|5u+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.