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Solution - Absolute value equations

Exact form: w=-8,23
w=-8 , \frac{2}{3}
Decimal form: w=8,0.667
w=-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
|4w7|=|5w+1|
without the absolute value bars:

|x|=|y||4w7|=|5w+1|
x=+y(4w7)=(5w+1)
x=y(4w7)=(5w+1)
+x=y(4w7)=(5w+1)
x=y(4w7)=(5w+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||4w7|=|5w+1|
x=+y , +x=y(4w7)=(5w+1)
x=y , x=y(4w7)=(5w+1)

2. Solve the two equations for w

10 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

w7=1

Add to both sides:

(-w-7)+7=1+7

Simplify the arithmetic:

w=1+7

Simplify the arithmetic:

w=8

Multiply both sides by :

-w·-1=8·-1

Remove the one(s):

w=8·-1

Simplify the arithmetic:

w=8

12 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

9w7=1

Add to both sides:

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

Simplify the arithmetic:

9w=1+7

Simplify the arithmetic:

9w=6

Divide both sides by :

(9w)9=69

Simplify the fraction:

w=69

Find the greatest common factor of the numerator and denominator:

w=(2·3)(3·3)

Factor out and cancel the greatest common factor:

w=23

3. List the solutions

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

4. Graph

Each line represents the function of one side of the equation:
y=|4w7|
y=|5w+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.