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Solution - Absolute value equations

Exact form: x=40,80173
x=40 , \frac{80}{173}
Decimal form: x=40,0.462
x=40 , 0.462

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
|98x500|=|75x+420|
without the absolute value bars:

|x|=|y||98x500|=|75x+420|
x=+y(98x500)=(75x+420)
x=y(98x500)=(75x+420)
+x=y(98x500)=(75x+420)
x=y(98x500)=(75x+420)

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||98x500|=|75x+420|
x=+y , +x=y(98x500)=(75x+420)
x=y , x=y(98x500)=(75x+420)

2. Solve the two equations for x

11 additional steps

(98x-500)=(75x+420)

Subtract from both sides:

(98x-500)-75x=(75x+420)-75x

Group like terms:

(98x-75x)-500=(75x+420)-75x

Simplify the arithmetic:

23x-500=(75x+420)-75x

Group like terms:

23x-500=(75x-75x)+420

Simplify the arithmetic:

23x500=420

Add to both sides:

(23x-500)+500=420+500

Simplify the arithmetic:

23x=420+500

Simplify the arithmetic:

23x=920

Divide both sides by :

(23x)23=92023

Simplify the fraction:

x=92023

Find the greatest common factor of the numerator and denominator:

x=(40·23)(1·23)

Factor out and cancel the greatest common factor:

x=40

10 additional steps

(98x-500)=-(75x+420)

Expand the parentheses:

(98x-500)=-75x-420

Add to both sides:

(98x-500)+75x=(-75x-420)+75x

Group like terms:

(98x+75x)-500=(-75x-420)+75x

Simplify the arithmetic:

173x-500=(-75x-420)+75x

Group like terms:

173x-500=(-75x+75x)-420

Simplify the arithmetic:

173x500=420

Add to both sides:

(173x-500)+500=-420+500

Simplify the arithmetic:

173x=420+500

Simplify the arithmetic:

173x=80

Divide both sides by :

(173x)173=80173

Simplify the fraction:

x=80173

3. List the solutions

x=40,80173
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|98x500|
y=|75x+420|
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.