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Solution - Absolute value equations

Exact form: x=15,19
x=\frac{1}{5} , \frac{1}{9}
Decimal form: x=0.2,0.111
x=0.2 , 0.111

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|2x||7x1|=0

Add |7x1| to both sides of the equation:

|2x||7x1|+|7x1|=|7x1|

Simplify the arithmetic

|2x|=|7x1|

2. 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
|2x|=|7x1|
without the absolute value bars:

|x|=|y||2x|=|7x1|
x=+y(2x)=(7x1)
x=y(2x)=((7x1))
+x=y(2x)=(7x1)
x=y(2x)=(7x1)

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||2x|=|7x1|
x=+y , +x=y(2x)=(7x1)
x=y , x=y(2x)=((7x1))

3. Solve the two equations for x

7 additional steps

2x=(7x-1)

Subtract from both sides:

(2x)-7x=(7x-1)-7x

Simplify the arithmetic:

-5x=(7x-1)-7x

Group like terms:

-5x=(7x-7x)-1

Simplify the arithmetic:

5x=1

Divide both sides by :

(-5x)-5=-1-5

Cancel out the negatives:

5x5=-1-5

Simplify the fraction:

x=-1-5

Cancel out the negatives:

x=15

6 additional steps

2x=-(7x-1)

Expand the parentheses:

2x=7x+1

Add to both sides:

(2x)+7x=(-7x+1)+7x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

9x=1

Divide both sides by :

(9x)9=19

Simplify the fraction:

x=19

4. List the solutions

x=15,19
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|2x|
y=|7x1|
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.