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Solution - Absolute value equations

Exact form: x=17,13
x=\frac{1}{7} , \frac{1}{3}
Decimal form: x=0.143,0.333
x=0.143 , 0.333

Other Ways to Solve

Absolute value equations

Step-by-step explanation

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

|2x|+|5x1|=0

Add |5x1| to both sides of the equation:

|2x|+|5x1||5x1|=|5x1|

Simplify the arithmetic

|2x|=|5x1|

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|=|5x1|
without the absolute value bars:

|x|=|y||2x|=|5x1|
x=+y(2x)=(5x1)
x=y(2x)=(5x1)
+x=y(2x)=(5x1)
x=y(2x)=(5x1)

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|=|5x1|
x=+y , +x=y(2x)=(5x1)
x=y , x=y(2x)=(5x1)

3. Solve the two equations for x

6 additional steps

2x=-(5x-1)

Expand the parentheses:

2x=5x+1

Add to both sides:

(2x)+5x=(-5x+1)+5x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

7x=1

Divide both sides by :

(7x)7=17

Simplify the fraction:

x=17

8 additional steps

2x=-(-(5x-1))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

2x=5x1

Subtract from both sides:

(2x)-5x=(5x-1)-5x

Simplify the arithmetic:

-3x=(5x-1)-5x

Group like terms:

-3x=(5x-5x)-1

Simplify the arithmetic:

3x=1

Divide both sides by :

(-3x)-3=-1-3

Cancel out the negatives:

3x3=-1-3

Simplify the fraction:

x=-1-3

Cancel out the negatives:

x=13

4. List the solutions

x=17,13
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|2x|
y=|5x1|
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.