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Solution - Absolute value equations

Exact form: x=2,-415
x=2 , -\frac{4}{15}
Decimal form: x=2,0.267
x=2 , -0.267

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
|10x3|=|5x+7|
without the absolute value bars:

|x|=|y||10x3|=|5x+7|
x=+y(10x3)=(5x+7)
x=y(10x3)=(5x+7)
+x=y(10x3)=(5x+7)
x=y(10x3)=(5x+7)

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||10x3|=|5x+7|
x=+y , +x=y(10x3)=(5x+7)
x=y , x=y(10x3)=(5x+7)

2. Solve the two equations for x

11 additional steps

(10x-3)=(5x+7)

Subtract from both sides:

(10x-3)-5x=(5x+7)-5x

Group like terms:

(10x-5x)-3=(5x+7)-5x

Simplify the arithmetic:

5x-3=(5x+7)-5x

Group like terms:

5x-3=(5x-5x)+7

Simplify the arithmetic:

5x3=7

Add to both sides:

(5x-3)+3=7+3

Simplify the arithmetic:

5x=7+3

Simplify the arithmetic:

5x=10

Divide both sides by :

(5x)5=105

Simplify the fraction:

x=105

Find the greatest common factor of the numerator and denominator:

x=(2·5)(1·5)

Factor out and cancel the greatest common factor:

x=2

10 additional steps

(10x-3)=-(5x+7)

Expand the parentheses:

(10x-3)=-5x-7

Add to both sides:

(10x-3)+5x=(-5x-7)+5x

Group like terms:

(10x+5x)-3=(-5x-7)+5x

Simplify the arithmetic:

15x-3=(-5x-7)+5x

Group like terms:

15x-3=(-5x+5x)-7

Simplify the arithmetic:

15x3=7

Add to both sides:

(15x-3)+3=-7+3

Simplify the arithmetic:

15x=7+3

Simplify the arithmetic:

15x=4

Divide both sides by :

(15x)15=-415

Simplify the fraction:

x=-415

3. List the solutions

x=2,-415
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|10x3|
y=|5x+7|
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.