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Solution - Absolute value equations

Exact form: x=2,-83
x=2 , -\frac{8}{3}
Mixed number form: x=2,-223
x=2 , -2\frac{2}{3}
Decimal form: x=2,2.667
x=2 , -2.667

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
|4x1|=|x+9|
without the absolute value bars:

|x|=|y||4x1|=|x+9|
x=+y(4x1)=(x+9)
x=y(4x1)=(x+9)
+x=y(4x1)=(x+9)
x=y(4x1)=(x+9)

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||4x1|=|x+9|
x=+y , +x=y(4x1)=(x+9)
x=y , x=y(4x1)=(x+9)

2. Solve the two equations for x

11 additional steps

(4x-1)=(-x+9)

Add to both sides:

(4x-1)+x=(-x+9)+x

Group like terms:

(4x+x)-1=(-x+9)+x

Simplify the arithmetic:

5x-1=(-x+9)+x

Group like terms:

5x-1=(-x+x)+9

Simplify the arithmetic:

5x1=9

Add to both sides:

(5x-1)+1=9+1

Simplify the arithmetic:

5x=9+1

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

(4x-1)=-(-x+9)

Expand the parentheses:

(4x-1)=x-9

Subtract from both sides:

(4x-1)-x=(x-9)-x

Group like terms:

(4x-x)-1=(x-9)-x

Simplify the arithmetic:

3x-1=(x-9)-x

Group like terms:

3x-1=(x-x)-9

Simplify the arithmetic:

3x1=9

Add to both sides:

(3x-1)+1=-9+1

Simplify the arithmetic:

3x=9+1

Simplify the arithmetic:

3x=8

Divide both sides by :

(3x)3=-83

Simplify the fraction:

x=-83

3. List the solutions

x=2,-83
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4x1|
y=|x+9|
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.