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Solution - Absolute value equations

Exact form: x=43,-67
x=\frac{4}{3} , -\frac{6}{7}
Mixed number form: x=113,-67
x=1\frac{1}{3} , -\frac{6}{7}
Decimal form: x=1.333,0.857
x=1.333 , -0.857

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
|8x3|=|x+9|
without the absolute value bars:

|x|=|y||8x3|=|x+9|
x=+y(8x3)=(x+9)
x=y(8x3)=(x+9)
+x=y(8x3)=(x+9)
x=y(8x3)=(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||8x3|=|x+9|
x=+y , +x=y(8x3)=(x+9)
x=y , x=y(8x3)=(x+9)

2. Solve the two equations for x

11 additional steps

(8x-3)=(-x+9)

Add to both sides:

(8x-3)+x=(-x+9)+x

Group like terms:

(8x+x)-3=(-x+9)+x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

9x3=9

Add to both sides:

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

Simplify the arithmetic:

9x=9+3

Simplify the arithmetic:

9x=12

Divide both sides by :

(9x)9=129

Simplify the fraction:

x=129

Find the greatest common factor of the numerator and denominator:

x=(4·3)(3·3)

Factor out and cancel the greatest common factor:

x=43

10 additional steps

(8x-3)=-(-x+9)

Expand the parentheses:

(8x-3)=x-9

Subtract from both sides:

(8x-3)-x=(x-9)-x

Group like terms:

(8x-x)-3=(x-9)-x

Simplify the arithmetic:

7x-3=(x-9)-x

Group like terms:

7x-3=(x-x)-9

Simplify the arithmetic:

7x3=9

Add to both sides:

(7x-3)+3=-9+3

Simplify the arithmetic:

7x=9+3

Simplify the arithmetic:

7x=6

Divide both sides by :

(7x)7=-67

Simplify the fraction:

x=-67

3. List the solutions

x=43,-67
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|8x3|
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.