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Solution - Absolute value equations

Exact form: x=53,1511
x=\frac{5}{3} , \frac{15}{11}
Mixed number form: x=123,1411
x=1\frac{2}{3} , 1\frac{4}{11}
Decimal form: x=1.667,1.364
x=1.667 , 1.364

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
5|2x3|=|x|
without the absolute value bars:

|x|=|y|5|2x3|=|x|
x=+y5(2x3)=(x)
x=y5(2x3)=(x)
+x=y5(2x3)=(x)
x=y5((2x3))=(x)

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|5|2x3|=|x|
x=+y , +x=y5(2x3)=(x)
x=y , x=y5(2x3)=(x)

2. Solve the two equations for x

13 additional steps

5·(2x-3)=x

Expand the parentheses:

5·2x+5·-3=x

Multiply the coefficients:

10x+5·-3=x

Simplify the arithmetic:

10x15=x

Subtract from both sides:

(10x-15)-x=x-x

Group like terms:

(10x-x)-15=x-x

Simplify the arithmetic:

9x15=xx

Simplify the arithmetic:

9x15=0

Add to both sides:

(9x-15)+15=0+15

Simplify the arithmetic:

9x=0+15

Simplify the arithmetic:

9x=15

Divide both sides by :

(9x)9=159

Simplify the fraction:

x=159

Find the greatest common factor of the numerator and denominator:

x=(5·3)(3·3)

Factor out and cancel the greatest common factor:

x=53

11 additional steps

5·(2x-3)=-x

Expand the parentheses:

5·2x+5·-3=-x

Multiply the coefficients:

10x+5·-3=-x

Simplify the arithmetic:

10x15=x

Add to both sides:

(10x-15)+x=-x+x

Group like terms:

(10x+x)-15=-x+x

Simplify the arithmetic:

11x15=x+x

Simplify the arithmetic:

11x15=0

Add to both sides:

(11x-15)+15=0+15

Simplify the arithmetic:

11x=0+15

Simplify the arithmetic:

11x=15

Divide both sides by :

(11x)11=1511

Simplify the fraction:

x=1511

3. List the solutions

x=53,1511
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=5|2x3|
y=|x|
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.