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Solution - Absolute value equations

Exact form: x=2,7
x=2 , -7

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
|5x1|=|3x+15|
without the absolute value bars:

|x|=|y||5x1|=|3x+15|
x=+y(5x1)=(3x+15)
x=y(5x1)=(3x+15)
+x=y(5x1)=(3x+15)
x=y(5x1)=(3x+15)

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||5x1|=|3x+15|
x=+y , +x=y(5x1)=(3x+15)
x=y , x=y(5x1)=(3x+15)

2. Solve the two equations for x

11 additional steps

(5x-1)=(-3x+15)

Add to both sides:

(5x-1)+3x=(-3x+15)+3x

Group like terms:

(5x+3x)-1=(-3x+15)+3x

Simplify the arithmetic:

8x-1=(-3x+15)+3x

Group like terms:

8x-1=(-3x+3x)+15

Simplify the arithmetic:

8x1=15

Add to both sides:

(8x-1)+1=15+1

Simplify the arithmetic:

8x=15+1

Simplify the arithmetic:

8x=16

Divide both sides by :

(8x)8=168

Simplify the fraction:

x=168

Find the greatest common factor of the numerator and denominator:

x=(2·8)(1·8)

Factor out and cancel the greatest common factor:

x=2

12 additional steps

(5x-1)=-(-3x+15)

Expand the parentheses:

(5x-1)=3x-15

Subtract from both sides:

(5x-1)-3x=(3x-15)-3x

Group like terms:

(5x-3x)-1=(3x-15)-3x

Simplify the arithmetic:

2x-1=(3x-15)-3x

Group like terms:

2x-1=(3x-3x)-15

Simplify the arithmetic:

2x1=15

Add to both sides:

(2x-1)+1=-15+1

Simplify the arithmetic:

2x=15+1

Simplify the arithmetic:

2x=14

Divide both sides by :

(2x)2=-142

Simplify the fraction:

x=-142

Find the greatest common factor of the numerator and denominator:

x=(-7·2)(1·2)

Factor out and cancel the greatest common factor:

x=7

3. List the solutions

x=2,7
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x1|
y=|3x+15|
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.