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Solution - Absolute value equations

Exact form: x=3,-173
x=3 , -\frac{17}{3}
Mixed number form: x=3,-523
x=3 , -5\frac{2}{3}
Decimal form: x=3,5.667
x=3 , -5.667

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
|5x2|=|2x+19|
without the absolute value bars:

|x|=|y||5x2|=|2x+19|
x=+y(5x2)=(2x+19)
x=y(5x2)=(2x+19)
+x=y(5x2)=(2x+19)
x=y(5x2)=(2x+19)

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||5x2|=|2x+19|
x=+y , +x=y(5x2)=(2x+19)
x=y , x=y(5x2)=(2x+19)

2. Solve the two equations for x

11 additional steps

(5x-2)=(-2x+19)

Add to both sides:

(5x-2)+2x=(-2x+19)+2x

Group like terms:

(5x+2x)-2=(-2x+19)+2x

Simplify the arithmetic:

7x-2=(-2x+19)+2x

Group like terms:

7x-2=(-2x+2x)+19

Simplify the arithmetic:

7x2=19

Add to both sides:

(7x-2)+2=19+2

Simplify the arithmetic:

7x=19+2

Simplify the arithmetic:

7x=21

Divide both sides by :

(7x)7=217

Simplify the fraction:

x=217

Find the greatest common factor of the numerator and denominator:

x=(3·7)(1·7)

Factor out and cancel the greatest common factor:

x=3

10 additional steps

(5x-2)=-(-2x+19)

Expand the parentheses:

(5x-2)=2x-19

Subtract from both sides:

(5x-2)-2x=(2x-19)-2x

Group like terms:

(5x-2x)-2=(2x-19)-2x

Simplify the arithmetic:

3x-2=(2x-19)-2x

Group like terms:

3x-2=(2x-2x)-19

Simplify the arithmetic:

3x2=19

Add to both sides:

(3x-2)+2=-19+2

Simplify the arithmetic:

3x=19+2

Simplify the arithmetic:

3x=17

Divide both sides by :

(3x)3=-173

Simplify the fraction:

x=-173

3. List the solutions

x=3,-173
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x2|
y=|2x+19|
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.