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Solution - Absolute value equations

Exact form:

x = \frac{8}{3}, - 12

x=\frac{8}{3} , -12

Mixed number form:

x = 2 \frac{2}{3}, - 12

x=2\frac{2}{3} , -12

Decimal form:

x = 2.667, - 12

x=2.667 , -12

See steps

Step-by-step explanation

1. Rewrite the equation without absolute value bars

Use the rules:
$| x | = | y |$ → $x = \pm y$ and $| x | = | y |$ → $\pm x = y$
to write all four options of the equation
$| - 2 x + 20 | = 2 | 2 x + 2 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| - 2 x + 20 \| = 2 \| 2 x + 2 \|$
$x = + y$	$(- 2 x + 20) = 2 (2 x + 2)$
$x = - y$	$(- 2 x + 20) = 2 (- (2 x + 2))$
$+ x = y$	$(- 2 x + 20) = 2 (2 x + 2)$
$- x = y$	$- (- 2 x + 20) = 2 (2 x + 2)$

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:

$\| x \| = \| y \|$	$\| - 2 x + 20 \| = 2 \| 2 x + 2 \|$
$x = + y$ , $+ x = y$	$(- 2 x + 20) = 2 (2 x + 2)$
$x = - y$ , $- x = y$	$(- 2 x + 20) = 2 (- (2 x + 2))$

2. Solve the two equations for x

16 additional steps

$(- 2 x + 20) = 2 \cdot (2 x + 2)$

Expand the parentheses:

$(- 2 x + 20) = 2 \cdot 2 x + 2 \cdot 2$

Multiply the coefficients:

$(- 2 x + 20) = 4 x + 2 \cdot 2$

Simplify the arithmetic:

$(- 2 x + 20) = 4 x + 4$

Subtract from both sides:

$(- 2 x + 20) - 4 x = (4 x + 4) - 4 x$

Group like terms:

$(- 2 x - 4 x) + 20 = (4 x + 4) - 4 x$

Simplify the arithmetic:

$- 6 x + 20 = (4 x + 4) - 4 x$

Group like terms:

$- 6 x + 20 = (4 x - 4 x) + 4$

Simplify the arithmetic:

$- 6 x + 20 = 4$

Subtract from both sides:

$(- 6 x + 20) - 20 = 4 - 20$

Simplify the arithmetic:

$- 6 x = 4 - 20$

Simplify the arithmetic:

$- 6 x = - 16$

Divide both sides by :

$\frac{(- 6 x)}{- 6} = \frac{- 16}{- 6}$

Cancel out the negatives:

$\frac{6 x}{6} = \frac{- 16}{- 6}$

Simplify the fraction:

$x = \frac{- 16}{- 6}$

Cancel out the negatives:

$x = \frac{16}{6}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(8 \cdot 2)}{(3 \cdot 2)}$

Factor out and cancel the greatest common factor:

$x = \frac{8}{3}$

15 additional steps

$(- 2 x + 20) = 2 \cdot (- (2 x + 2))$

Expand the parentheses:

$(- 2 x + 20) = 2 \cdot (- 2 x - 2)$

Expand the parentheses:

$(- 2 x + 20) = 2 \cdot - 2 x + 2 \cdot - 2$

Multiply the coefficients:

$(- 2 x + 20) = - 4 x + 2 \cdot - 2$

Simplify the arithmetic:

$(- 2 x + 20) = - 4 x - 4$

Add to both sides:

$(- 2 x + 20) + 4 x = (- 4 x - 4) + 4 x$

Group like terms:

$(- 2 x + 4 x) + 20 = (- 4 x - 4) + 4 x$

Simplify the arithmetic:

$2 x + 20 = (- 4 x - 4) + 4 x$

Group like terms:

$2 x + 20 = (- 4 x + 4 x) - 4$

Simplify the arithmetic:

$2 x + 20 = - 4$

Subtract from both sides:

$(2 x + 20) - 20 = - 4 - 20$

Simplify the arithmetic:

$2 x = - 4 - 20$

Simplify the arithmetic:

$2 x = - 24$

Divide both sides by :

$\frac{(2 x)}{2} = \frac{- 24}{2}$

Simplify the fraction:

$x = \frac{- 24}{2}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(- 12 \cdot 2)}{(1 \cdot 2)}$

Factor out and cancel the greatest common factor:

$x = - 12$

3. List the solutions

$x = \frac{8}{3}, - 12$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | - 2 x + 20 |$
$y = 2 | 2 x + 2 |$
The equation is true where the two lines cross.

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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.

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

2. Solve the two equations for x

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

2. Solve the two equations for x

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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