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Solution - Absolute value equations

Exact form:

x = \frac{2}{5}, \frac{2}{5}

x=\frac{2}{5} , \frac{2}{5}

Decimal form:

x = 0.4, 0.4

x=0.4 , 0.4

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
$6 | x - \frac{2}{5} | = | x - \frac{2}{5} |$
without the absolute value bars:

$\| x \| = \| y \|$	$6 \| x - \frac{2}{5} \| = \| x - \frac{2}{5} \|$
$x = + y$	$6 (x - \frac{2}{5}) = (x - \frac{2}{5})$
$x = - y$	$6 (x - \frac{2}{5}) = - (x - \frac{2}{5})$
$+ x = y$	$6 (x - \frac{2}{5}) = (x - \frac{2}{5})$
$- x = y$	$6 (- (x - \frac{2}{5})) = (x - \frac{2}{5})$

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:

$\| x \| = \| y \|$	$6 \| x - \frac{2}{5} \| = \| x - \frac{2}{5} \|$
$x = + y$ , $+ x = y$	$6 (x - \frac{2}{5}) = (x - \frac{2}{5})$
$x = - y$ , $- x = y$	$6 (x - \frac{2}{5}) = - (x - \frac{2}{5})$

2. Solve the two equations for x

17 additional steps

$6 \cdot (x + \frac{- 2}{5}) = (x + \frac{- 2}{5})$

Expand the parentheses:

$x \cdot 6 + \frac{(- 2 \cdot 6)}{5} = (x + \frac{- 2}{5})$

Simplify the arithmetic:

$6 x + \frac{- 12}{5} = (x + \frac{- 2}{5})$

Subtract from both sides:

$(6 x + \frac{- 12}{5}) - x = (x + \frac{- 2}{5}) - x$

Group like terms:

$(6 x - x) + \frac{- 12}{5} = (x + \frac{- 2}{5}) - x$

Simplify the arithmetic:

$5 x + \frac{- 12}{5} = (x + \frac{- 2}{5}) - x$

Group like terms:

$5 x + \frac{- 12}{5} = (x - x) + \frac{- 2}{5}$

Simplify the arithmetic:

$5 x + \frac{- 12}{5} = \frac{- 2}{5}$

Add to both sides:

$(5 x + \frac{- 12}{5}) + \frac{12}{5} = (\frac{- 2}{5}) + \frac{12}{5}$

Combine the fractions:

$5 x + \frac{(- 12 + 12)}{5} = (\frac{- 2}{5}) + \frac{12}{5}$

Combine the numerators:

$5 x + \frac{0}{5} = (\frac{- 2}{5}) + \frac{12}{5}$

Reduce the zero numerator:

$5 x + 0 = (\frac{- 2}{5}) + \frac{12}{5}$

Simplify the arithmetic:

$5 x = (\frac{- 2}{5}) + \frac{12}{5}$

Combine the fractions:

$5 x = \frac{(- 2 + 12)}{5}$

Combine the numerators:

$5 x = \frac{10}{5}$

Find the greatest common factor of the numerator and denominator:

$5 x = \frac{(2 \cdot 5)}{(1 \cdot 5)}$

Factor out and cancel the greatest common factor:

$5 x = 2$

Divide both sides by :

$\frac{(5 x)}{5} = \frac{2}{5}$

Simplify the fraction:

$x = \frac{2}{5}$

18 additional steps

$6 \cdot (x + \frac{- 2}{5}) = - (x + \frac{- 2}{5})$

Expand the parentheses:

$x \cdot 6 + \frac{(- 2 \cdot 6)}{5} = - (x + \frac{- 2}{5})$

Simplify the arithmetic:

$6 x + \frac{- 12}{5} = - (x + \frac{- 2}{5})$

Expand the parentheses:

$6 x + \frac{- 12}{5} = - x + \frac{2}{5}$

Add to both sides:

$(6 x + \frac{- 12}{5}) + x = (- x + \frac{2}{5}) + x$

Group like terms:

$(6 x + x) + \frac{- 12}{5} = (- x + \frac{2}{5}) + x$

Simplify the arithmetic:

$7 x + \frac{- 12}{5} = (- x + \frac{2}{5}) + x$

Group like terms:

$7 x + \frac{- 12}{5} = (- x + x) + \frac{2}{5}$

Simplify the arithmetic:

$7 x + \frac{- 12}{5} = \frac{2}{5}$

Add to both sides:

$(7 x + \frac{- 12}{5}) + \frac{12}{5} = (\frac{2}{5}) + \frac{12}{5}$

Combine the fractions:

$7 x + \frac{(- 12 + 12)}{5} = (\frac{2}{5}) + \frac{12}{5}$

Combine the numerators:

$7 x + \frac{0}{5} = (\frac{2}{5}) + \frac{12}{5}$

Reduce the zero numerator:

$7 x + 0 = (\frac{2}{5}) + \frac{12}{5}$

Simplify the arithmetic:

$7 x = (\frac{2}{5}) + \frac{12}{5}$

Combine the fractions:

$7 x = \frac{(2 + 12)}{5}$

Combine the numerators:

$7 x = \frac{14}{5}$

Divide both sides by :

$\frac{(7 x)}{7} = \frac{(\frac{14}{5})}{7}$

Simplify the fraction:

$x = \frac{(\frac{14}{5})}{7}$

Simplify the arithmetic:

$x = \frac{14}{(5 \cdot 7)}$

$x = \frac{2}{5}$

3. List the solutions

$x = \frac{2}{5}, \frac{2}{5}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = 6 | x - \frac{2}{5} |$
$y = | x - \frac{2}{5} |$
The equation is true where the two lines cross.

How did we do?

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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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