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Solution - Absolute value equations

Exact form:

x = 0, - \frac{4}{27}

x=0 , -\frac{4}{27}

Decimal form:

x = 0, - 0.148

x=0 , -0.148

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
$| x + \frac{1}{12} | = | \frac{1}{8} x + \frac{1}{12} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| x + \frac{1}{12} \| = \| \frac{1}{8} x + \frac{1}{12} \|$
$x = + y$	$(x + \frac{1}{12}) = (\frac{1}{8} x + \frac{1}{12})$
$x = - y$	$(x + \frac{1}{12}) = - (\frac{1}{8} x + \frac{1}{12})$
$+ x = y$	$(x + \frac{1}{12}) = (\frac{1}{8} x + \frac{1}{12})$
$- x = y$	$- (x + \frac{1}{12}) = (\frac{1}{8} x + \frac{1}{12})$

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:

$\| x \| = \| y \|$	$\| x + \frac{1}{12} \| = \| \frac{1}{8} x + \frac{1}{12} \|$
$x = + y$ , $+ x = y$	$(x + \frac{1}{12}) = (\frac{1}{8} x + \frac{1}{12})$
$x = - y$ , $- x = y$	$(x + \frac{1}{12}) = - (\frac{1}{8} x + \frac{1}{12})$

2. Solve the two equations for x

19 additional steps

$(x + \frac{1}{12}) = (\frac{1}{8} x + \frac{1}{12})$

Subtract from both sides:

$(x + \frac{1}{12}) - \frac{1}{8} \cdot x = (\frac{1}{8} x + \frac{1}{12}) - \frac{1}{8} x$

Group like terms:

$(x + \frac{- 1}{8} \cdot x) + \frac{1}{12} = (\frac{1}{8} \cdot x + \frac{1}{12}) - \frac{1}{8} x$

Group the coefficients:

$(1 + \frac{- 1}{8}) x + \frac{1}{12} = (\frac{1}{8} \cdot x + \frac{1}{12}) - \frac{1}{8} x$

Convert the integer into a fraction:

$(\frac{8}{8} + \frac{- 1}{8}) x + \frac{1}{12} = (\frac{1}{8} \cdot x + \frac{1}{12}) - \frac{1}{8} x$

Combine the fractions:

$\frac{(8 - 1)}{8} \cdot x + \frac{1}{12} = (\frac{1}{8} \cdot x + \frac{1}{12}) - \frac{1}{8} x$

Combine the numerators:

$\frac{7}{8} \cdot x + \frac{1}{12} = (\frac{1}{8} \cdot x + \frac{1}{12}) - \frac{1}{8} x$

Group like terms:

$\frac{7}{8} \cdot x + \frac{1}{12} = (\frac{1}{8} \cdot x + \frac{- 1}{8} x) + \frac{1}{12}$

Combine the fractions:

$\frac{7}{8} \cdot x + \frac{1}{12} = \frac{(1 - 1)}{8} x + \frac{1}{12}$

Combine the numerators:

$\frac{7}{8} \cdot x + \frac{1}{12} = \frac{0}{8} x + \frac{1}{12}$

Reduce the zero numerator:

$\frac{7}{8} x + \frac{1}{12} = 0 x + \frac{1}{12}$

Simplify the arithmetic:

$\frac{7}{8} x + \frac{1}{12} = \frac{1}{12}$

Subtract from both sides:

$(\frac{7}{8} x + \frac{1}{12}) - \frac{1}{12} = (\frac{1}{12}) - \frac{1}{12}$

Combine the fractions:

$\frac{7}{8} x + \frac{(1 - 1)}{12} = (\frac{1}{12}) - \frac{1}{12}$

Combine the numerators:

$\frac{7}{8} x + \frac{0}{12} = (\frac{1}{12}) - \frac{1}{12}$

Reduce the zero numerator:

$\frac{7}{8} x + 0 = (\frac{1}{12}) - \frac{1}{12}$

Simplify the arithmetic:

$\frac{7}{8} x = (\frac{1}{12}) - \frac{1}{12}$

Combine the fractions:

$\frac{7}{8} x = \frac{(1 - 1)}{12}$

Combine the numerators:

$\frac{7}{8} x = \frac{0}{12}$

Reduce the zero numerator:

$\frac{7}{8} x = 0$

Divide both sides by the coefficient:

$x = 0$

27 additional steps

$(x + \frac{1}{12}) = - (\frac{1}{8} x + \frac{1}{12})$

Expand the parentheses:

$(x + \frac{1}{12}) = \frac{- 1}{8} x + \frac{- 1}{12}$

Add to both sides:

$(x + \frac{1}{12}) + \frac{1}{8} \cdot x = (\frac{- 1}{8} x + \frac{- 1}{12}) + \frac{1}{8} x$

Group like terms:

$(x + \frac{1}{8} \cdot x) + \frac{1}{12} = (\frac{- 1}{8} \cdot x + \frac{- 1}{12}) + \frac{1}{8} x$

Group the coefficients:

$(1 + \frac{1}{8}) x + \frac{1}{12} = (\frac{- 1}{8} \cdot x + \frac{- 1}{12}) + \frac{1}{8} x$

Convert the integer into a fraction:

$(\frac{8}{8} + \frac{1}{8}) x + \frac{1}{12} = (\frac{- 1}{8} \cdot x + \frac{- 1}{12}) + \frac{1}{8} x$

Combine the fractions:

$\frac{(8 + 1)}{8} \cdot x + \frac{1}{12} = (\frac{- 1}{8} \cdot x + \frac{- 1}{12}) + \frac{1}{8} x$

Combine the numerators:

$\frac{9}{8} \cdot x + \frac{1}{12} = (\frac{- 1}{8} \cdot x + \frac{- 1}{12}) + \frac{1}{8} x$

Group like terms:

$\frac{9}{8} \cdot x + \frac{1}{12} = (\frac{- 1}{8} \cdot x + \frac{1}{8} x) + \frac{- 1}{12}$

Combine the fractions:

$\frac{9}{8} \cdot x + \frac{1}{12} = \frac{(- 1 + 1)}{8} x + \frac{- 1}{12}$

Combine the numerators:

$\frac{9}{8} \cdot x + \frac{1}{12} = \frac{0}{8} x + \frac{- 1}{12}$

Reduce the zero numerator:

$\frac{9}{8} x + \frac{1}{12} = 0 x + \frac{- 1}{12}$

Simplify the arithmetic:

$\frac{9}{8} x + \frac{1}{12} = \frac{- 1}{12}$

Subtract from both sides:

$(\frac{9}{8} x + \frac{1}{12}) - \frac{1}{12} = (\frac{- 1}{12}) - \frac{1}{12}$

Combine the fractions:

$\frac{9}{8} x + \frac{(1 - 1)}{12} = (\frac{- 1}{12}) - \frac{1}{12}$

Combine the numerators:

$\frac{9}{8} x + \frac{0}{12} = (\frac{- 1}{12}) - \frac{1}{12}$

Reduce the zero numerator:

$\frac{9}{8} x + 0 = (\frac{- 1}{12}) - \frac{1}{12}$

Simplify the arithmetic:

$\frac{9}{8} x = (\frac{- 1}{12}) - \frac{1}{12}$

Combine the fractions:

$\frac{9}{8} x = \frac{(- 1 - 1)}{12}$

Combine the numerators:

$\frac{9}{8} x = \frac{- 2}{12}$

Find the greatest common factor of the numerator and denominator:

$\frac{9}{8} x = \frac{(- 1 \cdot 2)}{(6 \cdot 2)}$

Factor out and cancel the greatest common factor:

$\frac{9}{8} x = \frac{- 1}{6}$

Multiply both sides by inverse fraction :

$(\frac{9}{8} x) \cdot \frac{8}{9} = (\frac{- 1}{6}) \cdot \frac{8}{9}$

Group like terms:

$(\frac{9}{8} \cdot \frac{8}{9}) x = (\frac{- 1}{6}) \cdot \frac{8}{9}$

Multiply the coefficients:

$\frac{(9 \cdot 8)}{(8 \cdot 9)} x = (\frac{- 1}{6}) \cdot \frac{8}{9}$

Simplify the fraction:

$x = (\frac{- 1}{6}) \cdot \frac{8}{9}$

Multiply the fraction(s):

$x = \frac{(- 1 \cdot 8)}{(6 \cdot 9)}$

Simplify the arithmetic:

$x = \frac{- 4}{(3 \cdot 9)}$

$x = \frac{- 4}{27}$

3. List the solutions

$x = 0, - \frac{4}{27}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | x + \frac{1}{12} |$
$y = | \frac{1}{8} x + \frac{1}{12} |$
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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