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Solution - Absolute value equations

Exact form:

x = \frac{13}{15}, \frac{7}{45}

x=\frac{13}{15} , \frac{7}{45}

Decimal form:

x = 0.867, 0.156

x=0.867 , 0.156

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
$| x + \frac{1}{5} | = | 2 x - \frac{2}{3} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| x + \frac{1}{5} \| = \| 2 x - \frac{2}{3} \|$
$x = + y$	$(x + \frac{1}{5}) = (2 x - \frac{2}{3})$
$x = - y$	$(x + \frac{1}{5}) = - (2 x - \frac{2}{3})$
$+ x = y$	$(x + \frac{1}{5}) = (2 x - \frac{2}{3})$
$- x = y$	$- (x + \frac{1}{5}) = (2 x - \frac{2}{3})$

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:

$\| x \| = \| y \|$	$\| x + \frac{1}{5} \| = \| 2 x - \frac{2}{3} \|$
$x = + y$ , $+ x = y$	$(x + \frac{1}{5}) = (2 x - \frac{2}{3})$
$x = - y$ , $- x = y$	$(x + \frac{1}{5}) = - (2 x - \frac{2}{3})$

2. Solve the two equations for x

17 additional steps

$(x + \frac{1}{5}) = (2 x + \frac{- 2}{3})$

Subtract from both sides:

$(x + \frac{1}{5}) - 2 x = (2 x + \frac{- 2}{3}) - 2 x$

Group like terms:

$(x - 2 x) + \frac{1}{5} = (2 x + \frac{- 2}{3}) - 2 x$

Simplify the arithmetic:

$- x + \frac{1}{5} = (2 x + \frac{- 2}{3}) - 2 x$

Group like terms:

$- x + \frac{1}{5} = (2 x - 2 x) + \frac{- 2}{3}$

Simplify the arithmetic:

$- x + \frac{1}{5} = \frac{- 2}{3}$

Subtract from both sides:

$(- x + \frac{1}{5}) - \frac{1}{5} = (\frac{- 2}{3}) - \frac{1}{5}$

Combine the fractions:

$- x + \frac{(1 - 1)}{5} = (\frac{- 2}{3}) - \frac{1}{5}$

Combine the numerators:

$- x + \frac{0}{5} = (\frac{- 2}{3}) - \frac{1}{5}$

Reduce the zero numerator:

$- x + 0 = (\frac{- 2}{3}) - \frac{1}{5}$

Simplify the arithmetic:

$- x = (\frac{- 2}{3}) - \frac{1}{5}$

Find the lowest common denominator:

$- x = \frac{(- 2 \cdot 5)}{(3 \cdot 5)} + \frac{(- 1 \cdot 3)}{(5 \cdot 3)}$

Multiply the denominators:

$- x = \frac{(- 2 \cdot 5)}{15} + \frac{(- 1 \cdot 3)}{15}$

Multiply the numerators:

$- x = \frac{- 10}{15} + \frac{- 3}{15}$

Combine the fractions:

$- x = \frac{(- 10 - 3)}{15}$

Combine the numerators:

$- x = \frac{- 13}{15}$

Multiply both sides by :

$- x \cdot - 1 = (\frac{- 13}{15}) \cdot - 1$

Remove the one(s):

$x = (\frac{- 13}{15}) \cdot - 1$

Remove the one(s):

$x = \frac{13}{15}$

19 additional steps

$(x + \frac{1}{5}) = - (2 x + \frac{- 2}{3})$

Expand the parentheses:

$(x + \frac{1}{5}) = - 2 x + \frac{2}{3}$

Add to both sides:

$(x + \frac{1}{5}) + 2 x = (- 2 x + \frac{2}{3}) + 2 x$

Group like terms:

$(x + 2 x) + \frac{1}{5} = (- 2 x + \frac{2}{3}) + 2 x$

Simplify the arithmetic:

$3 x + \frac{1}{5} = (- 2 x + \frac{2}{3}) + 2 x$

Group like terms:

$3 x + \frac{1}{5} = (- 2 x + 2 x) + \frac{2}{3}$

Simplify the arithmetic:

$3 x + \frac{1}{5} = \frac{2}{3}$

Subtract from both sides:

$(3 x + \frac{1}{5}) - \frac{1}{5} = (\frac{2}{3}) - \frac{1}{5}$

Combine the fractions:

$3 x + \frac{(1 - 1)}{5} = (\frac{2}{3}) - \frac{1}{5}$

Combine the numerators:

$3 x + \frac{0}{5} = (\frac{2}{3}) - \frac{1}{5}$

Reduce the zero numerator:

$3 x + 0 = (\frac{2}{3}) - \frac{1}{5}$

Simplify the arithmetic:

$3 x = (\frac{2}{3}) - \frac{1}{5}$

Find the lowest common denominator:

$3 x = \frac{(2 \cdot 5)}{(3 \cdot 5)} + \frac{(- 1 \cdot 3)}{(5 \cdot 3)}$

Multiply the denominators:

$3 x = \frac{(2 \cdot 5)}{15} + \frac{(- 1 \cdot 3)}{15}$

Multiply the numerators:

$3 x = \frac{10}{15} + \frac{- 3}{15}$

Combine the fractions:

$3 x = \frac{(10 - 3)}{15}$

Combine the numerators:

$3 x = \frac{7}{15}$

Divide both sides by :

$\frac{(3 x)}{3} = \frac{(\frac{7}{15})}{3}$

Simplify the fraction:

$x = \frac{(\frac{7}{15})}{3}$

Simplify the arithmetic:

$x = \frac{7}{(15 \cdot 3)}$

$x = \frac{7}{45}$

3. List the solutions

$x = \frac{13}{15}, \frac{7}{45}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | x + \frac{1}{5} |$
$y = | 2 x - \frac{2}{3} |$
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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