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Solution - Absolute value equations

Exact form:

x = - \frac{216}{11}, - \frac{54}{13}

x=-\frac{216}{11} , -\frac{54}{13}

Mixed number form:

x = - 19 \frac{7}{11}, - 4 \frac{2}{13}

x=-19\frac{7}{11} , -4\frac{2}{13}

Decimal form:

x = - 19.636, - 4.154

x=-19.636 , -4.154

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
$| \frac{4}{9} x + 5 | = | \frac{1}{27} x - 3 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{4}{9} x + 5 \| = \| \frac{1}{27} x - 3 \|$
$x = + y$	$(\frac{4}{9} x + 5) = (\frac{1}{27} x - 3)$
$x = - y$	$(\frac{4}{9} x + 5) = - (\frac{1}{27} x - 3)$
$+ x = y$	$(\frac{4}{9} x + 5) = (\frac{1}{27} x - 3)$
$- x = y$	$- (\frac{4}{9} x + 5) = (\frac{1}{27} x - 3)$

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:

$\| x \| = \| y \|$	$\| \frac{4}{9} x + 5 \| = \| \frac{1}{27} x - 3 \|$
$x = + y$ , $+ x = y$	$(\frac{4}{9} x + 5) = (\frac{1}{27} x - 3)$
$x = - y$ , $- x = y$	$(\frac{4}{9} x + 5) = - (\frac{1}{27} x - 3)$

2. Solve the two equations for x

21 additional steps

$(\frac{4}{9} \cdot x + 5) = (\frac{1}{27} x - 3)$

Subtract from both sides:

$(\frac{4}{9} x + 5) - \frac{1}{27} \cdot x = (\frac{1}{27} x - 3) - \frac{1}{27} x$

Group like terms:

$(\frac{4}{9} \cdot x + \frac{- 1}{27} \cdot x) + 5 = (\frac{1}{27} \cdot x - 3) - \frac{1}{27} x$

Group the coefficients:

$(\frac{4}{9} + \frac{- 1}{27}) x + 5 = (\frac{1}{27} \cdot x - 3) - \frac{1}{27} x$

Find the lowest common denominator:

$(\frac{(4 \cdot 3)}{(9 \cdot 3)} + \frac{- 1}{27}) x + 5 = (\frac{1}{27} \cdot x - 3) - \frac{1}{27} x$

Multiply the denominators:

$(\frac{(4 \cdot 3)}{27} + \frac{- 1}{27}) x + 5 = (\frac{1}{27} \cdot x - 3) - \frac{1}{27} x$

Multiply the numerators:

$(\frac{12}{27} + \frac{- 1}{27}) x + 5 = (\frac{1}{27} \cdot x - 3) - \frac{1}{27} x$

Combine the fractions:

$\frac{(12 - 1)}{27} \cdot x + 5 = (\frac{1}{27} \cdot x - 3) - \frac{1}{27} x$

Combine the numerators:

$\frac{11}{27} \cdot x + 5 = (\frac{1}{27} \cdot x - 3) - \frac{1}{27} x$

Group like terms:

$\frac{11}{27} \cdot x + 5 = (\frac{1}{27} \cdot x + \frac{- 1}{27} x) - 3$

Combine the fractions:

$\frac{11}{27} \cdot x + 5 = \frac{(1 - 1)}{27} x - 3$

Combine the numerators:

$\frac{11}{27} \cdot x + 5 = \frac{0}{27} x - 3$

Reduce the zero numerator:

$\frac{11}{27} x + 5 = 0 x - 3$

Simplify the arithmetic:

$\frac{11}{27} x + 5 = - 3$

Subtract from both sides:

$(\frac{11}{27} x + 5) - 5 = - 3 - 5$

Simplify the arithmetic:

$\frac{11}{27} x = - 3 - 5$

Simplify the arithmetic:

$\frac{11}{27} x = - 8$

Multiply both sides by inverse fraction :

$(\frac{11}{27} x) \cdot \frac{27}{11} = - 8 \cdot \frac{27}{11}$

Group like terms:

$(\frac{11}{27} \cdot \frac{27}{11}) x = - 8 \cdot \frac{27}{11}$

Multiply the coefficients:

$\frac{(11 \cdot 27)}{(27 \cdot 11)} x = - 8 \cdot \frac{27}{11}$

Simplify the fraction:

$x = - 8 \cdot \frac{27}{11}$

Multiply the fraction(s):

$x = \frac{(- 8 \cdot 27)}{11}$

Simplify the arithmetic:

$x = \frac{- 216}{11}$

22 additional steps

$(\frac{4}{9} x + 5) = - (\frac{1}{27} x - 3)$

Expand the parentheses:

$(\frac{4}{9} \cdot x + 5) = \frac{- 1}{27} x + 3$

Add to both sides:

$(\frac{4}{9} x + 5) + \frac{1}{27} \cdot x = (\frac{- 1}{27} x + 3) + \frac{1}{27} x$

Group like terms:

$(\frac{4}{9} \cdot x + \frac{1}{27} \cdot x) + 5 = (\frac{- 1}{27} \cdot x + 3) + \frac{1}{27} x$

Group the coefficients:

$(\frac{4}{9} + \frac{1}{27}) x + 5 = (\frac{- 1}{27} \cdot x + 3) + \frac{1}{27} x$

Find the lowest common denominator:

$(\frac{(4 \cdot 3)}{(9 \cdot 3)} + \frac{1}{27}) x + 5 = (\frac{- 1}{27} \cdot x + 3) + \frac{1}{27} x$

Multiply the denominators:

$(\frac{(4 \cdot 3)}{27} + \frac{1}{27}) x + 5 = (\frac{- 1}{27} \cdot x + 3) + \frac{1}{27} x$

Multiply the numerators:

$(\frac{12}{27} + \frac{1}{27}) x + 5 = (\frac{- 1}{27} \cdot x + 3) + \frac{1}{27} x$

Combine the fractions:

$\frac{(12 + 1)}{27} \cdot x + 5 = (\frac{- 1}{27} \cdot x + 3) + \frac{1}{27} x$

Combine the numerators:

$\frac{13}{27} \cdot x + 5 = (\frac{- 1}{27} \cdot x + 3) + \frac{1}{27} x$

Group like terms:

$\frac{13}{27} \cdot x + 5 = (\frac{- 1}{27} \cdot x + \frac{1}{27} x) + 3$

Combine the fractions:

$\frac{13}{27} \cdot x + 5 = \frac{(- 1 + 1)}{27} x + 3$

Combine the numerators:

$\frac{13}{27} \cdot x + 5 = \frac{0}{27} x + 3$

Reduce the zero numerator:

$\frac{13}{27} x + 5 = 0 x + 3$

Simplify the arithmetic:

$\frac{13}{27} x + 5 = 3$

Subtract from both sides:

$(\frac{13}{27} x + 5) - 5 = 3 - 5$

Simplify the arithmetic:

$\frac{13}{27} x = 3 - 5$

Simplify the arithmetic:

$\frac{13}{27} x = - 2$

Multiply both sides by inverse fraction :

$(\frac{13}{27} x) \cdot \frac{27}{13} = - 2 \cdot \frac{27}{13}$

Group like terms:

$(\frac{13}{27} \cdot \frac{27}{13}) x = - 2 \cdot \frac{27}{13}$

Multiply the coefficients:

$\frac{(13 \cdot 27)}{(27 \cdot 13)} x = - 2 \cdot \frac{27}{13}$

Simplify the fraction:

$x = - 2 \cdot \frac{27}{13}$

Multiply the fraction(s):

$x = \frac{(- 2 \cdot 27)}{13}$

Simplify the arithmetic:

$x = \frac{- 54}{13}$

3. List the solutions

$x = - \frac{216}{11}, - \frac{54}{13}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{4}{9} x + 5 |$
$y = | \frac{1}{27} x - 3 |$
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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