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Solution - Absolute value equations

Exact form:

x = - \frac{2}{3}, - \frac{6}{5}

x=-\frac{2}{3} , -\frac{6}{5}

Mixed number form:

x = - \frac{2}{3}, - 1 \frac{1}{5}

x=-\frac{2}{3} , -1\frac{1}{5}

Decimal form:

x = - 0.667, - 1.2

x=-0.667 , -1.2

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
$| \frac{1}{2} x + \frac{2}{3} | = | \frac{3}{4} x + \frac{5}{6} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{2} x + \frac{2}{3} \| = \| \frac{3}{4} x + \frac{5}{6} \|$
$x = + y$	$(\frac{1}{2} x + \frac{2}{3}) = (\frac{3}{4} x + \frac{5}{6})$
$x = - y$	$(\frac{1}{2} x + \frac{2}{3}) = - (\frac{3}{4} x + \frac{5}{6})$
$+ x = y$	$(\frac{1}{2} x + \frac{2}{3}) = (\frac{3}{4} x + \frac{5}{6})$
$- x = y$	$- (\frac{1}{2} x + \frac{2}{3}) = (\frac{3}{4} x + \frac{5}{6})$

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:

$\| x \| = \| y \|$	$\| \frac{1}{2} x + \frac{2}{3} \| = \| \frac{3}{4} x + \frac{5}{6} \|$
$x = + y$ , $+ x = y$	$(\frac{1}{2} x + \frac{2}{3}) = (\frac{3}{4} x + \frac{5}{6})$
$x = - y$ , $- x = y$	$(\frac{1}{2} x + \frac{2}{3}) = - (\frac{3}{4} x + \frac{5}{6})$

2. Solve the two equations for x

30 additional steps

$(\frac{1}{2} \cdot x + \frac{2}{3}) = (\frac{3}{4} x + \frac{5}{6})$

Subtract from both sides:

$(\frac{1}{2} x + \frac{2}{3}) - \frac{3}{4} \cdot x = (\frac{3}{4} x + \frac{5}{6}) - \frac{3}{4} x$

Group like terms:

$(\frac{1}{2} \cdot x + \frac{- 3}{4} \cdot x) + \frac{2}{3} = (\frac{3}{4} \cdot x + \frac{5}{6}) - \frac{3}{4} x$

Group the coefficients:

$(\frac{1}{2} + \frac{- 3}{4}) x + \frac{2}{3} = (\frac{3}{4} \cdot x + \frac{5}{6}) - \frac{3}{4} x$

Find the lowest common denominator:

$(\frac{(1 \cdot 2)}{(2 \cdot 2)} + \frac{- 3}{4}) x + \frac{2}{3} = (\frac{3}{4} \cdot x + \frac{5}{6}) - \frac{3}{4} x$

Multiply the denominators:

$(\frac{(1 \cdot 2)}{4} + \frac{- 3}{4}) x + \frac{2}{3} = (\frac{3}{4} \cdot x + \frac{5}{6}) - \frac{3}{4} x$

Multiply the numerators:

$(\frac{2}{4} + \frac{- 3}{4}) x + \frac{2}{3} = (\frac{3}{4} \cdot x + \frac{5}{6}) - \frac{3}{4} x$

Combine the fractions:

$\frac{(2 - 3)}{4} \cdot x + \frac{2}{3} = (\frac{3}{4} \cdot x + \frac{5}{6}) - \frac{3}{4} x$

Combine the numerators:

$\frac{- 1}{4} \cdot x + \frac{2}{3} = (\frac{3}{4} \cdot x + \frac{5}{6}) - \frac{3}{4} x$

Group like terms:

$\frac{- 1}{4} \cdot x + \frac{2}{3} = (\frac{3}{4} \cdot x + \frac{- 3}{4} x) + \frac{5}{6}$

Combine the fractions:

$\frac{- 1}{4} \cdot x + \frac{2}{3} = \frac{(3 - 3)}{4} x + \frac{5}{6}$

Combine the numerators:

$\frac{- 1}{4} \cdot x + \frac{2}{3} = \frac{0}{4} x + \frac{5}{6}$

Reduce the zero numerator:

$\frac{- 1}{4} x + \frac{2}{3} = 0 x + \frac{5}{6}$

Simplify the arithmetic:

$\frac{- 1}{4} x + \frac{2}{3} = \frac{5}{6}$

Subtract from both sides:

$(\frac{- 1}{4} x + \frac{2}{3}) - \frac{2}{3} = (\frac{5}{6}) - \frac{2}{3}$

Combine the fractions:

$\frac{- 1}{4} x + \frac{(2 - 2)}{3} = (\frac{5}{6}) - \frac{2}{3}$

Combine the numerators:

$\frac{- 1}{4} x + \frac{0}{3} = (\frac{5}{6}) - \frac{2}{3}$

Reduce the zero numerator:

$\frac{- 1}{4} x + 0 = (\frac{5}{6}) - \frac{2}{3}$

Simplify the arithmetic:

$\frac{- 1}{4} x = (\frac{5}{6}) - \frac{2}{3}$

Find the lowest common denominator:

$\frac{- 1}{4} x = \frac{5}{6} + \frac{(- 2 \cdot 2)}{(3 \cdot 2)}$

Multiply the denominators:

$\frac{- 1}{4} x = \frac{5}{6} + \frac{(- 2 \cdot 2)}{6}$

Multiply the numerators:

$\frac{- 1}{4} x = \frac{5}{6} + \frac{- 4}{6}$

Combine the fractions:

$\frac{- 1}{4} x = \frac{(5 - 4)}{6}$

Combine the numerators:

$\frac{- 1}{4} x = \frac{1}{6}$

Multiply both sides by inverse fraction :

$(\frac{- 1}{4} x) \cdot \frac{4}{- 1} = (\frac{1}{6}) \cdot \frac{4}{- 1}$

Group like terms:

$(\frac{- 1}{4} \cdot - 4) x = (\frac{1}{6}) \cdot \frac{4}{- 1}$

Multiply the coefficients:

$\frac{(- 1 \cdot - 4)}{4} x = (\frac{1}{6}) \cdot \frac{4}{- 1}$

Simplify the arithmetic:

$1 x = (\frac{1}{6}) \cdot \frac{4}{- 1}$

$x = (\frac{1}{6}) \cdot \frac{4}{- 1}$

Multiply the fraction(s):

$x = \frac{(1 \cdot - 4)}{6}$

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

$x = \frac{- 2}{3}$

31 additional steps

$(\frac{1}{2} x + \frac{2}{3}) = - (\frac{3}{4} x + \frac{5}{6})$

Expand the parentheses:

$(\frac{1}{2} \cdot x + \frac{2}{3}) = \frac{- 3}{4} x + \frac{- 5}{6}$

Add to both sides:

$(\frac{1}{2} x + \frac{2}{3}) + \frac{3}{4} \cdot x = (\frac{- 3}{4} x + \frac{- 5}{6}) + \frac{3}{4} x$

Group like terms:

$(\frac{1}{2} \cdot x + \frac{3}{4} \cdot x) + \frac{2}{3} = (\frac{- 3}{4} \cdot x + \frac{- 5}{6}) + \frac{3}{4} x$

Group the coefficients:

$(\frac{1}{2} + \frac{3}{4}) x + \frac{2}{3} = (\frac{- 3}{4} \cdot x + \frac{- 5}{6}) + \frac{3}{4} x$

Find the lowest common denominator:

$(\frac{(1 \cdot 2)}{(2 \cdot 2)} + \frac{3}{4}) x + \frac{2}{3} = (\frac{- 3}{4} \cdot x + \frac{- 5}{6}) + \frac{3}{4} x$

Multiply the denominators:

$(\frac{(1 \cdot 2)}{4} + \frac{3}{4}) x + \frac{2}{3} = (\frac{- 3}{4} \cdot x + \frac{- 5}{6}) + \frac{3}{4} x$

Multiply the numerators:

$(\frac{2}{4} + \frac{3}{4}) x + \frac{2}{3} = (\frac{- 3}{4} \cdot x + \frac{- 5}{6}) + \frac{3}{4} x$

Combine the fractions:

$\frac{(2 + 3)}{4} \cdot x + \frac{2}{3} = (\frac{- 3}{4} \cdot x + \frac{- 5}{6}) + \frac{3}{4} x$

Combine the numerators:

$\frac{5}{4} \cdot x + \frac{2}{3} = (\frac{- 3}{4} \cdot x + \frac{- 5}{6}) + \frac{3}{4} x$

Group like terms:

$\frac{5}{4} \cdot x + \frac{2}{3} = (\frac{- 3}{4} \cdot x + \frac{3}{4} x) + \frac{- 5}{6}$

Combine the fractions:

$\frac{5}{4} \cdot x + \frac{2}{3} = \frac{(- 3 + 3)}{4} x + \frac{- 5}{6}$

Combine the numerators:

$\frac{5}{4} \cdot x + \frac{2}{3} = \frac{0}{4} x + \frac{- 5}{6}$

Reduce the zero numerator:

$\frac{5}{4} x + \frac{2}{3} = 0 x + \frac{- 5}{6}$

Simplify the arithmetic:

$\frac{5}{4} x + \frac{2}{3} = \frac{- 5}{6}$

Subtract from both sides:

$(\frac{5}{4} x + \frac{2}{3}) - \frac{2}{3} = (\frac{- 5}{6}) - \frac{2}{3}$

Combine the fractions:

$\frac{5}{4} x + \frac{(2 - 2)}{3} = (\frac{- 5}{6}) - \frac{2}{3}$

Combine the numerators:

$\frac{5}{4} x + \frac{0}{3} = (\frac{- 5}{6}) - \frac{2}{3}$

Reduce the zero numerator:

$\frac{5}{4} x + 0 = (\frac{- 5}{6}) - \frac{2}{3}$

Simplify the arithmetic:

$\frac{5}{4} x = (\frac{- 5}{6}) - \frac{2}{3}$

Find the lowest common denominator:

$\frac{5}{4} x = \frac{- 5}{6} + \frac{(- 2 \cdot 2)}{(3 \cdot 2)}$

Multiply the denominators:

$\frac{5}{4} x = \frac{- 5}{6} + \frac{(- 2 \cdot 2)}{6}$

Multiply the numerators:

$\frac{5}{4} x = \frac{- 5}{6} + \frac{- 4}{6}$

Combine the fractions:

$\frac{5}{4} x = \frac{(- 5 - 4)}{6}$

Combine the numerators:

$\frac{5}{4} x = \frac{- 9}{6}$

Find the greatest common factor of the numerator and denominator:

$\frac{5}{4} x = \frac{(- 3 \cdot 3)}{(2 \cdot 3)}$

Factor out and cancel the greatest common factor:

$\frac{5}{4} x = \frac{- 3}{2}$

Multiply both sides by inverse fraction :

$(\frac{5}{4} x) \cdot \frac{4}{5} = (\frac{- 3}{2}) \cdot \frac{4}{5}$

Group like terms:

$(\frac{5}{4} \cdot \frac{4}{5}) x = (\frac{- 3}{2}) \cdot \frac{4}{5}$

Multiply the coefficients:

$\frac{(5 \cdot 4)}{(4 \cdot 5)} x = (\frac{- 3}{2}) \cdot \frac{4}{5}$

Simplify the fraction:

$x = (\frac{- 3}{2}) \cdot \frac{4}{5}$

Multiply the fraction(s):

$x = \frac{(- 3 \cdot 4)}{(2 \cdot 5)}$

Simplify the arithmetic:

$x = \frac{- 6}{5}$

3. List the solutions

$x = - \frac{2}{3}, - \frac{6}{5}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{2} x + \frac{2}{3} |$
$y = | \frac{3}{4} x + \frac{5}{6} |$
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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