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Solution - Absolute value equations

Exact form:

x = 9, \frac{33}{2}

x=9 , \frac{33}{2}

Mixed number form:

x = 9, 16 \frac{1}{2}

x=9 , 16\frac{1}{2}

Decimal form:

x = 9, 16.5

x=9 , 16.5

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
$| \frac{1}{2} x - 7 | = | \frac{1}{6} x - 4 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{2} x - 7 \| = \| \frac{1}{6} x - 4 \|$
$x = + y$	$(\frac{1}{2} x - 7) = (\frac{1}{6} x - 4)$
$x = - y$	$(\frac{1}{2} x - 7) = - (\frac{1}{6} x - 4)$
$+ x = y$	$(\frac{1}{2} x - 7) = (\frac{1}{6} x - 4)$
$- x = y$	$- (\frac{1}{2} x - 7) = (\frac{1}{6} x - 4)$

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:

$\| x \| = \| y \|$	$\| \frac{1}{2} x - 7 \| = \| \frac{1}{6} x - 4 \|$
$x = + y$ , $+ x = y$	$(\frac{1}{2} x - 7) = (\frac{1}{6} x - 4)$
$x = - y$ , $- x = y$	$(\frac{1}{2} x - 7) = - (\frac{1}{6} x - 4)$

2. Solve the two equations for x

22 additional steps

$(\frac{1}{2} \cdot x - 7) = (\frac{1}{6} x - 4)$

Subtract from both sides:

$(\frac{1}{2} x - 7) - \frac{1}{6} \cdot x = (\frac{1}{6} x - 4) - \frac{1}{6} x$

Group like terms:

$(\frac{1}{2} \cdot x + \frac{- 1}{6} \cdot x) - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Group the coefficients:

$(\frac{1}{2} + \frac{- 1}{6}) x - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Find the lowest common denominator:

$(\frac{(1 \cdot 3)}{(2 \cdot 3)} + \frac{- 1}{6}) x - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Multiply the denominators:

$(\frac{(1 \cdot 3)}{6} + \frac{- 1}{6}) x - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Multiply the numerators:

$(\frac{3}{6} + \frac{- 1}{6}) x - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Combine the fractions:

$\frac{(3 - 1)}{6} \cdot x - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Combine the numerators:

$\frac{2}{6} \cdot x - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Find the greatest common factor of the numerator and denominator:

$\frac{(1 \cdot 2)}{(3 \cdot 2)} \cdot x - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Factor out and cancel the greatest common factor:

$\frac{1}{3} \cdot x - 7 = (\frac{1}{6} \cdot x - 4) - \frac{1}{6} x$

Group like terms:

$\frac{1}{3} \cdot x - 7 = (\frac{1}{6} \cdot x + \frac{- 1}{6} x) - 4$

Combine the fractions:

$\frac{1}{3} \cdot x - 7 = \frac{(1 - 1)}{6} x - 4$

Combine the numerators:

$\frac{1}{3} \cdot x - 7 = \frac{0}{6} x - 4$

Reduce the zero numerator:

$\frac{1}{3} x - 7 = 0 x - 4$

Simplify the arithmetic:

$\frac{1}{3} x - 7 = - 4$

Add to both sides:

$(\frac{1}{3} x - 7) + 7 = - 4 + 7$

Simplify the arithmetic:

$\frac{1}{3} x = - 4 + 7$

Simplify the arithmetic:

$\frac{1}{3} x = 3$

Multiply both sides by inverse fraction :

$(\frac{1}{3} x) \cdot \frac{3}{1} = 3 \cdot \frac{3}{1}$

Group like terms:

$(\frac{1}{3} \cdot 3) x = 3 \cdot \frac{3}{1}$

Multiply the coefficients:

$\frac{(1 \cdot 3)}{3} x = 3 \cdot \frac{3}{1}$

Simplify the fraction:

$x = 3 \cdot \frac{3}{1}$

Simplify the arithmetic:

$x = 9$

24 additional steps

$(\frac{1}{2} x - 7) = - (\frac{1}{6} x - 4)$

Expand the parentheses:

$(\frac{1}{2} \cdot x - 7) = \frac{- 1}{6} x + 4$

Add to both sides:

$(\frac{1}{2} x - 7) + \frac{1}{6} \cdot x = (\frac{- 1}{6} x + 4) + \frac{1}{6} x$

Group like terms:

$(\frac{1}{2} \cdot x + \frac{1}{6} \cdot x) - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Group the coefficients:

$(\frac{1}{2} + \frac{1}{6}) x - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Find the lowest common denominator:

$(\frac{(1 \cdot 3)}{(2 \cdot 3)} + \frac{1}{6}) x - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Multiply the denominators:

$(\frac{(1 \cdot 3)}{6} + \frac{1}{6}) x - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Multiply the numerators:

$(\frac{3}{6} + \frac{1}{6}) x - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Combine the fractions:

$\frac{(3 + 1)}{6} \cdot x - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Combine the numerators:

$\frac{4}{6} \cdot x - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Find the greatest common factor of the numerator and denominator:

$\frac{(2 \cdot 2)}{(3 \cdot 2)} \cdot x - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Factor out and cancel the greatest common factor:

$\frac{2}{3} \cdot x - 7 = (\frac{- 1}{6} \cdot x + 4) + \frac{1}{6} x$

Group like terms:

$\frac{2}{3} \cdot x - 7 = (\frac{- 1}{6} \cdot x + \frac{1}{6} x) + 4$

Combine the fractions:

$\frac{2}{3} \cdot x - 7 = \frac{(- 1 + 1)}{6} x + 4$

Combine the numerators:

$\frac{2}{3} \cdot x - 7 = \frac{0}{6} x + 4$

Reduce the zero numerator:

$\frac{2}{3} x - 7 = 0 x + 4$

Simplify the arithmetic:

$\frac{2}{3} x - 7 = 4$

Add to both sides:

$(\frac{2}{3} x - 7) + 7 = 4 + 7$

Simplify the arithmetic:

$\frac{2}{3} x = 4 + 7$

Simplify the arithmetic:

$\frac{2}{3} x = 11$

Multiply both sides by inverse fraction :

$(\frac{2}{3} x) \cdot \frac{3}{2} = 11 \cdot \frac{3}{2}$

Group like terms:

$(\frac{2}{3} \cdot \frac{3}{2}) x = 11 \cdot \frac{3}{2}$

Multiply the coefficients:

$\frac{(2 \cdot 3)}{(3 \cdot 2)} x = 11 \cdot \frac{3}{2}$

Simplify the fraction:

$x = 11 \cdot \frac{3}{2}$

Multiply the fraction(s):

$x = \frac{(11 \cdot 3)}{2}$

Simplify the arithmetic:

$x = \frac{33}{2}$

3. List the solutions

$x = 9, \frac{33}{2}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{2} x - 7 |$
$y = | \frac{1}{6} x - 4 |$
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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