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Solution - Absolute value equations

Exact form:

x = 32, \frac{8}{3}

x=32 , \frac{8}{3}

Mixed number form:

x = 32, 2 \frac{2}{3}

x=32 , 2\frac{2}{3}

Decimal form:

x = 32, 2.667

x=32 , 2.667

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

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

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:

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

2. Solve the two equations for x

20 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Group the coefficients:

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

Find the lowest common denominator:

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

Multiply the denominators:

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

Multiply the numerators:

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

Combine the fractions:

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

Combine the numerators:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

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

Reduce the zero numerator:

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

Simplify the arithmetic:

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

Add to both sides:

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

Simplify the arithmetic:

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

Simplify the arithmetic:

$\frac{1}{4} x = 8$

Multiply both sides by inverse fraction :

$(\frac{1}{4} x) \cdot \frac{4}{1} = 8 \cdot \frac{4}{1}$

Group like terms:

$(\frac{1}{4} \cdot 4) x = 8 \cdot \frac{4}{1}$

Multiply the coefficients:

$\frac{(1 \cdot 4)}{4} x = 8 \cdot \frac{4}{1}$

Simplify the fraction:

$x = 8 \cdot \frac{4}{1}$

Simplify the arithmetic:

$x = 32$

22 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Group the coefficients:

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

Find the lowest common denominator:

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

Multiply the denominators:

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

Multiply the numerators:

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

Combine the fractions:

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

Combine the numerators:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

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

Reduce the zero numerator:

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

Simplify the arithmetic:

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

Add to both sides:

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

Simplify the arithmetic:

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

Simplify the arithmetic:

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

Multiply both sides by inverse fraction :

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

Group like terms:

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

Multiply the coefficients:

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

Simplify the fraction:

$x = 2 \cdot \frac{4}{3}$

Multiply the fraction(s):

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

Simplify the arithmetic:

$x = \frac{8}{3}$

3. List the solutions

$x = 32, \frac{8}{3}$
(2 solution(s))

4. Graph

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