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Solution - Absolute value equations

Exact form:

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

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

Mixed number form:

x = - 6, 1 \frac{1}{5}

x=-6 , 1\frac{1}{5}

Decimal form:

x = - 6, 1.2

x=-6 , 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
$\frac{1}{3} | x - 3 | = \frac{1}{2} | x |$
without the absolute value bars:

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

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:

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

2. Solve the two equations for x

24 additional steps

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

Multiply the fraction(s):

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

Break up the fraction:

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

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

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

Subtract from both sides:

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

Group like terms:

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

Group the coefficients:

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

Find the lowest common denominator:

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

Multiply the denominators:

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

Multiply the numerators:

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

Combine the fractions:

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

Combine the numerators:

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

Combine the fractions:

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

Combine the numerators:

$\frac{- 1}{6} \cdot x - 1 = \frac{0}{2} x$

Reduce the zero numerator:

$\frac{- 1}{6} x - 1 = 0 x$

Simplify the arithmetic:

$\frac{- 1}{6} x - 1 = 0$

Add to both sides:

$(\frac{- 1}{6} x - 1) + 1 = 0 + 1$

Simplify the arithmetic:

$\frac{- 1}{6} x = 0 + 1$

Simplify the arithmetic:

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

Multiply both sides by inverse fraction :

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

Group like terms:

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

Multiply the coefficients:

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

Simplify the arithmetic:

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

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

Simplify the arithmetic:

$x = - 6$

26 additional steps

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

Multiply the fraction(s):

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

Break up the fraction:

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

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

$\frac{x}{3} - 1 = \frac{1}{2} \cdot - x$

Group like terms:

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

Multiply the coefficients:

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

Simplify the arithmetic:

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

Add to both sides:

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

Group like terms:

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

Group the coefficients:

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

Find the lowest common denominator:

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

Multiply the denominators:

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

Multiply the numerators:

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

Combine the fractions:

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

Combine the numerators:

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

Combine the fractions:

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

Combine the numerators:

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

Reduce the zero numerator:

$\frac{5}{6} x - 1 = 0 x$

Simplify the arithmetic:

$\frac{5}{6} x - 1 = 0$

Add to both sides:

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

Simplify the arithmetic:

$\frac{5}{6} x = 0 + 1$

Simplify the arithmetic:

$\frac{5}{6} x = 1$

Multiply both sides by inverse fraction :

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

Group like terms:

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

Multiply the coefficients:

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

Simplify the fraction:

$x = 1 \cdot \frac{6}{5}$

Remove the one(s):

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

3. List the solutions

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

4. Graph

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