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Solution - Absolute value equations

Exact form:

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

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

Mixed number form:

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

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

Decimal form:

x = - 3, 1.667

x=-3 , 1.667

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

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

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:

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

2. Solve the two equations for x

24 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Group the coefficients:

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

Convert the integer into a fraction:

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

Combine the fractions:

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

Combine the numerators:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

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

Reduce the zero numerator:

$\frac{1}{2} x + \frac{- 1}{2} = 0 x - 2$

Simplify the arithmetic:

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

Add to both sides:

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

Combine the fractions:

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

Combine the numerators:

$\frac{1}{2} x + \frac{0}{2} = - 2 + \frac{1}{2}$

Reduce the zero numerator:

$\frac{1}{2} x + 0 = - 2 + \frac{1}{2}$

Simplify the arithmetic:

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

Convert the integer into a fraction:

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

Combine the fractions:

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

Combine the numerators:

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

Multiply both sides by inverse fraction :

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

Group like terms:

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

Multiply the coefficients:

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

Simplify the fraction:

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

Multiply the fraction(s):

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

Simplify the arithmetic:

$x = - 3$

25 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Group the coefficients:

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

Convert the integer into a fraction:

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

Combine the fractions:

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

Combine the numerators:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

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

Reduce the zero numerator:

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

Simplify the arithmetic:

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

Add to both sides:

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

Combine the fractions:

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

Combine the numerators:

$\frac{3}{2} x + \frac{0}{2} = 2 + \frac{1}{2}$

Reduce the zero numerator:

$\frac{3}{2} x + 0 = 2 + \frac{1}{2}$

Simplify the arithmetic:

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

Convert the integer into a fraction:

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

Combine the fractions:

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

Combine the numerators:

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

Multiply both sides by inverse fraction :

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

Group like terms:

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

Multiply the coefficients:

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

Simplify the fraction:

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

Multiply the fraction(s):

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

Simplify the arithmetic:

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

3. List the solutions

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

4. Graph

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