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Solution - Absolute value equations

Exact form:

x = 4, - 2

x=4 , -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
$| \frac{1}{10} x + \frac{1}{2} | = | \frac{1}{5} x + \frac{1}{10} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{10} x + \frac{1}{2} \| = \| \frac{1}{5} x + \frac{1}{10} \|$
$x = + y$	$(\frac{1}{10} x + \frac{1}{2}) = (\frac{1}{5} x + \frac{1}{10})$
$x = - y$	$(\frac{1}{10} x + \frac{1}{2}) = - (\frac{1}{5} x + \frac{1}{10})$
$+ x = y$	$(\frac{1}{10} x + \frac{1}{2}) = (\frac{1}{5} x + \frac{1}{10})$
$- x = y$	$- (\frac{1}{10} x + \frac{1}{2}) = (\frac{1}{5} x + \frac{1}{10})$

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:

$\| x \| = \| y \|$	$\| \frac{1}{10} x + \frac{1}{2} \| = \| \frac{1}{5} x + \frac{1}{10} \|$
$x = + y$ , $+ x = y$	$(\frac{1}{10} x + \frac{1}{2}) = (\frac{1}{5} x + \frac{1}{10})$
$x = - y$ , $- x = y$	$(\frac{1}{10} x + \frac{1}{2}) = - (\frac{1}{5} x + \frac{1}{10})$

2. Solve the two equations for x

31 additional steps

$(\frac{1}{10} \cdot x + \frac{1}{2}) = (\frac{1}{5} x + \frac{1}{10})$

Subtract from both sides:

$(\frac{1}{10} x + \frac{1}{2}) - \frac{1}{5} \cdot x = (\frac{1}{5} x + \frac{1}{10}) - \frac{1}{5} x$

Group like terms:

$(\frac{1}{10} \cdot x + \frac{- 1}{5} \cdot x) + \frac{1}{2} = (\frac{1}{5} \cdot x + \frac{1}{10}) - \frac{1}{5} x$

Group the coefficients:

$(\frac{1}{10} + \frac{- 1}{5}) x + \frac{1}{2} = (\frac{1}{5} \cdot x + \frac{1}{10}) - \frac{1}{5} x$

Find the lowest common denominator:

$(\frac{1}{10} + \frac{(- 1 \cdot 2)}{(5 \cdot 2)}) x + \frac{1}{2} = (\frac{1}{5} \cdot x + \frac{1}{10}) - \frac{1}{5} x$

Multiply the denominators:

$(\frac{1}{10} + \frac{(- 1 \cdot 2)}{10}) x + \frac{1}{2} = (\frac{1}{5} \cdot x + \frac{1}{10}) - \frac{1}{5} x$

Multiply the numerators:

$(\frac{1}{10} + \frac{- 2}{10}) x + \frac{1}{2} = (\frac{1}{5} \cdot x + \frac{1}{10}) - \frac{1}{5} x$

Combine the fractions:

$\frac{(1 - 2)}{10} \cdot x + \frac{1}{2} = (\frac{1}{5} \cdot x + \frac{1}{10}) - \frac{1}{5} x$

Combine the numerators:

$\frac{- 1}{10} \cdot x + \frac{1}{2} = (\frac{1}{5} \cdot x + \frac{1}{10}) - \frac{1}{5} x$

Group like terms:

$\frac{- 1}{10} \cdot x + \frac{1}{2} = (\frac{1}{5} \cdot x + \frac{- 1}{5} x) + \frac{1}{10}$

Combine the fractions:

$\frac{- 1}{10} \cdot x + \frac{1}{2} = \frac{(1 - 1)}{5} x + \frac{1}{10}$

Combine the numerators:

$\frac{- 1}{10} \cdot x + \frac{1}{2} = \frac{0}{5} x + \frac{1}{10}$

Reduce the zero numerator:

$\frac{- 1}{10} x + \frac{1}{2} = 0 x + \frac{1}{10}$

Simplify the arithmetic:

$\frac{- 1}{10} x + \frac{1}{2} = \frac{1}{10}$

Subtract from both sides:

$(\frac{- 1}{10} x + \frac{1}{2}) - \frac{1}{2} = (\frac{1}{10}) - \frac{1}{2}$

Combine the fractions:

$\frac{- 1}{10} x + \frac{(1 - 1)}{2} = (\frac{1}{10}) - \frac{1}{2}$

Combine the numerators:

$\frac{- 1}{10} x + \frac{0}{2} = (\frac{1}{10}) - \frac{1}{2}$

Reduce the zero numerator:

$\frac{- 1}{10} x + 0 = (\frac{1}{10}) - \frac{1}{2}$

Simplify the arithmetic:

$\frac{- 1}{10} x = (\frac{1}{10}) - \frac{1}{2}$

Find the lowest common denominator:

$\frac{- 1}{10} x = \frac{1}{10} + \frac{(- 1 \cdot 5)}{(2 \cdot 5)}$

Multiply the denominators:

$\frac{- 1}{10} x = \frac{1}{10} + \frac{(- 1 \cdot 5)}{10}$

Multiply the numerators:

$\frac{- 1}{10} x = \frac{1}{10} + \frac{- 5}{10}$

Combine the fractions:

$\frac{- 1}{10} x = \frac{(1 - 5)}{10}$

Combine the numerators:

$\frac{- 1}{10} x = \frac{- 4}{10}$

Find the greatest common factor of the numerator and denominator:

$\frac{- 1}{10} x = \frac{(- 2 \cdot 2)}{(5 \cdot 2)}$

Factor out and cancel the greatest common factor:

$\frac{- 1}{10} x = \frac{- 2}{5}$

Multiply both sides by inverse fraction :

$(\frac{- 1}{10} x) \cdot \frac{10}{- 1} = (\frac{- 2}{5}) \cdot \frac{10}{- 1}$

Group like terms:

$(\frac{- 1}{10} \cdot - 10) x = (\frac{- 2}{5}) \cdot \frac{10}{- 1}$

Multiply the coefficients:

$\frac{(- 1 \cdot - 10)}{10} x = (\frac{- 2}{5}) \cdot \frac{10}{- 1}$

Simplify the arithmetic:

$1 x = (\frac{- 2}{5}) \cdot \frac{10}{- 1}$

$x = (\frac{- 2}{5}) \cdot \frac{10}{- 1}$

Multiply the fraction(s):

$x = \frac{(- 2 \cdot - 10)}{5}$

Simplify the arithmetic:

$x = 4$

31 additional steps

$(\frac{1}{10} x + \frac{1}{2}) = - (\frac{1}{5} x + \frac{1}{10})$

Expand the parentheses:

$(\frac{1}{10} \cdot x + \frac{1}{2}) = \frac{- 1}{5} x + \frac{- 1}{10}$

Add to both sides:

$(\frac{1}{10} x + \frac{1}{2}) + \frac{1}{5} \cdot x = (\frac{- 1}{5} x + \frac{- 1}{10}) + \frac{1}{5} x$

Group like terms:

$(\frac{1}{10} \cdot x + \frac{1}{5} \cdot x) + \frac{1}{2} = (\frac{- 1}{5} \cdot x + \frac{- 1}{10}) + \frac{1}{5} x$

Group the coefficients:

$(\frac{1}{10} + \frac{1}{5}) x + \frac{1}{2} = (\frac{- 1}{5} \cdot x + \frac{- 1}{10}) + \frac{1}{5} x$

Find the lowest common denominator:

$(\frac{1}{10} + \frac{(1 \cdot 2)}{(5 \cdot 2)}) x + \frac{1}{2} = (\frac{- 1}{5} \cdot x + \frac{- 1}{10}) + \frac{1}{5} x$

Multiply the denominators:

$(\frac{1}{10} + \frac{(1 \cdot 2)}{10}) x + \frac{1}{2} = (\frac{- 1}{5} \cdot x + \frac{- 1}{10}) + \frac{1}{5} x$

Multiply the numerators:

$(\frac{1}{10} + \frac{2}{10}) x + \frac{1}{2} = (\frac{- 1}{5} \cdot x + \frac{- 1}{10}) + \frac{1}{5} x$

Combine the fractions:

$\frac{(1 + 2)}{10} \cdot x + \frac{1}{2} = (\frac{- 1}{5} \cdot x + \frac{- 1}{10}) + \frac{1}{5} x$

Combine the numerators:

$\frac{3}{10} \cdot x + \frac{1}{2} = (\frac{- 1}{5} \cdot x + \frac{- 1}{10}) + \frac{1}{5} x$

Group like terms:

$\frac{3}{10} \cdot x + \frac{1}{2} = (\frac{- 1}{5} \cdot x + \frac{1}{5} x) + \frac{- 1}{10}$

Combine the fractions:

$\frac{3}{10} \cdot x + \frac{1}{2} = \frac{(- 1 + 1)}{5} x + \frac{- 1}{10}$

Combine the numerators:

$\frac{3}{10} \cdot x + \frac{1}{2} = \frac{0}{5} x + \frac{- 1}{10}$

Reduce the zero numerator:

$\frac{3}{10} x + \frac{1}{2} = 0 x + \frac{- 1}{10}$

Simplify the arithmetic:

$\frac{3}{10} x + \frac{1}{2} = \frac{- 1}{10}$

Subtract from both sides:

$(\frac{3}{10} x + \frac{1}{2}) - \frac{1}{2} = (\frac{- 1}{10}) - \frac{1}{2}$

Combine the fractions:

$\frac{3}{10} x + \frac{(1 - 1)}{2} = (\frac{- 1}{10}) - \frac{1}{2}$

Combine the numerators:

$\frac{3}{10} x + \frac{0}{2} = (\frac{- 1}{10}) - \frac{1}{2}$

Reduce the zero numerator:

$\frac{3}{10} x + 0 = (\frac{- 1}{10}) - \frac{1}{2}$

Simplify the arithmetic:

$\frac{3}{10} x = (\frac{- 1}{10}) - \frac{1}{2}$

Find the lowest common denominator:

$\frac{3}{10} x = \frac{- 1}{10} + \frac{(- 1 \cdot 5)}{(2 \cdot 5)}$

Multiply the denominators:

$\frac{3}{10} x = \frac{- 1}{10} + \frac{(- 1 \cdot 5)}{10}$

Multiply the numerators:

$\frac{3}{10} x = \frac{- 1}{10} + \frac{- 5}{10}$

Combine the fractions:

$\frac{3}{10} x = \frac{(- 1 - 5)}{10}$

Combine the numerators:

$\frac{3}{10} x = \frac{- 6}{10}$

Find the greatest common factor of the numerator and denominator:

$\frac{3}{10} x = \frac{(- 3 \cdot 2)}{(5 \cdot 2)}$

Factor out and cancel the greatest common factor:

$\frac{3}{10} x = \frac{- 3}{5}$

Multiply both sides by inverse fraction :

$(\frac{3}{10} x) \cdot \frac{10}{3} = (\frac{- 3}{5}) \cdot \frac{10}{3}$

Group like terms:

$(\frac{3}{10} \cdot \frac{10}{3}) x = (\frac{- 3}{5}) \cdot \frac{10}{3}$

Multiply the coefficients:

$\frac{(3 \cdot 10)}{(10 \cdot 3)} x = (\frac{- 3}{5}) \cdot \frac{10}{3}$

Simplify the fraction:

$x = (\frac{- 3}{5}) \cdot \frac{10}{3}$

Multiply the fraction(s):

$x = \frac{(- 3 \cdot 10)}{(5 \cdot 3)}$

Simplify the arithmetic:

$x = - 2$

3. List the solutions

$x = 4, - 2$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{10} x + \frac{1}{2} |$
$y = | \frac{1}{5} x + \frac{1}{10} |$
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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