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Solution - Absolute value equations

Exact form:

x = \frac{7}{2}, \frac{7}{4}

x=\frac{7}{2} , \frac{7}{4}

Mixed number form:

x = 3 \frac{1}{2}, 1 \frac{3}{4}

x=3\frac{1}{2} , 1\frac{3}{4}

Decimal form:

x = 3.5, 1.75

x=3.5 , 1.75

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
$| \frac{1}{2} x | = | \frac{3}{2} x - \frac{7}{2} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{2} x \| = \| \frac{3}{2} x - \frac{7}{2} \|$
$x = + y$	$(\frac{1}{2} x) = (\frac{3}{2} x - \frac{7}{2})$
$x = - y$	$(\frac{1}{2} x) = - (\frac{3}{2} x - \frac{7}{2})$
$+ x = y$	$(\frac{1}{2} x) = (\frac{3}{2} x - \frac{7}{2})$
$- x = y$	$- (\frac{1}{2} x) = (\frac{3}{2} x - \frac{7}{2})$

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:

$\| x \| = \| y \|$	$\| \frac{1}{2} x \| = \| \frac{3}{2} x - \frac{7}{2} \|$
$x = + y$ , $+ x = y$	$(\frac{1}{2} x) = (\frac{3}{2} x - \frac{7}{2})$
$x = - y$ , $- x = y$	$(\frac{1}{2} x) = - (\frac{3}{2} x - \frac{7}{2})$

2. Solve the two equations for x

13 additional steps

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

Subtract from both sides:

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

Combine the fractions:

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

Combine the numerators:

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

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

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

Simplify the arithmetic:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

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

Reduce the zero numerator:

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

Simplify the arithmetic:

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

Multiply both sides by :

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

Remove the one(s):

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

Remove the one(s):

$x = \frac{7}{2}$

14 additional steps

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

Expand the parentheses:

$\frac{1}{2} \cdot x = \frac{- 3}{2} x + \frac{7}{2}$

Add to both sides:

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

Combine the fractions:

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

Combine the numerators:

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

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

$2 x = \frac{0}{2} x + \frac{7}{2}$

Reduce the zero numerator:

$2 x = 0 x + \frac{7}{2}$

Simplify the arithmetic:

$2 x = \frac{7}{2}$

Divide both sides by :

$\frac{(2 x)}{2} = \frac{(\frac{7}{2})}{2}$

Simplify the fraction:

$x = \frac{(\frac{7}{2})}{2}$

Simplify the arithmetic:

$x = \frac{7}{(2 \cdot 2)}$

$x = \frac{7}{4}$

3. List the solutions

$x = \frac{7}{2}, \frac{7}{4}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{2} x |$
$y = | \frac{3}{2} x - \frac{7}{2} |$
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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