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Solution - Absolute value equations

Exact form:

x = \frac{20}{3}, - \frac{36}{5}

x=\frac{20}{3} , -\frac{36}{5}

Mixed number form:

x = 6 \frac{2}{3}, - 7 \frac{1}{5}

x=6\frac{2}{3} , -7\frac{1}{5}

Decimal form:

x = 6.667, - 7.2

x=6.667 , -7.2

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
$| 2 x + 4 | = | \frac{1}{2} x + 14 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 2 x + 4 \| = \| \frac{1}{2} x + 14 \|$
$x = + y$	$(2 x + 4) = (\frac{1}{2} x + 14)$
$x = - y$	$(2 x + 4) = - (\frac{1}{2} x + 14)$
$+ x = y$	$(2 x + 4) = (\frac{1}{2} x + 14)$
$- x = y$	$- (2 x + 4) = (\frac{1}{2} x + 14)$

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:

$\| x \| = \| y \|$	$\| 2 x + 4 \| = \| \frac{1}{2} x + 14 \|$
$x = + y$ , $+ x = y$	$(2 x + 4) = (\frac{1}{2} x + 14)$
$x = - y$ , $- x = y$	$(2 x + 4) = - (\frac{1}{2} x + 14)$

2. Solve the two equations for x

19 additional steps

$(2 x + 4) = (\frac{1}{2} x + 14)$

Subtract from both sides:

$(2 x + 4) - \frac{1}{2} \cdot x = (\frac{1}{2} x + 14) - \frac{1}{2} x$

Group like terms:

$(2 x + \frac{- 1}{2} \cdot x) + 4 = (\frac{1}{2} \cdot x + 14) - \frac{1}{2} x$

Group the coefficients:

$(2 + \frac{- 1}{2}) x + 4 = (\frac{1}{2} \cdot x + 14) - \frac{1}{2} x$

Convert the integer into a fraction:

$(\frac{4}{2} + \frac{- 1}{2}) x + 4 = (\frac{1}{2} \cdot x + 14) - \frac{1}{2} x$

Combine the fractions:

$\frac{(4 - 1)}{2} \cdot x + 4 = (\frac{1}{2} \cdot x + 14) - \frac{1}{2} x$

Combine the numerators:

$\frac{3}{2} \cdot x + 4 = (\frac{1}{2} \cdot x + 14) - \frac{1}{2} x$

Group like terms:

$\frac{3}{2} \cdot x + 4 = (\frac{1}{2} \cdot x + \frac{- 1}{2} x) + 14$

Combine the fractions:

$\frac{3}{2} \cdot x + 4 = \frac{(1 - 1)}{2} x + 14$

Combine the numerators:

$\frac{3}{2} \cdot x + 4 = \frac{0}{2} x + 14$

Reduce the zero numerator:

$\frac{3}{2} x + 4 = 0 x + 14$

Simplify the arithmetic:

$\frac{3}{2} x + 4 = 14$

Subtract from both sides:

$(\frac{3}{2} x + 4) - 4 = 14 - 4$

Simplify the arithmetic:

$\frac{3}{2} x = 14 - 4$

Simplify the arithmetic:

$\frac{3}{2} x = 10$

Multiply both sides by inverse fraction :

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

Group like terms:

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

Multiply the coefficients:

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

Simplify the fraction:

$x = 10 \cdot \frac{2}{3}$

Multiply the fraction(s):

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

Simplify the arithmetic:

$x = \frac{20}{3}$

20 additional steps

$(2 x + 4) = - (\frac{1}{2} x + 14)$

Expand the parentheses:

$(2 x + 4) = \frac{- 1}{2} x - 14$

Add to both sides:

$(2 x + 4) + \frac{1}{2} \cdot x = (\frac{- 1}{2} x - 14) + \frac{1}{2} x$

Group like terms:

$(2 x + \frac{1}{2} \cdot x) + 4 = (\frac{- 1}{2} \cdot x - 14) + \frac{1}{2} x$

Group the coefficients:

$(2 + \frac{1}{2}) x + 4 = (\frac{- 1}{2} \cdot x - 14) + \frac{1}{2} x$

Convert the integer into a fraction:

$(\frac{4}{2} + \frac{1}{2}) x + 4 = (\frac{- 1}{2} \cdot x - 14) + \frac{1}{2} x$

Combine the fractions:

$\frac{(4 + 1)}{2} \cdot x + 4 = (\frac{- 1}{2} \cdot x - 14) + \frac{1}{2} x$

Combine the numerators:

$\frac{5}{2} \cdot x + 4 = (\frac{- 1}{2} \cdot x - 14) + \frac{1}{2} x$

Group like terms:

$\frac{5}{2} \cdot x + 4 = (\frac{- 1}{2} \cdot x + \frac{1}{2} x) - 14$

Combine the fractions:

$\frac{5}{2} \cdot x + 4 = \frac{(- 1 + 1)}{2} x - 14$

Combine the numerators:

$\frac{5}{2} \cdot x + 4 = \frac{0}{2} x - 14$

Reduce the zero numerator:

$\frac{5}{2} x + 4 = 0 x - 14$

Simplify the arithmetic:

$\frac{5}{2} x + 4 = - 14$

Subtract from both sides:

$(\frac{5}{2} x + 4) - 4 = - 14 - 4$

Simplify the arithmetic:

$\frac{5}{2} x = - 14 - 4$

Simplify the arithmetic:

$\frac{5}{2} x = - 18$

Multiply both sides by inverse fraction :

$(\frac{5}{2} x) \cdot \frac{2}{5} = - 18 \cdot \frac{2}{5}$

Group like terms:

$(\frac{5}{2} \cdot \frac{2}{5}) x = - 18 \cdot \frac{2}{5}$

Multiply the coefficients:

$\frac{(5 \cdot 2)}{(2 \cdot 5)} x = - 18 \cdot \frac{2}{5}$

Simplify the fraction:

$x = - 18 \cdot \frac{2}{5}$

Multiply the fraction(s):

$x = \frac{(- 18 \cdot 2)}{5}$

Simplify the arithmetic:

$x = \frac{- 36}{5}$

3. List the solutions

$x = \frac{20}{3}, - \frac{36}{5}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 2 x + 4 |$
$y = | \frac{1}{2} x + 14 |$
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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