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Solution - Absolute value equations

Exact form:

w = 4, - \frac{28}{5}

w=4 , -\frac{28}{5}

Mixed number form:

w = 4, - 5 \frac{3}{5}

w=4 , -5\frac{3}{5}

Decimal form:

w = 4, - 5.6

w=4 , -5.6

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
$| \frac{1}{4} w + 5 | = | w + 2 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{4} w + 5 \| = \| w + 2 \|$
$x = + y$	$(\frac{1}{4} w + 5) = (w + 2)$
$x = - y$	$(\frac{1}{4} w + 5) = - (w + 2)$
$+ x = y$	$(\frac{1}{4} w + 5) = (w + 2)$
$- x = y$	$- (\frac{1}{4} w + 5) = (w + 2)$

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:

$\| x \| = \| y \|$	$\| \frac{1}{4} w + 5 \| = \| w + 2 \|$
$x = + y$ , $+ x = y$	$(\frac{1}{4} w + 5) = (w + 2)$
$x = - y$ , $- x = y$	$(\frac{1}{4} w + 5) = - (w + 2)$

2. Solve the two equations for w

19 additional steps

$(\frac{1}{4} w + 5) = (w + 2)$

Subtract from both sides:

$(\frac{1}{4} w + 5) - w = (w + 2) - w$

Group like terms:

$(\frac{1}{4} w - w) + 5 = (w + 2) - w$

Group the coefficients:

$(\frac{1}{4} - 1) w + 5 = (w + 2) - w$

Convert the integer into a fraction:

$(\frac{1}{4} + \frac{- 4}{4}) w + 5 = (w + 2) - w$

Combine the fractions:

$\frac{(1 - 4)}{4} w + 5 = (w + 2) - w$

Combine the numerators:

$\frac{- 3}{4} w + 5 = (w + 2) - w$

Group like terms:

$\frac{- 3}{4} w + 5 = (w - w) + 2$

Simplify the arithmetic:

$\frac{- 3}{4} w + 5 = 2$

Subtract from both sides:

$(\frac{- 3}{4} w + 5) - 5 = 2 - 5$

Simplify the arithmetic:

$\frac{- 3}{4} w = 2 - 5$

Simplify the arithmetic:

$\frac{- 3}{4} w = - 3$

Multiply both sides by inverse fraction :

$(\frac{- 3}{4} w) \cdot \frac{4}{- 3} = - 3 \cdot \frac{4}{- 3}$

Move the negative sign from the denominator to the numerator:

$\frac{- 3}{4} w \cdot \frac{- 4}{3} = - 3 \cdot \frac{4}{- 3}$

Group like terms:

$(\frac{- 3}{4} \cdot \frac{- 4}{3}) w = - 3 \cdot \frac{4}{- 3}$

Multiply the coefficients:

$\frac{(- 3 \cdot - 4)}{(4 \cdot 3)} w = - 3 \cdot \frac{4}{- 3}$

Simplify the arithmetic:

$1 w = - 3 \cdot \frac{4}{- 3}$

$w = - 3 \cdot \frac{4}{- 3}$

Move the negative sign from the denominator to the numerator:

$w = - 3 \cdot \frac{- 4}{3}$

Multiply the fraction(s):

$w = \frac{(- 3 \cdot - 4)}{3}$

Simplify the arithmetic:

$w = 4$

17 additional steps

$(\frac{1}{4} w + 5) = - (w + 2)$

Expand the parentheses:

$(\frac{1}{4} w + 5) = - w - 2$

Add to both sides:

$(\frac{1}{4} w + 5) + w = (- w - 2) + w$

Group like terms:

$(\frac{1}{4} w + w) + 5 = (- w - 2) + w$

Group the coefficients:

$(\frac{1}{4} + 1) w + 5 = (- w - 2) + w$

Convert the integer into a fraction:

$(\frac{1}{4} + \frac{4}{4}) w + 5 = (- w - 2) + w$

Combine the fractions:

$\frac{(1 + 4)}{4} w + 5 = (- w - 2) + w$

Combine the numerators:

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

Group like terms:

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

Simplify the arithmetic:

$\frac{5}{4} w + 5 = - 2$

Subtract from both sides:

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

Simplify the arithmetic:

$\frac{5}{4} w = - 2 - 5$

Simplify the arithmetic:

$\frac{5}{4} w = - 7$

Multiply both sides by inverse fraction :

$(\frac{5}{4} w) \cdot \frac{4}{5} = - 7 \cdot \frac{4}{5}$

Group like terms:

$(\frac{5}{4} \cdot \frac{4}{5}) w = - 7 \cdot \frac{4}{5}$

Multiply the coefficients:

$\frac{(5 \cdot 4)}{(4 \cdot 5)} w = - 7 \cdot \frac{4}{5}$

Simplify the fraction:

$w = - 7 \cdot \frac{4}{5}$

Multiply the fraction(s):

$w = \frac{(- 7 \cdot 4)}{5}$

Simplify the arithmetic:

$w = \frac{- 28}{5}$

3. List the solutions

$w = 4, - \frac{28}{5}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{4} w + 5 |$
$y = | w + 2 |$
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 w

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 w

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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