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Solution - Absolute value equations

Exact form:

w = - \frac{1}{2}, \frac{1}{10}

w=-\frac{1}{2} , \frac{1}{10}

Decimal form:

w = - 0.5, 0.1

w=-0.5 , 0.1

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
$2 | 4 w - 1 | = 3 | 4 w |$
without the absolute value bars:

$\| x \| = \| y \|$	$2 \| 4 w - 1 \| = 3 \| 4 w \|$
$x = + y$	$2 (4 w - 1) = 3 (4 w)$
$x = - y$	$2 (4 w - 1) = 3 (- (4 w))$
$+ x = y$	$2 (4 w - 1) = 3 (4 w)$
$- x = y$	$2 (- (4 w - 1)) = 3 (4 w)$

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:

$\| x \| = \| y \|$	$2 \| 4 w - 1 \| = 3 \| 4 w \|$
$x = + y$ , $+ x = y$	$2 (4 w - 1) = 3 (4 w)$
$x = - y$ , $- x = y$	$2 (4 w - 1) = 3 (- (4 w))$

2. Solve the two equations for w

16 additional steps

$2 \cdot (4 w - 1) = 3 \cdot 4 w$

Expand the parentheses:

$2 \cdot 4 w + 2 \cdot - 1 = 3 \cdot 4 w$

Multiply the coefficients:

$8 w + 2 \cdot - 1 = 3 \cdot 4 w$

Simplify the arithmetic:

$8 w - 2 = 3 \cdot 4 w$

Multiply the coefficients:

$8 w - 2 = 12 w$

Subtract from both sides:

$(8 w - 2) - 12 w = (12 w) - 12 w$

Group like terms:

$(8 w - 12 w) - 2 = (12 w) - 12 w$

Simplify the arithmetic:

$- 4 w - 2 = (12 w) - 12 w$

Simplify the arithmetic:

$- 4 w - 2 = 0$

Add to both sides:

$(- 4 w - 2) + 2 = 0 + 2$

Simplify the arithmetic:

$- 4 w = 0 + 2$

Simplify the arithmetic:

$- 4 w = 2$

Divide both sides by :

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

Cancel out the negatives:

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

Simplify the fraction:

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

Move the negative sign from the denominator to the numerator:

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

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

$w = \frac{- 1}{2}$

14 additional steps

$2 \cdot (4 w - 1) = 3 \cdot - (4 w)$

Expand the parentheses:

$2 \cdot 4 w + 2 \cdot - 1 = 3 \cdot - (4 w)$

Multiply the coefficients:

$8 w + 2 \cdot - 1 = 3 \cdot - (4 w)$

Simplify the arithmetic:

$8 w - 2 = 3 \cdot - (4 w)$

Multiply the coefficients:

$8 w - 2 = - 12 w$

Add to both sides:

$(8 w - 2) + 12 w = (- 12 w) + 12 w$

Group like terms:

$(8 w + 12 w) - 2 = (- 12 w) + 12 w$

Simplify the arithmetic:

$20 w - 2 = (- 12 w) + 12 w$

Simplify the arithmetic:

$20 w - 2 = 0$

Add to both sides:

$(20 w - 2) + 2 = 0 + 2$

Simplify the arithmetic:

$20 w = 0 + 2$

Simplify the arithmetic:

$20 w = 2$

Divide both sides by :

$\frac{(20 w)}{20} = \frac{2}{20}$

Simplify the fraction:

$w = \frac{2}{20}$

Find the greatest common factor of the numerator and denominator:

$w = \frac{(1 \cdot 2)}{(10 \cdot 2)}$

Factor out and cancel the greatest common factor:

$w = \frac{1}{10}$

3. List the solutions

$w = - \frac{1}{2}, \frac{1}{10}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = 2 | 4 w - 1 |$
$y = 3 | 4 w |$
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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