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Solution - Absolute value equations

Exact form:

b = 28, 12

b=28 , 12

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
$| \frac{1}{2} b - 8 | = | \frac{1}{4} b - 1 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{2} b - 8 \| = \| \frac{1}{4} b - 1 \|$
$x = + y$	$(\frac{1}{2} b - 8) = (\frac{1}{4} b - 1)$
$x = - y$	$(\frac{1}{2} b - 8) = - (\frac{1}{4} b - 1)$
$+ x = y$	$(\frac{1}{2} b - 8) = (\frac{1}{4} b - 1)$
$- x = y$	$- (\frac{1}{2} b - 8) = (\frac{1}{4} b - 1)$

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:

$\| x \| = \| y \|$	$\| \frac{1}{2} b - 8 \| = \| \frac{1}{4} b - 1 \|$
$x = + y$ , $+ x = y$	$(\frac{1}{2} b - 8) = (\frac{1}{4} b - 1)$
$x = - y$ , $- x = y$	$(\frac{1}{2} b - 8) = - (\frac{1}{4} b - 1)$

2. Solve the two equations for b

20 additional steps

$(\frac{1}{2} \cdot b - 8) = (\frac{1}{4} b - 1)$

Subtract from both sides:

$(\frac{1}{2} b - 8) - \frac{1}{4} \cdot b = (\frac{1}{4} b - 1) - \frac{1}{4} b$

Group like terms:

$(\frac{1}{2} \cdot b + \frac{- 1}{4} \cdot b) - 8 = (\frac{1}{4} \cdot b - 1) - \frac{1}{4} b$

Group the coefficients:

$(\frac{1}{2} + \frac{- 1}{4}) b - 8 = (\frac{1}{4} \cdot b - 1) - \frac{1}{4} b$

Find the lowest common denominator:

$(\frac{(1 \cdot 2)}{(2 \cdot 2)} + \frac{- 1}{4}) b - 8 = (\frac{1}{4} \cdot b - 1) - \frac{1}{4} b$

Multiply the denominators:

$(\frac{(1 \cdot 2)}{4} + \frac{- 1}{4}) b - 8 = (\frac{1}{4} \cdot b - 1) - \frac{1}{4} b$

Multiply the numerators:

$(\frac{2}{4} + \frac{- 1}{4}) b - 8 = (\frac{1}{4} \cdot b - 1) - \frac{1}{4} b$

Combine the fractions:

$\frac{(2 - 1)}{4} \cdot b - 8 = (\frac{1}{4} \cdot b - 1) - \frac{1}{4} b$

Combine the numerators:

$\frac{1}{4} \cdot b - 8 = (\frac{1}{4} \cdot b - 1) - \frac{1}{4} b$

Group like terms:

$\frac{1}{4} \cdot b - 8 = (\frac{1}{4} \cdot b + \frac{- 1}{4} b) - 1$

Combine the fractions:

$\frac{1}{4} \cdot b - 8 = \frac{(1 - 1)}{4} b - 1$

Combine the numerators:

$\frac{1}{4} \cdot b - 8 = \frac{0}{4} b - 1$

Reduce the zero numerator:

$\frac{1}{4} b - 8 = 0 b - 1$

Simplify the arithmetic:

$\frac{1}{4} b - 8 = - 1$

Add to both sides:

$(\frac{1}{4} b - 8) + 8 = - 1 + 8$

Simplify the arithmetic:

$\frac{1}{4} b = - 1 + 8$

Simplify the arithmetic:

$\frac{1}{4} b = 7$

Multiply both sides by inverse fraction :

$(\frac{1}{4} b) \cdot \frac{4}{1} = 7 \cdot \frac{4}{1}$

Group like terms:

$(\frac{1}{4} \cdot 4) b = 7 \cdot \frac{4}{1}$

Multiply the coefficients:

$\frac{(1 \cdot 4)}{4} b = 7 \cdot \frac{4}{1}$

Simplify the fraction:

$b = 7 \cdot \frac{4}{1}$

Simplify the arithmetic:

$b = 28$

22 additional steps

$(\frac{1}{2} b - 8) = - (\frac{1}{4} b - 1)$

Expand the parentheses:

$(\frac{1}{2} \cdot b - 8) = \frac{- 1}{4} b + 1$

Add to both sides:

$(\frac{1}{2} b - 8) + \frac{1}{4} \cdot b = (\frac{- 1}{4} b + 1) + \frac{1}{4} b$

Group like terms:

$(\frac{1}{2} \cdot b + \frac{1}{4} \cdot b) - 8 = (\frac{- 1}{4} \cdot b + 1) + \frac{1}{4} b$

Group the coefficients:

$(\frac{1}{2} + \frac{1}{4}) b - 8 = (\frac{- 1}{4} \cdot b + 1) + \frac{1}{4} b$

Find the lowest common denominator:

$(\frac{(1 \cdot 2)}{(2 \cdot 2)} + \frac{1}{4}) b - 8 = (\frac{- 1}{4} \cdot b + 1) + \frac{1}{4} b$

Multiply the denominators:

$(\frac{(1 \cdot 2)}{4} + \frac{1}{4}) b - 8 = (\frac{- 1}{4} \cdot b + 1) + \frac{1}{4} b$

Multiply the numerators:

$(\frac{2}{4} + \frac{1}{4}) b - 8 = (\frac{- 1}{4} \cdot b + 1) + \frac{1}{4} b$

Combine the fractions:

$\frac{(2 + 1)}{4} \cdot b - 8 = (\frac{- 1}{4} \cdot b + 1) + \frac{1}{4} b$

Combine the numerators:

$\frac{3}{4} \cdot b - 8 = (\frac{- 1}{4} \cdot b + 1) + \frac{1}{4} b$

Group like terms:

$\frac{3}{4} \cdot b - 8 = (\frac{- 1}{4} \cdot b + \frac{1}{4} b) + 1$

Combine the fractions:

$\frac{3}{4} \cdot b - 8 = \frac{(- 1 + 1)}{4} b + 1$

Combine the numerators:

$\frac{3}{4} \cdot b - 8 = \frac{0}{4} b + 1$

Reduce the zero numerator:

$\frac{3}{4} b - 8 = 0 b + 1$

Simplify the arithmetic:

$\frac{3}{4} b - 8 = 1$

Add to both sides:

$(\frac{3}{4} b - 8) + 8 = 1 + 8$

Simplify the arithmetic:

$\frac{3}{4} b = 1 + 8$

Simplify the arithmetic:

$\frac{3}{4} b = 9$

Multiply both sides by inverse fraction :

$(\frac{3}{4} b) \cdot \frac{4}{3} = 9 \cdot \frac{4}{3}$

Group like terms:

$(\frac{3}{4} \cdot \frac{4}{3}) b = 9 \cdot \frac{4}{3}$

Multiply the coefficients:

$\frac{(3 \cdot 4)}{(4 \cdot 3)} b = 9 \cdot \frac{4}{3}$

Simplify the fraction:

$b = 9 \cdot \frac{4}{3}$

Multiply the fraction(s):

$b = \frac{(9 \cdot 4)}{3}$

Simplify the arithmetic:

$b = 12$

3. List the solutions

$b = 28, 12$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{2} b - 8 |$
$y = | \frac{1}{4} b - 1 |$
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 b

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 b

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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