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Solution - Absolute value equations

Exact form:

b = - \frac{1}{11}, - \frac{5}{13}

b=-\frac{1}{11} , -\frac{5}{13}

Decimal form:

b = - 0.091, - 0.385

b=-0.091 , -0.385

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
$| b + \frac{1}{4} | = | \frac{1}{12} b + \frac{1}{6} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| b + \frac{1}{4} \| = \| \frac{1}{12} b + \frac{1}{6} \|$
$x = + y$	$(b + \frac{1}{4}) = (\frac{1}{12} b + \frac{1}{6})$
$x = - y$	$(b + \frac{1}{4}) = - (\frac{1}{12} b + \frac{1}{6})$
$+ x = y$	$(b + \frac{1}{4}) = (\frac{1}{12} b + \frac{1}{6})$
$- x = y$	$- (b + \frac{1}{4}) = (\frac{1}{12} b + \frac{1}{6})$

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:

$\| x \| = \| y \|$	$\| b + \frac{1}{4} \| = \| \frac{1}{12} b + \frac{1}{6} \|$
$x = + y$ , $+ x = y$	$(b + \frac{1}{4}) = (\frac{1}{12} b + \frac{1}{6})$
$x = - y$ , $- x = y$	$(b + \frac{1}{4}) = - (\frac{1}{12} b + \frac{1}{6})$

2. Solve the two equations for b

26 additional steps

$(b + \frac{1}{4}) = (\frac{1}{12} b + \frac{1}{6})$

Subtract from both sides:

$(b + \frac{1}{4}) - \frac{1}{12} \cdot b = (\frac{1}{12} b + \frac{1}{6}) - \frac{1}{12} b$

Group like terms:

$(b + \frac{- 1}{12} \cdot b) + \frac{1}{4} = (\frac{1}{12} \cdot b + \frac{1}{6}) - \frac{1}{12} b$

Group the coefficients:

$(1 + \frac{- 1}{12}) b + \frac{1}{4} = (\frac{1}{12} \cdot b + \frac{1}{6}) - \frac{1}{12} b$

Convert the integer into a fraction:

$(\frac{12}{12} + \frac{- 1}{12}) b + \frac{1}{4} = (\frac{1}{12} \cdot b + \frac{1}{6}) - \frac{1}{12} b$

Combine the fractions:

$\frac{(12 - 1)}{12} \cdot b + \frac{1}{4} = (\frac{1}{12} \cdot b + \frac{1}{6}) - \frac{1}{12} b$

Combine the numerators:

$\frac{11}{12} \cdot b + \frac{1}{4} = (\frac{1}{12} \cdot b + \frac{1}{6}) - \frac{1}{12} b$

Group like terms:

$\frac{11}{12} \cdot b + \frac{1}{4} = (\frac{1}{12} \cdot b + \frac{- 1}{12} b) + \frac{1}{6}$

Combine the fractions:

$\frac{11}{12} \cdot b + \frac{1}{4} = \frac{(1 - 1)}{12} b + \frac{1}{6}$

Combine the numerators:

$\frac{11}{12} \cdot b + \frac{1}{4} = \frac{0}{12} b + \frac{1}{6}$

Reduce the zero numerator:

$\frac{11}{12} b + \frac{1}{4} = 0 b + \frac{1}{6}$

Simplify the arithmetic:

$\frac{11}{12} b + \frac{1}{4} = \frac{1}{6}$

Subtract from both sides:

$(\frac{11}{12} b + \frac{1}{4}) - \frac{1}{4} = (\frac{1}{6}) - \frac{1}{4}$

Combine the fractions:

$\frac{11}{12} b + \frac{(1 - 1)}{4} = (\frac{1}{6}) - \frac{1}{4}$

Combine the numerators:

$\frac{11}{12} b + \frac{0}{4} = (\frac{1}{6}) - \frac{1}{4}$

Reduce the zero numerator:

$\frac{11}{12} b + 0 = (\frac{1}{6}) - \frac{1}{4}$

Simplify the arithmetic:

$\frac{11}{12} b = (\frac{1}{6}) - \frac{1}{4}$

Find the lowest common denominator:

$\frac{11}{12} b = \frac{(1 \cdot 2)}{(6 \cdot 2)} + \frac{(- 1 \cdot 3)}{(4 \cdot 3)}$

Multiply the denominators:

$\frac{11}{12} b = \frac{(1 \cdot 2)}{12} + \frac{(- 1 \cdot 3)}{12}$

Multiply the numerators:

$\frac{11}{12} b = \frac{2}{12} + \frac{- 3}{12}$

Combine the fractions:

$\frac{11}{12} b = \frac{(2 - 3)}{12}$

Combine the numerators:

$\frac{11}{12} b = \frac{- 1}{12}$

Multiply both sides by inverse fraction :

$(\frac{11}{12} b) \cdot \frac{12}{11} = (\frac{- 1}{12}) \cdot \frac{12}{11}$

Group like terms:

$(\frac{11}{12} \cdot \frac{12}{11}) b = (\frac{- 1}{12}) \cdot \frac{12}{11}$

Multiply the coefficients:

$\frac{(11 \cdot 12)}{(12 \cdot 11)} b = (\frac{- 1}{12}) \cdot \frac{12}{11}$

Simplify the fraction:

$b = (\frac{- 1}{12}) \cdot \frac{12}{11}$

Multiply the fraction(s):

$b = \frac{(- 1 \cdot 12)}{(12 \cdot 11)}$

Simplify the arithmetic:

$b = \frac{- 1}{11}$

27 additional steps

$(b + \frac{1}{4}) = - (\frac{1}{12} b + \frac{1}{6})$

Expand the parentheses:

$(b + \frac{1}{4}) = \frac{- 1}{12} b + \frac{- 1}{6}$

Add to both sides:

$(b + \frac{1}{4}) + \frac{1}{12} \cdot b = (\frac{- 1}{12} b + \frac{- 1}{6}) + \frac{1}{12} b$

Group like terms:

$(b + \frac{1}{12} \cdot b) + \frac{1}{4} = (\frac{- 1}{12} \cdot b + \frac{- 1}{6}) + \frac{1}{12} b$

Group the coefficients:

$(1 + \frac{1}{12}) b + \frac{1}{4} = (\frac{- 1}{12} \cdot b + \frac{- 1}{6}) + \frac{1}{12} b$

Convert the integer into a fraction:

$(\frac{12}{12} + \frac{1}{12}) b + \frac{1}{4} = (\frac{- 1}{12} \cdot b + \frac{- 1}{6}) + \frac{1}{12} b$

Combine the fractions:

$\frac{(12 + 1)}{12} \cdot b + \frac{1}{4} = (\frac{- 1}{12} \cdot b + \frac{- 1}{6}) + \frac{1}{12} b$

Combine the numerators:

$\frac{13}{12} \cdot b + \frac{1}{4} = (\frac{- 1}{12} \cdot b + \frac{- 1}{6}) + \frac{1}{12} b$

Group like terms:

$\frac{13}{12} \cdot b + \frac{1}{4} = (\frac{- 1}{12} \cdot b + \frac{1}{12} b) + \frac{- 1}{6}$

Combine the fractions:

$\frac{13}{12} \cdot b + \frac{1}{4} = \frac{(- 1 + 1)}{12} b + \frac{- 1}{6}$

Combine the numerators:

$\frac{13}{12} \cdot b + \frac{1}{4} = \frac{0}{12} b + \frac{- 1}{6}$

Reduce the zero numerator:

$\frac{13}{12} b + \frac{1}{4} = 0 b + \frac{- 1}{6}$

Simplify the arithmetic:

$\frac{13}{12} b + \frac{1}{4} = \frac{- 1}{6}$

Subtract from both sides:

$(\frac{13}{12} b + \frac{1}{4}) - \frac{1}{4} = (\frac{- 1}{6}) - \frac{1}{4}$

Combine the fractions:

$\frac{13}{12} b + \frac{(1 - 1)}{4} = (\frac{- 1}{6}) - \frac{1}{4}$

Combine the numerators:

$\frac{13}{12} b + \frac{0}{4} = (\frac{- 1}{6}) - \frac{1}{4}$

Reduce the zero numerator:

$\frac{13}{12} b + 0 = (\frac{- 1}{6}) - \frac{1}{4}$

Simplify the arithmetic:

$\frac{13}{12} b = (\frac{- 1}{6}) - \frac{1}{4}$

Find the lowest common denominator:

$\frac{13}{12} b = \frac{(- 1 \cdot 2)}{(6 \cdot 2)} + \frac{(- 1 \cdot 3)}{(4 \cdot 3)}$

Multiply the denominators:

$\frac{13}{12} b = \frac{(- 1 \cdot 2)}{12} + \frac{(- 1 \cdot 3)}{12}$

Multiply the numerators:

$\frac{13}{12} b = \frac{- 2}{12} + \frac{- 3}{12}$

Combine the fractions:

$\frac{13}{12} b = \frac{(- 2 - 3)}{12}$

Combine the numerators:

$\frac{13}{12} b = \frac{- 5}{12}$

Multiply both sides by inverse fraction :

$(\frac{13}{12} b) \cdot \frac{12}{13} = (\frac{- 5}{12}) \cdot \frac{12}{13}$

Group like terms:

$(\frac{13}{12} \cdot \frac{12}{13}) b = (\frac{- 5}{12}) \cdot \frac{12}{13}$

Multiply the coefficients:

$\frac{(13 \cdot 12)}{(12 \cdot 13)} b = (\frac{- 5}{12}) \cdot \frac{12}{13}$

Simplify the fraction:

$b = (\frac{- 5}{12}) \cdot \frac{12}{13}$

Multiply the fraction(s):

$b = \frac{(- 5 \cdot 12)}{(12 \cdot 13)}$

Simplify the arithmetic:

$b = \frac{- 5}{13}$

3. List the solutions

$b = - \frac{1}{11}, - \frac{5}{13}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | b + \frac{1}{4} |$
$y = | \frac{1}{12} b + \frac{1}{6} |$
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 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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