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Solution - Absolute value equations

Exact form:

f = \frac{7}{12}

f=\frac{7}{12}

Decimal form:

f = 0.583

f=0.583

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
$| f - \frac{4}{3} | = | f + \frac{1}{6} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| f - \frac{4}{3} \| = \| f + \frac{1}{6} \|$
$x = + y$	$(f - \frac{4}{3}) = (f + \frac{1}{6})$
$x = - y$	$(f - \frac{4}{3}) = - (f + \frac{1}{6})$
$+ x = y$	$(f - \frac{4}{3}) = (f + \frac{1}{6})$
$- x = y$	$- (f - \frac{4}{3}) = (f + \frac{1}{6})$

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:

$\| x \| = \| y \|$	$\| f - \frac{4}{3} \| = \| f + \frac{1}{6} \|$
$x = + y$ , $+ x = y$	$(f - \frac{4}{3}) = (f + \frac{1}{6})$
$x = - y$ , $- x = y$	$(f - \frac{4}{3}) = - (f + \frac{1}{6})$

2. Solve the two equations for f

5 additional steps

$(f + \frac{- 4}{3}) = (f + \frac{1}{6})$

Subtract from both sides:

$(f + \frac{- 4}{3}) - f = (f + \frac{1}{6}) - f$

Group like terms:

$(f - f) + \frac{- 4}{3} = (f + \frac{1}{6}) - f$

Simplify the arithmetic:

$\frac{- 4}{3} = (f + \frac{1}{6}) - f$

Group like terms:

$\frac{- 4}{3} = (f - f) + \frac{1}{6}$

Simplify the arithmetic:

$\frac{- 4}{3} = \frac{1}{6}$

The statement is false:

$\frac{- 4}{3} = \frac{1}{6}$

The equation is false so it has no solution.

19 additional steps

$(f + \frac{- 4}{3}) = - (f + \frac{1}{6})$

Expand the parentheses:

$(f + \frac{- 4}{3}) = - f + \frac{- 1}{6}$

Add to both sides:

$(f + \frac{- 4}{3}) + f = (- f + \frac{- 1}{6}) + f$

Group like terms:

$(f + f) + \frac{- 4}{3} = (- f + \frac{- 1}{6}) + f$

Simplify the arithmetic:

$2 f + \frac{- 4}{3} = (- f + \frac{- 1}{6}) + f$

Group like terms:

$2 f + \frac{- 4}{3} = (- f + f) + \frac{- 1}{6}$

Simplify the arithmetic:

$2 f + \frac{- 4}{3} = \frac{- 1}{6}$

Add to both sides:

$(2 f + \frac{- 4}{3}) + \frac{4}{3} = (\frac{- 1}{6}) + \frac{4}{3}$

Combine the fractions:

$2 f + \frac{(- 4 + 4)}{3} = (\frac{- 1}{6}) + \frac{4}{3}$

Combine the numerators:

$2 f + \frac{0}{3} = (\frac{- 1}{6}) + \frac{4}{3}$

Reduce the zero numerator:

$2 f + 0 = (\frac{- 1}{6}) + \frac{4}{3}$

Simplify the arithmetic:

$2 f = (\frac{- 1}{6}) + \frac{4}{3}$

Find the lowest common denominator:

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

Multiply the denominators:

$2 f = \frac{- 1}{6} + \frac{(4 \cdot 2)}{6}$

Multiply the numerators:

$2 f = \frac{- 1}{6} + \frac{8}{6}$

Combine the fractions:

$2 f = \frac{(- 1 + 8)}{6}$

Combine the numerators:

$2 f = \frac{7}{6}$

Divide both sides by :

$\frac{(2 f)}{2} = \frac{(\frac{7}{6})}{2}$

Simplify the fraction:

$f = \frac{(\frac{7}{6})}{2}$

Simplify the arithmetic:

$f = \frac{7}{(6 \cdot 2)}$

$f = \frac{7}{12}$

3. Graph

Each line represents the function of one side of the equation:
$y = | f - \frac{4}{3} |$
$y = | f + \frac{1}{6} |$
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 f

3. 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 f

3. Graph

Why learn this

Terms and topics

Related links

Latest Related Drills Solved