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Solution - Absolute value equations

Exact form:

r = - \frac{23}{6}, \frac{13}{18}

r=-\frac{23}{6} , \frac{13}{18}

Mixed number form:

r = - 3 \frac{5}{6}, \frac{13}{18}

r=-3\frac{5}{6} , \frac{13}{18}

Decimal form:

r = - 3.833, 0.722

r=-3.833 , 0.722

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
$| 2 r + \frac{5}{6} | = | r - 3 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 2 r + \frac{5}{6} \| = \| r - 3 \|$
$x = + y$	$(2 r + \frac{5}{6}) = (r - 3)$
$x = - y$	$(2 r + \frac{5}{6}) = - (r - 3)$
$+ x = y$	$(2 r + \frac{5}{6}) = (r - 3)$
$- x = y$	$- (2 r + \frac{5}{6}) = (r - 3)$

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:

$\| x \| = \| y \|$	$\| 2 r + \frac{5}{6} \| = \| r - 3 \|$
$x = + y$ , $+ x = y$	$(2 r + \frac{5}{6}) = (r - 3)$
$x = - y$ , $- x = y$	$(2 r + \frac{5}{6}) = - (r - 3)$

2. Solve the two equations for r

12 additional steps

$(2 r + \frac{5}{6}) = (r - 3)$

Subtract from both sides:

$(2 r + \frac{5}{6}) - r = (r - 3) - r$

Group like terms:

$(2 r - r) + \frac{5}{6} = (r - 3) - r$

Simplify the arithmetic:

$r + \frac{5}{6} = (r - 3) - r$

Group like terms:

$r + \frac{5}{6} = (r - r) - 3$

Simplify the arithmetic:

$r + \frac{5}{6} = - 3$

Subtract from both sides:

$(r + \frac{5}{6}) - \frac{5}{6} = - 3 - \frac{5}{6}$

Combine the fractions:

$r + \frac{(5 - 5)}{6} = - 3 - \frac{5}{6}$

Combine the numerators:

$r + \frac{0}{6} = - 3 - \frac{5}{6}$

Reduce the zero numerator:

$r + 0 = - 3 - \frac{5}{6}$

Simplify the arithmetic:

$r = - 3 - \frac{5}{6}$

Convert the integer into a fraction:

$r = \frac{- 18}{6} + \frac{- 5}{6}$

Combine the fractions:

$r = \frac{(- 18 - 5)}{6}$

Combine the numerators:

$r = \frac{- 23}{6}$

17 additional steps

$(2 r + \frac{5}{6}) = - (r - 3)$

Expand the parentheses:

$(2 r + \frac{5}{6}) = - r + 3$

Add to both sides:

$(2 r + \frac{5}{6}) + r = (- r + 3) + r$

Group like terms:

$(2 r + r) + \frac{5}{6} = (- r + 3) + r$

Simplify the arithmetic:

$3 r + \frac{5}{6} = (- r + 3) + r$

Group like terms:

$3 r + \frac{5}{6} = (- r + r) + 3$

Simplify the arithmetic:

$3 r + \frac{5}{6} = 3$

Subtract from both sides:

$(3 r + \frac{5}{6}) - \frac{5}{6} = 3 - \frac{5}{6}$

Combine the fractions:

$3 r + \frac{(5 - 5)}{6} = 3 - \frac{5}{6}$

Combine the numerators:

$3 r + \frac{0}{6} = 3 - \frac{5}{6}$

Reduce the zero numerator:

$3 r + 0 = 3 - \frac{5}{6}$

Simplify the arithmetic:

$3 r = 3 - \frac{5}{6}$

Convert the integer into a fraction:

$3 r = \frac{18}{6} + \frac{- 5}{6}$

Combine the fractions:

$3 r = \frac{(18 - 5)}{6}$

Combine the numerators:

$3 r = \frac{13}{6}$

Divide both sides by :

$\frac{(3 r)}{3} = \frac{(\frac{13}{6})}{3}$

Simplify the fraction:

$r = \frac{(\frac{13}{6})}{3}$

Simplify the arithmetic:

$r = \frac{13}{(6 \cdot 3)}$

$r = \frac{13}{18}$

3. List the solutions

$r = - \frac{23}{6}, \frac{13}{18}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 2 r + \frac{5}{6} |$
$y = | r - 3 |$
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 r

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 r

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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