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Solution - Absolute value equations

Exact form:

r = \frac{32}{3}, 2

r=\frac{32}{3} , 2

Mixed number form:

r = 10 \frac{2}{3}, 2

r=10\frac{2}{3} , 2

Decimal form:

r = 10.667, 2

r=10.667 , 2

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
$| 4 r + 5 | = | 7 r - 27 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 4 r + 5 \| = \| 7 r - 27 \|$
$x = + y$	$(4 r + 5) = (7 r - 27)$
$x = - y$	$(4 r + 5) = - (7 r - 27)$
$+ x = y$	$(4 r + 5) = (7 r - 27)$
$- x = y$	$- (4 r + 5) = (7 r - 27)$

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:

$\| x \| = \| y \|$	$\| 4 r + 5 \| = \| 7 r - 27 \|$
$x = + y$ , $+ x = y$	$(4 r + 5) = (7 r - 27)$
$x = - y$ , $- x = y$	$(4 r + 5) = - (7 r - 27)$

2. Solve the two equations for r

11 additional steps

$(4 r + 5) = (7 r - 27)$

Subtract from both sides:

$(4 r + 5) - 7 r = (7 r - 27) - 7 r$

Group like terms:

$(4 r - 7 r) + 5 = (7 r - 27) - 7 r$

Simplify the arithmetic:

$- 3 r + 5 = (7 r - 27) - 7 r$

Group like terms:

$- 3 r + 5 = (7 r - 7 r) - 27$

Simplify the arithmetic:

$- 3 r + 5 = - 27$

Subtract from both sides:

$(- 3 r + 5) - 5 = - 27 - 5$

Simplify the arithmetic:

$- 3 r = - 27 - 5$

Simplify the arithmetic:

$- 3 r = - 32$

Divide both sides by :

$\frac{(- 3 r)}{- 3} = \frac{- 32}{- 3}$

Cancel out the negatives:

$\frac{3 r}{3} = \frac{- 32}{- 3}$

Simplify the fraction:

$r = \frac{- 32}{- 3}$

Cancel out the negatives:

$r = \frac{32}{3}$

12 additional steps

$(4 r + 5) = - (7 r - 27)$

Expand the parentheses:

$(4 r + 5) = - 7 r + 27$

Add to both sides:

$(4 r + 5) + 7 r = (- 7 r + 27) + 7 r$

Group like terms:

$(4 r + 7 r) + 5 = (- 7 r + 27) + 7 r$

Simplify the arithmetic:

$11 r + 5 = (- 7 r + 27) + 7 r$

Group like terms:

$11 r + 5 = (- 7 r + 7 r) + 27$

Simplify the arithmetic:

$11 r + 5 = 27$

Subtract from both sides:

$(11 r + 5) - 5 = 27 - 5$

Simplify the arithmetic:

$11 r = 27 - 5$

Simplify the arithmetic:

$11 r = 22$

Divide both sides by :

$\frac{(11 r)}{11} = \frac{22}{11}$

Simplify the fraction:

$r = \frac{22}{11}$

Find the greatest common factor of the numerator and denominator:

$r = \frac{(2 \cdot 11)}{(1 \cdot 11)}$

Factor out and cancel the greatest common factor:

$r = 2$

3. List the solutions

$r = \frac{32}{3}, 2$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 4 r + 5 |$
$y = | 7 r - 27 |$
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 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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