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Solution - Absolute value equations

Exact form:

r = \frac{37}{5}, 3

r=\frac{37}{5} , 3

Mixed number form:

r = 7 \frac{2}{5}, 3

r=7\frac{2}{5} , 3

Decimal form:

r = 7.4, 3

r=7.4 , 3

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
$| 2 r + 5 | = | 7 r - 32 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 2 r + 5 \| = \| 7 r - 32 \|$
$x = + y$	$(2 r + 5) = (7 r - 32)$
$x = - y$	$(2 r + 5) = - (7 r - 32)$
$+ x = y$	$(2 r + 5) = (7 r - 32)$
$- x = y$	$- (2 r + 5) = (7 r - 32)$

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:

$\| x \| = \| y \|$	$\| 2 r + 5 \| = \| 7 r - 32 \|$
$x = + y$ , $+ x = y$	$(2 r + 5) = (7 r - 32)$
$x = - y$ , $- x = y$	$(2 r + 5) = - (7 r - 32)$

2. Solve the two equations for r

11 additional steps

$(2 r + 5) = (7 r - 32)$

Subtract from both sides:

$(2 r + 5) - 7 r = (7 r - 32) - 7 r$

Group like terms:

$(2 r - 7 r) + 5 = (7 r - 32) - 7 r$

Simplify the arithmetic:

$- 5 r + 5 = (7 r - 32) - 7 r$

Group like terms:

$- 5 r + 5 = (7 r - 7 r) - 32$

Simplify the arithmetic:

$- 5 r + 5 = - 32$

Subtract from both sides:

$(- 5 r + 5) - 5 = - 32 - 5$

Simplify the arithmetic:

$- 5 r = - 32 - 5$

Simplify the arithmetic:

$- 5 r = - 37$

Divide both sides by :

$\frac{(- 5 r)}{- 5} = \frac{- 37}{- 5}$

Cancel out the negatives:

$\frac{5 r}{5} = \frac{- 37}{- 5}$

Simplify the fraction:

$r = \frac{- 37}{- 5}$

Cancel out the negatives:

$r = \frac{37}{5}$

12 additional steps

$(2 r + 5) = - (7 r - 32)$

Expand the parentheses:

$(2 r + 5) = - 7 r + 32$

Add to both sides:

$(2 r + 5) + 7 r = (- 7 r + 32) + 7 r$

Group like terms:

$(2 r + 7 r) + 5 = (- 7 r + 32) + 7 r$

Simplify the arithmetic:

$9 r + 5 = (- 7 r + 32) + 7 r$

Group like terms:

$9 r + 5 = (- 7 r + 7 r) + 32$

Simplify the arithmetic:

$9 r + 5 = 32$

Subtract from both sides:

$(9 r + 5) - 5 = 32 - 5$

Simplify the arithmetic:

$9 r = 32 - 5$

Simplify the arithmetic:

$9 r = 27$

Divide both sides by :

$\frac{(9 r)}{9} = \frac{27}{9}$

Simplify the fraction:

$r = \frac{27}{9}$

Find the greatest common factor of the numerator and denominator:

$r = \frac{(3 \cdot 9)}{(1 \cdot 9)}$

Factor out and cancel the greatest common factor:

$r = 3$

3. List the solutions

$r = \frac{37}{5}, 3$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 2 r + 5 |$
$y = | 7 r - 32 |$
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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