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Elementary Algebra

Lynn Marecek, MaryAnne Anthony-Smith

$Math Studyset 91影视 Explanations$ Math

2nd Edition 路 1240 pages

ISBN: 9780998625713

Chapter 8: Q. 340 (page 982)

In the following exercises, solve.
$\frac{r}{3 r - 15} - \frac{1}{4 r + 20}$ $= \frac{3 r^{2} + 17 r + 40}{12 r^{2} - 300}$

Short Answer

Expert verified

The solution is $r = 5, - 5 .$

Step by step solution

01

Step 1. Given Information

The given rational expression is

$\frac{r}{3 r - 15} - \frac{1}{4 r + 20} = \frac{3 r^{2} + 17 r + 40}{12 r^{2} - 300}$

02

Step 2. Least Common Denominator(LCD)

On Simplifying the given equation,

$\frac{r}{3 r - 15} - \frac{1}{4 r + 20} = \frac{3 r^{2} + 17 r + 40}{12 r^{2} - 300} \frac{r}{3 (r - 5)} - \frac{1}{4 (r + 5)} = \frac{3 r^{2} + 17 r + 40}{12 (r^{2} - 25)}$

Therefore,

The LCD is $12 (r + 5) (r - 5)$

03

Step 3. Clearing fractions

Multiply all terms by LCD,

$12 (r + 5) (r - 5) [\frac{r}{3 (r - 5)} - \frac{1}{4 (r + 5)}] = \frac{3 r^{2} + 17 r + 40}{12 (r^{2} - 25)} 脳 12 (r + 5) (r - 5) 4 (r + 5) (r) - 3 (r - 5) = 3 r^{2} + 17 r + 40 4 r^{2} + 20 r - 3 r + 15 = 3 r^{2} + 17 r + 40 r^{2} - 25 = 0 (r + 5) (r - 5) = 0$

04

Step 4. Solution of the equation

Using the Zero product property,

$r + 5 = 0, r - 5 = 0 r = - 5, r = 5$

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