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Q. 105

Expert-verified

Intermediate Algebra

Found in: Page 668

Book edition OER 2017

Author(s) OPENSTAX

Pages 1346 pages

ISBN 9780998625720

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Short Answer

In the following exercises, (a) find the LCD for the given rational expressions (b) rewrite them as equivalent rational expressions with the lowest common denominator.
$\frac{2}{3 d^{2} + 14 d - 5}, \frac{5 d}{3 d^{2} - 19 d + 6}$

(a) The LCD is $(3 d - 1) (d + 5) (d - 6)$ .

(b) The expressions with LCD are $\frac{2 (d - 6)}{(3 d - 1) (d + 5) (d - 6)}, \frac{5 d (d + 5)}{(3 d - 1) (d - 6) (d + 5)}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part (a) Step 1. Given Information

The given expressions are $\frac{2}{3 d^{2} + 14 d - 5}, \frac{5 d}{3 d^{2} - 19 d + 6}$ .

Part (a) Step 2. Find LCD

Factor the denominators of both the given rational expressions.

$3 d^{2} + 14 d - 5 = (3 d - 1) (d + 5) 3 d^{2} - 19 d + 6 = (3 d - 1) (d - 6)$

So, the LCD is $(3 d - 1) (d + 5) (d - 6)$ .

Part (b) Step 1. Given Information

The given expressions are $\frac{2}{3 d^{2} + 14 d - 5}, \frac{5 d}{3 d^{2} - 19 d + 6}$

Part (b) Step 2. Express the expressions with LCD

Multiply and divide the expressions by the missing factor to make the denominator equal to LCD.

$\frac{2}{3 d^{2} + 14 d - 5} = \frac{2 (d - 6)}{(3 d - 1) (d + 5) (d - 6)} \frac{5 d}{3 d^{2} - 19 d + 6} = \frac{5 d (d + 5)}{(3 d - 1) (d - 6) (d + 5)}$

So, the expressions with equal denominators are role="math" localid="1645521296602" $\frac{2 (d - 6)}{(3 d - 1) (d + 5) (d - 6)}, \frac{5 d (d + 5)}{(3 d - 1) (d - 6) (d + 5)}$ .

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