Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 427: 44


$\displaystyle \frac{7x^{2}+12x-12}{(x-3)(x+2)(x+3)}$

Work Step by Step

1. Find the LCD . 1st denominator = $(x-3)(x+2)$ 2nd denominator = $(x-3)(x+3)$ List factors of the 1st denominator. From each next denominator, add only those factors that do not yet appear in the list. LCD = $(x-3)(x+2)(x+3)$ 2. Rewrite each rational expression with the the LCDas the denominator = $\displaystyle \frac{(3x-2)(x+3)}{(x-3)(x+2)(x+3)}+\frac{(4x-3)(x+2)}{(x-3)(x+3)(x+2)}$ 3. Add or subtract numerators, placing the resulting expression over the LCD. = $\displaystyle \frac{(3x-2)(x+3)+(4x-3)(x+2)}{(x-3)(x+2)(x+3)}$ 4. If possible, simplify. = $\displaystyle \frac{3x^{2}+9x-2x-6+4x^{2}+8x-3x-6}{(x-3)(x+2)(x+3)} $ = $\displaystyle \frac{7x^{2}+12x-12}{(x-3)(x+2)(x+3)}$ The numerator has the form $ax^{2}+bx+c.$ To factor, find factors of $ac$ whose sum is $b.$ If successful, rewrite $bx$ and factor in pairs. ... we can't find factors of $-84$ whose sum is $+12$ ... we can't factor the numerator, so we leave it as is.
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