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: 34

Answer

$\displaystyle \frac{7x^{2}-4x+12}{(x+4)(x-2)(x-1)}$

Work Step by Step

1. Find the LCD . 1st denominator = $(x+4)(x-2)$ 2nd denominator = $(x-2)(x-1)$ List factors of the 1st denominator. From each next denominator, add only those factors that do not yet appear in the list. LCD = $(x+4)(x-2)(x-1)$ $ \displaystyle \frac{7x}{(x+4)(x-2)}+\frac{3}{(x-2)(x-1)}=$ 2. Rewrite each rational expression with the the LCDas the denominator = $\displaystyle \frac{7x(x-1)}{(x+4)(x-2)(x-1)}+\frac{3(x+4)}{(x-2)(x-1)(x+4)}$ 3. Add or subtract numerators, placing the resulting expression over the LCD. = $\displaystyle \frac{7x(x-1)+3(x+4)}{(x+4)(x-2)(x-1)}$ 4. If possible, simplify. = $\displaystyle \frac{7x^{2}-7x+3x+12}{(x+4)(x-2)(x-1)} $ = $\displaystyle \frac{7x^{2}-4x+12}{(x+4)(x-2)(x-1)}$ The numerator is of 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 $-4$ ... we can't factor the numerator, so we leave it as it is.
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