Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 51

Answer

$\frac{2x+7}{(x+3)(x+4)}$

Work Step by Step

$\frac{2}{x+3}-\frac{1}{x^{2}+7x+12}$ Factorise and replace the term $x^{2}+7x+12$: $=\frac{2}{x+3}-\frac{1}{(x+3)(x+4)}$ Find the lowest common denominator for the two fractions (i.e. $(x+3)(x+4)$) and adjust accordingly: $=\frac{2\times (x+4)}{(x+3)\times (x+4)}-\frac{1}{(x+3)(x+4)}$ $=\frac{2x+8}{(x+3)(x+4)}-\frac{1}{(x+3)(x+4)}$ Combine the fractions: $=\frac{2x+8-1}{(x+3)(x+4)}$ Simplify: $=\frac{2x+7}{(x+3)(x+4)}$
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