Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 11 - Rational Expressions and Functions - 11-2 Multiplying and Dividing Rational Expressions - Practice and Problem-Solving Exercises - Page 674: 50


$ \frac{2c-1}{3c(c+2)}, with$ $c\ne-2$ $and$ $c\ne0$

Work Step by Step

Given : $\frac{\frac{c+4}{c^{2}+5c+6}}{\frac{3c^{2}+12c}{2c^{2}+5c-3}}$ This becomes : $\frac{c+4}{(c+3)(c+2)} \times \frac{(2c-1)(c+3)}{3c(c+4)}$ $= \frac{2c-1}{3c(c+2)}$ (After dividing out the common factors (c+3) and (c+4))
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