Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.3 Multiplying and Dividing Rational Expressions - 7.3 Exercises: 39

Answer

$= 1$

Work Step by Step

$\frac{6t^{2}+t-35}{10t^{2}+17t-20} \div \frac{12t^{2}-t-63}{20t^{2}+29t-36}$ $= \frac{6t^{2}+t-35}{10t^{2}+17t-20} \times \frac{20t^{2}+29t-36}{12t^{2}-t-63}$ $= \frac{6t^{2}-14t+15t-35}{10t^{2}+25t-8t-20} \times \frac{20t^{2}+45t-16t-36}{12t^{2}-28t+27t-63}$ $= \frac{2t(3t-7)+5(3t-7)}{5t(2t+5)-4(2t+5)} \times \frac{5t(4t+9)-4(4t+9)}{4t(3t-7)+9(3t-7)}$ $= \frac{(2t+5)(3t-7)}{(5t-4)(2t+5)} \times \frac{(5t-4)(4t+9)}{(4t+9)(3t-7)}$ $= \frac{(3t-7)}{(5t-4)} \times \frac{(5t-4)(4t+9)}{(4t+9)(3t-7)}$ $= \frac{(3t-7)}{(5t-4)} \times \frac{(5t-4)}{(3t-7)}$ $= \frac{1}{(5t-4)} \times \frac{(5t-4)}{1}$ $= \frac{1}{1} \times \frac{1}{1}$ $= 1$
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