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: 29

Answer

$\frac{1}{t^2+9}$

Work Step by Step

$\frac{t-3}{t^2+9}\times\frac{t+3}{t^2-9}$ Factorise the fraction $\frac{t-3}{t^2-9}$: $=\frac{t-3}{t^2+9}\times\frac{t+3}{(t-3)(t+3)}$ Multiply the fractions: $=\frac{(t-3)(t+3)}{(t^2+9)(t-3)(t+3)}$ Simplify the fraction: $=\frac{1}{t^2+9}$
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