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 - Page 43: 28



Work Step by Step

$=\frac{x^2+2x-3}{x^2-2x-3}\times\frac{3-x}{3+x}$ Simplify the fraction $\frac{x^2+2x-3}{x^2-2x-3}$: $=\frac{(x+3)(x-1)}{(x+1)(x-3)}$ Multiply the fractions: $=\frac{(x+3)(x-1)}{(x+1)(x-3)}\times\frac{3-x}{3+x}$ $=\frac{(x+3)(x-1)(-x+3)}{(x+1)(x-3)(x+3)}$ Extract the negative sign from $-x+3$: $=\frac{-(x-3)(x-1)(x+3)}{(x+1)(x-3)(x+3)}$ Simplify the fraction: $=-\frac{x-1}{x+1}$
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