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: 45

Answer

$\frac{10x-18}{(2x-3)^{2}}$

Work Step by Step

$\frac{5}{2x-3}-\frac{3}{(2x-3)^{2}}$ Find the lowest common denominator (i.e. $(2x-3)$) and adjust the fractions: $=\frac{5\times (2x-3)}{(2x-3)^{2}}-\frac{3}{(2x-3)^{2}}$ Expand any brackets: $=\frac{10x-15}{(2x-3)^{2}}-\frac{3}{(2x-3)^{2}}$ Combine the fractions: $=\frac{10x-15-3}{(2x-3)^{2}}$ Collect like terms: $=\frac{10x-18}{(2x-3)^{2}}$
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