Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 429: 96

Answer

It makes sense

Work Step by Step

The first addition is solved in $3$ steps: 1) Place the sum over the common denominator $$E_1=\dfrac{2x^3+11x^2+5x^3+4x^2}{x+3}.$$ 2) Reduce like terms: $$E_1=\dfrac{7x^3+15x^2}{x+3}.$$ 3) Factor the numerator: $$E_1=\dfrac{x^2(7x+15)}{x+3}.$$ The second addition is solved in $5$ steps: 1) Determine the Least Common Denominator: $$LCD=(x+3)(x-3).$$ 2) Multiply the first fraction by $\dfrac{x-3}{x-3}$ and the second by $\dfrac{x+3}{x+3}$: $$E_2=\dfrac{2(x-3)}{(x+3)(x-3)}+\dfrac{5(x+3)}{(x+3)(x-3)}.$$ 3) Place the sum over the common denominator: $$E_2=\dfrac{2(x-3)+5(x+3)}{(x+3)(x-3)}.$$ 4) Clear parenthesis: $$E_2=\dfrac{2x-6+5x+15}{(x+3)(x-3)}.$$ 5) Reduce like terms: $$E_2=\dfrac{7x+9}{(x+3)(x-3)}.$$ So the given statement makes sense.
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