Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set: 108

Answer

$$ \frac{3 - 16x}{48x^2}$$

Work Step by Step

To simplify, we remember the rules for exponents: $$ a^b \times a ^c = a ^ {b+c}$$ $$ \frac {a^b}{a^c} = a ^ {b-c}$$ $$ a ^ {b^c} = a^{bc}$$ This gives: $$ \frac {1} {16x^2} - \frac{1}{3x}$$ We multiply both terms to create like denominators: $$ (\frac {1} {16x^2} \times \frac{3}{3}) - (\frac{1}{3x} \times \frac{16x}{16x})$$ Now, we use PEMDAS, a method for simplification. This acronym reminds us that we must first address parentheses, and then exponents. Next, we address multiplication and division, going from left to right. Lastly, we simplify addition and subtraction from left to right. This gives: $$ \frac{3 - 16x}{48x^2}$$
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