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: 6

Answer

(a) Yes (b) No

Work Step by Step

(a) $\frac{3+a}{3}=\frac{1}{3}(3+a)=1+\frac{a}{3}$ hence $\frac{3+a}{3}=1+\frac{a}{3}$ (b) When $x=2$, $\frac{2}{4+x}=\frac{2}{6}=\frac{1}{3}$ however $\frac{1}{2}+\frac{2}{x}=\frac{1}{2}+\frac{2}{2}=\frac{1}{2}+1=\frac{3}{2} $ hence$ \frac{1}{3}\ne \frac{3}{2}$ so LHS $\ne$ RHS hence $\frac{2}{4+x}\ne \frac{1}{2}+\frac{2}{x}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.