Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises: 4

Answer

$k= -\displaystyle \frac{3}{8}$

Work Step by Step

The aim is, step by step, to isolate $k$ on one side of the equation. By multiplying both sides with the LCD of all the fractions, we obtain an equation with no fractions: (LCD=24) $\displaystyle \frac{2}{3}k- k+\displaystyle \frac{3}{8}=\frac{1}{2}\qquad.../\times 24$ $ 24\displaystyle \cdot\frac{2}{3}k-24\cdot k+24\displaystyle \cdot\frac{3}{8}=24\cdot\frac{1}{2}\qquad $simplify $16k-24k+9=12$ Subtract $9$ from both sides and the left side will contain only$ k^{\prime}s.$ $ 16k-24k=12-9 \qquad$... simplify $-8k=3$ Dividing both sides with $(-8)$ isolates k on the left side. $-8k=3 \qquad$...$/\div(-8)$ $k= -\displaystyle \frac{3}{8}$
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