Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1-2 - Cumulative Review: 18

Answer

$p = \frac{248}{35}$

Work Step by Step

$p + 3(\frac{1}{4}p-4) = \frac{2}{5}$ $p + 3(\frac{p}{4}-4) = \frac{2}{5}$ $p + \frac{3p}{4}-12= \frac{2}{5}$ $4p + 3p - 4(12) = 4(\frac{2}{5})$ $7p - 48 = \frac{8}{5}$ $5(7p) - 5(48) = 8$ $35p - 240 = 8$ $35p - 240 - 8 = 0$ $35p - 248 = 0$ $35p = 248$ $p = \frac{248}{35}$ Check: $\frac{248}{35} + 3(\frac{1}{4}(\frac{248}{35})-4) \overset{?}{=} \frac{2}{5}$ $\frac{248}{35} + 3(-\frac{78}{35}) \overset{?}{=} \frac{2}{5}$ $\frac{248}{35} -\frac{234}{35} \overset{?}{=} \frac{2}{5}$ $\frac{14}{35} \overset{?}{=} \frac{2}{5}$ $\frac{2}{5} = \frac{2}{5}$
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