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: 26

Answer

$c = 22$ $d = 4$

Work Step by Step

$c = 3d+10$ $c = 2d + 14$ 1. Rewrite one equation to isolate for $d$ $c = 2d + 14$ $c - 14 = 2d$ $d = \frac{c-14}{2}$ 2. Substitute the previous equation for $d$ into the first equation. Solve for $c$. $c = 3(\frac{c-14}{2}) + 10$ $c = \frac{3c-42}{2} + 10$ $2c = 3c-42 + 20$ $2c - 3c = - 42 + 20$ $-c = - 22$ $c = 22$ 3. Substitute $c$ back into any of the equations to solve for $d$ $c = 2d + 14$ $c - 14 = 2d$ $d = \frac{(22)-14}{2}$ $d = \frac{8}{2}$ $d = 4$ Check: $c = 3d + 10$ $22 \overset{?}{=} 3(4) + 10$ $22 \overset{?}{=} 12 + 10$ $22 = 22$
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