Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.5 - Problem Solving and Using Formulas - Exercise Set - Page 66: 13

Answer

$x=8$

Work Step by Step

$y_{1}=10(2x - 1)$ equation 1 $y_{2}=2x + 1$ equation 2 $y_{1}$ is 14 more than 8 times $y_{2}$ --> $y_{1} = 14+8y_{2}$ equation 3 Substitute equations 1 & 2 to equation 3, $y_{1} = 14+8y_{2}$: $10(2x - 1)= 14+8(2x + 1)$ Expand $10(2x - 1)$ = $20x-10$ Expand $8(2x + 1)$ = $16x+8$ Rewrite the equation: $20x-10 = 14 + 16x+8$ $20x-10 = 22 + 16x$ Add 10 to both sides: $20x-10+10 = 22 +10 + 16x$ $20x = 32 + 16x$ Simplify: $4x = 32$ Divide both sides by 4: $x=8$
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