College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Section 8.1 - Systems of Linear Equations: Substitution and Elimination - 8.1 Assess Your Understanding - Page 557: 76

Answer

$r=5$% $Y=1200$

Work Step by Step

$0.05Y - 1000r = 10$ $0.05Y + 800r = 100$ Multiply first equation by -1 and add to second equation to eliminate Y $-0.05Y + 1000r = -10$ $0.05Y + 800r = 100$ $1000r + 800r = -10 + 100$ $1800r = 90$ $r=0.05$ $0.05Y - 1000 \times 0.05 = 10$ $0.05Y - 50 = 10$ $0.05Y = 50 + 10$ $0.05Y = 60$ $Y=1200$
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