Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 9 - 9.2 - Two-Variable Linear Systems - 9.2 Exercises - Page 646: 18



Work Step by Step

Multiplying equation 1 by -8, we get: $-16r-32s=-40$ and $16r+50s=55$ Adding both equations, we get $18s=15$ and thus $s=\frac{5}{6}$ (Elimination) Substituting the value of s in equation 2: $16r + 50\times \frac{5}{6}=55$ This becomes $16r=55-\frac{250}{6}=\frac{80}{6}$ and $r=\frac{80}{96}=\frac{5}{6}$ Thus, we get $(\frac{5}{6},\frac{5}{6})$ as a solution.
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