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: 24



Work Step by Step

Multiplying equation 1 by $200$ and equation 2 by $300$, we get: $10x-6y=42$ and $21x+6y=48$. Adding both equations, we get $31x=90$ and thus $x=\frac{90}{31}$ (Elimination) Substituting the value of x in equation 1: $10\times\frac{90}{31}-6y=42$ This becomes $6y=\frac{900}{31}-42$ and $y=(900−1302)/186=\frac{-67}{31}$ Thus, we get $(\frac{90}{31},\frac{-67}{31})$ as a solution.
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