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: 15



Work Step by Step

Multiplying equation 2 by 3, we get: $5x+3y=6$ and $9x-3y=15$ Adding both equations, we get $14x=21$ and thus $x=\frac{3}{2}$ (Elimination) Substituting the value of x in equation 1: $5\times \frac{3}{2} + 3y=6$ This becomes $3y=6-\frac{15}{2}=\frac{-3}{2}$ and $y=\frac{-1}{2}$ Thus, we get $(\frac{3}{2},\frac{-1}{2})$ as a solution.
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