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: 30



Work Step by Step

Multiplying equation 1 by $6$, we get: $3x+2y=23$ and $x-2y=5$. Adding both equations, we get $4x=28$ and thus $x=\frac{28}{4}=7$ (Elimination) Substituting the value of x in equation 2 : $7-2y=5$ This becomes $2y=7-5=2$ and $y=(2)/2=1$ Thus, we get $(7,1)$ as a solution.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.