Algebra and Trigonometry 10th Edition

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

Chapter 9 - 9.1 - Linear and Nonlinear Systems of Equations - 9.1 Exercises - Page 635: 18



Work Step by Step

Here, we have $6x-3y=4 \implies x=4-2y$ Now, plug $x=4-2y$ into the second equation. $6(4-2y)-3y=4$ $ 24-12y-3y=4$ $15y=20 \implies y=\dfrac{4}{3}$ Use the $y$ value and the equation above to solve for $x$: $ x=4-2(\dfrac{4}{3})=\dfrac{4}{3}$ Hence, $(x,y)=(\dfrac{4}{3},\dfrac{4}{3})$
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.