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

Answer

$(\dfrac{4}{3},\dfrac{4}{3})$

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, $(x,y)=(\dfrac{4}{3},\dfrac{4}{3})$
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