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: 19



Work Step by Step

Here, we have $15x+8y=23 \implies x=\dfrac{1}{3}+\dfrac{2}{3} y$ ...(1) and $3x-2y=1 \implies 3x=1+2y$ ...(2) Now, plug $x=\dfrac{1}{3}+\dfrac{2}{3} y$ into the second equation. $15(\dfrac{1}{3}+\dfrac{2}{3} y)+8y=23$ or, $15y+8y=23 \implies y=1$ Use the $y$ value in the second equation and solve for $x$. $3x=1+2(1) \implies x=1$ Hence, $(x,y)=(1,1)$
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