Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 10 - Section 10.8 - Systems of Nonlinear Equations - 10.8 Exercises - Page 754: 31

Answer

$x=\dfrac{1}{5}$ and $y=\dfrac{1}{3}$

Work Step by Step

Consider our first equation: $\dfrac{2}{x}-\dfrac{3}{y}=1$ This gives: $xy=2y-3x$ ...(1) Next, consider our second equation: $\dfrac{-4}{x}+\dfrac{7}{y}=1$ This gives: $xy=-4y-7x$ ...(2) Now, equate both equations (1) and (2), we get $2y-3x=-4y-7x$ or, $3y=5x \implies y=\dfrac{5x}{3}$ Plug $y=\dfrac{5x}{3}$ in the first equation , we get $\dfrac{2}{x}-\dfrac{3}{\dfrac{5x}{3}}=1 \implies \dfrac{10-9}{5x}=1$ This gives: $5x =1 \implies x=\dfrac{1}{5}$ Thus, $y=\dfrac{5(\dfrac{1}{5})}{3}=\dfrac{1}{3}$
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