Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 26


{$(\dfrac{4 \sqrt 2}{3},\dfrac{2}{3}),(-\dfrac{4 \sqrt 2}{3},\dfrac{2}{3})$}

Work Step by Step

Let us multiply the second equation by $-1$.After adding the two equations, we get $y^2-(y-3)^2=-5$ or, $6y=4$; or $y=\dfrac{2}{3}$ From first equation $x^2+y^2=4$ when $y=\dfrac{2}{3}$, we have $x=\pm \dfrac{4 \sqrt 2}{3}$ Thus, solution set is {$(\dfrac{4 \sqrt 2}{3},\dfrac{2}{3}),(-\dfrac{4 \sqrt 2}{3},\dfrac{2}{3})$}
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