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: 31

Answer

{$(1,4),(-1,-4),(2\sqrt 2,\sqrt 2),(-2\sqrt 2,-\sqrt 2)$}

Work Step by Step

Re-write the second equation as: $x=\dfrac{4}{y}$ First equation yields: $2(\dfrac{4}{y})^2+y^2=18$ or, $y^4-18y^2+32=0$ or,$(y^2-16)(y^2-2)=0$ or, $y=${$4,-4,\sqrt 2,- \sqrt 2$} From first equation when $y=4$, we have $x=1$ From first equation when $y=-4$, we have $x=-1$ From first equation when $y=\sqrt 2$, we have $x=2 \sqrt 2$ From first equation when $y=-\sqrt 2$, we have $x=-2 \sqrt 2$ Thus, solution set is: {$(1,4),(-1,-4),(2\sqrt 2,\sqrt 2),(-2\sqrt 2,-\sqrt 2)$}
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