Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 27


$(0,-2)$, $(0,2)$, $(-1,-\sqrt 3)$, $(-1,\sqrt 3)$

Work Step by Step

Here, we have $-1(y^2-x)=(-1)(4) \implies x-y^2=-4$ Now, $(x-y^2) +(x^2+y^2) =-4+4$ This gives: $x^2+x=0 \implies x(x+1)=0$ when $x=0$ then we have $y^2-0=4 \implies y=-2$ and $y=2$ when $x=-1$ then we have $y^2-(-1)=4 \implies y^2=3$ and $y=\sqrt 3; -\sqrt 3$ Hence, our answers are: $(0,-2)$, $(0,2)$, $(-1,-\sqrt 3)$, $(-1,\sqrt 3)$
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