Precalculus (6th Edition) Blitzer

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

Chapter 7 - Review Exercises - Page 877: 28


$(-3,- \sqrt 6)$, $(-3,\sqrt 6)$, $(3,- \sqrt 6)$, $(3,\sqrt 6)$

Work Step by Step

Re-arrange the given two equations and then use the substitution method to get $x^2+y^2=15$ and $-x^2-y^2=-15$ Now, $(2x^2+2y^2)+(-x^2-y^2)=24 -15$ This gives: $x^2=9 \implies x=3,-3$ when $x=3$ then we have $(3)^2+y^2=15 \implies y =\sqrt 6, -\sqrt 6$ when $x=-3$ then we have $(-3)^2+y^2=15 \implies y =\sqrt 6, -\sqrt 6$ Hence, our answers are: $(-3,- \sqrt 6)$, $(-3,\sqrt 6)$, $(3,- \sqrt 6)$, $(3,\sqrt 6)$
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