Precalculus (6th Edition) Blitzer

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

Chapter 7 - Mid-Chapter Check Point - Page 853: 11


$(-2,-2)$, $(2,-2)$, $(-\sqrt 7,1)$, $(\sqrt 7,1)$

Work Step by Step

Re-arrange the given equations to get $-x^2+y=-6$ $(-x^2+y)+(x^2+y^2)=-6+8$ $ \implies y^2+y-2=0$ This gives: $ (y+2)(y-1)=0$ when $ y=-2$ gives: $x^2=4 \implies x=2,-2$ when $ y=-1$ gives: $x^2=7 \implies x=-\sqrt 7, \sqrt 7$ Hence, our answers are: $(-2,-2)$, $(2,-2)$, $(-\sqrt 7,1)$, $(\sqrt 7,1)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.