Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 662: 35



Work Step by Step

When squaring $(1-y^2)$ he got an incorrect result; the correct result is: $1-2y^2+y^4$, then our equation is: $1-2y^2+y^4+y^2-2+2y^2-2y=-1\\y^4+y^2-2y=0\\y(y^3+y-2)=0\\y(y-1)(y^2+y+2)=0$ The third term has a positive discriminant, so it cannot be $0$. Thus, $y=0$ or $y=1$. Thus the solutions are: $(1,0),(0,1)$
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.