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 - Guided Practice for Examples 3 and 4 - Page 661: 8



Work Step by Step

Adding the first and the second equation we get: $2x^2-16x+30=0\\x^2-8x+15=0\\(x-3)(x-5)=0$ Thus $x=3$ or $x=5$. When $x=3$ then $y=\pm\sqrt{3^2-9}=\pm0=0$ When $x=5$ then $y=\pm\sqrt{5^2-9}=\pm\sqrt{16}=\pm4$ Thus the solutions are: $(3,0),(5,4),(5,-4)$
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.