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: 29



Work Step by Step

Adding the first and the second equation we get: $2x^2-16x=0\\2x(x-8)=0$ Thus $x=0$ or $x=8$ Plugging this into the second equation we get: if $x=0$: $0-y^2-0+8y=24\\y^2-8y+24=0$ This has a negative discriminant, thus there is no solution here. if $x=8$: $64-y^2-64+8y=-24\\y^2-8y-24=0$ Using the quadratic formula $y=4\pm2\sqrt{10}$ Thus the solutions are: $(8,4\pm2\sqrt{10})$
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.