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: 30



Work Step by Step

Adding the first and the second equation we get: $16x^2-48x-160=0\\x^2-3x-10=0\\(x-5)(x+2)=0$ Thus $x=5$ or $x=-2$ Plugging this into the second equation we get: if $x=5$: $y^2-240-16y-32=0\\y^2-16y-272=0$ Using the quadratic formula $y=8\pm4\sqrt{21}$ if $x=-2$: $y^2+96-16y-32=0\\y^2-16y+64=0\\(y-8)^2=0$ Thus $y=8$ Thus the solutions are: $(5,8\pm4\sqrt{21}),(-2,8)$
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.