College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Section 8.6 - Systems of Nonlinear Equations - 8.6 Assess Your Understanding - Page 614: 14

Answer

$(-2,-2)$ and $(2,-2)$
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Work Step by Step

$\left\{\begin{aligned}x^{2}+y^{2}&=8\\x^{2}+y^{2}+4y&=0\end{aligned}\right.$ Graphed with desmos.com/calculator. Substitute 8 for $x^{2}+y^{2}$ in the second equation: $8+4y=0$ $4y=-8$ $y=-2$ Back-substituting, $x^{2}+(-2)^{2}=8$ $x^{2}=4$ $x=\pm 2$ Intersections: $(-2,-2)$ and $(2,-2)$
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