Answer
$y=-3,x=-1$ or $y=1,x=3$.
Work Step by Step
Our equations according to the exercise are: $x-y=2, x^2+y^2=10$.
So $x=y+2$, thus $(y+2)^2+y^2=10\\y^2+4y+4+y^2=10\\2y^2+4y=6\\y^2+2y-3=0\\(y+3)(y-1)=0$
Thus $y=-3,x=-1$ or $y=1,x=3$.
College Algebra (10th Edition)
by Sullivan, Michael
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 616: 71
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