SOLUTION: Question: How many real solutions does this nonlinear system have? x^2+y2=13 =x^2+2y^2=14 (a) 0 (b) 1 (c) 2 (d) 3 (e) 4

Question 30336: Question:
How many real solutions does this nonlinear system have?
x^2+y2=13
=x^2+2y^2=14
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4
Found 2 solutions by sdmmadam@yahoo.com, Nate:
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
x^2+y2=13 ----(1)
x^2+2y^2=14 ----(2)
That is (x^2+y^2)+y^2 = 14 ----(2)
putting (1) in (2)
13+y^2 =14
y^2 = 14-13
y^2 =1
y = +1 or y = -1 for which we get x^2 = 12
And x^2 = 12 implies x = +sqrt(12) and x = -sqrt(12)
Therefore there are four solutions
which is your choice (e) 4

Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
the equations show us a circle and an ellipse; since the ellipse is larger than the circle, there are four points
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