# 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

Algebra ->  Algebra  -> College  -> Linear Algebra -> 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      Log On

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 Click here to see ALL problems on Linear Algebra 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) 4Found 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