Hi, Mario!
Is the system supposed to be the following?
x^{2} + 2y^{2} = 2
x  y = 2
If so, graphing the system shows an ellipse with a line outside of it; they do not intersect, therefore no solution.
To solve it algebraically, solve the linear equation for either variable: x = y + 2
Substitute y+2 into the first equation:
(y + 2)^{2} + 2y^{2} = 2
y^{2} + 4y + 4 + 2y^{2} = 2
3y^{2} + 4y + 2 = 0
Using the quadratic formula results in a negative number under the square root sign; therefore, there is no real number solution.
Hope this helps!
Kathye P.
5/7/2014

Kathye P.