+9519 We express x^2 + y^2 and x^3 + y^3 in terms of (x + y) and (xy).
\(x^2 + y^2 = (x + y)^2 - 2(xy)\\ x^3 + y^3 = (x + y)((x + y)^2 - 3(xy))\)
Now, let x + y = t, xy = u.
\(\begin{cases}t^2 - 2u = 7\\t(t^2 - 3u)=10\end{cases}\)
Substituting the first equation into the second,
\(\begin{cases}t^2 - 2u = 7\\t(7 - u)=10\end{cases}\)
Manipulating the second equation, \(u = 7 - \dfrac{10}t\)
Substituting,
\(t^2 - 14 + \dfrac{20}t = 7\\ t^3 - 21t + 20 = 0\)
By factor theorem, (t - 1) is a factor of the left hand side.
\((t - 1)(t^2 + t - 20) = 0\\ (t - 1)(t - 4)(t + 5) = 0\\ t = 1 \text{ or }t = 4 \text{ or } t= -5\)
Therefore, the possible values of (x + y) are 1, 4, and -5.
