jmc

Let $s=x+y$ and $p=xy.$ We can write the given quantities in terms of $s$ and $p$ as follows: $$\begin{array}{rl}7& =x^2+y^2=(x+y{)}^2-2xy=s^2-2p,\\ 10& =x^3+y^3=(x+y{)}^3-3xy(x+y)=s^3-3sp.\end{array}$$Now we solve for $s$ and $p.$ The first equation gives $p=\frac{s^2-7}{2}.$ Substituting this into the second equation, we have $$10=s^3-3s\left(\frac{s^2-7}{2}\right)$$which simplifies to $s^3-21s+20=0.$ This factors as $(s-1)(s-4)(s+5)=0,$ so the possible values for $s$ are $\boxed{-5,1,4}.$