Answer: 4

Small Hint:

Let $s=x+y$ and $p=xy$

Big Hint:

Use $x^2+y^2=s^2-2p$ to eliminate $p$ from $x^3+y^3=s^3-3ps$

Solution:

Let $s=x+y$ and $p=xy.$ From $x^2+y^2=s^2-2p=7,$ we get $p=\frac{s^2-7}{2}.$ Also, $$\begin{aligned} 10&=x^3+y^3\\ &=s^3-3ps\\ &=s^3-\frac{3s(s^2-7)}2. \end{aligned}$$ Hence $$\begin{aligned} s^3-21s+20&=0,\\ (s-1)(s-4)(s+5)&=0. \end{aligned}$$ Each root gives a possible pair of complex roots of $t^2-st+p=0,$ so the largest real possible value of $s=x+y$ is $4.$