solve the quintic non-linear system $x^5+y^5=33$ and $x+y=3$
I tried a variety of things here, all too ugly to post.  the most promising was doing long division $(x^5+y^5)/(x+y), $but it still looked kind of ugly and not easily factorable.  I am looking to solve this with simple high school level algebra techniques.  Can someone give me a hint?  Much appreciated.
 A: The first thing I try is plug some random small integers to the equations and see
whether they work. I already get two "obvious" solutions $(x,y) = (2,1)$ and $(1,2)$. 
To see this exhaust all solutions, introduce a parameter $u$ such that $x = \frac{3+u}{2}$ and $y = \frac{3-u}{2}$.
In terms of $u$, we have
$$\begin{align}
& x^5 + y^5 = 33\\
\iff & (3+u)^5 + (3-u)^5 = 33\cdot 2^5\\
\iff & 2(3^5 + 10\cdot 3^3 u^2 + 5\cdot 3 u^4) = 33\cdot 2^5\\
\iff & 10\cdot 3^2 u^2 + 5 u^4 = \frac13\left(\frac12\cdot 33\cdot 2^5 - 3^5\right)
= 11\cdot16 - 81 = 85\\
\iff & u^4 + 18u^2 -19 = 0\\
\iff & (u^2-1)(u^2+19) = 0\\
\implies & u = \pm 1\end{align}$$
This means the equations have two and only two real solutions: $(x,y) = (2,1)$ and $(1,2)$.
A: From the second $y=3-x$. Plug in the first equation and expand
$$x^5+y^5=x^5+(3-x)^5=15 x^4-90 x^3+270 x^2-405 x+243$$
making the equation to be
$$15 x^4-90 x^3+270 x^2-405 x+210=0$$ Divide all terms by $15$ to get
$$x^4-6 x^3+18 x^2-27 x+14=0$$ which, by inspection, shows two simple roots.
Just continue.
