Two 3-cycles generate $A_5$ I want to solve the following exercise, from Dummit & Foote's Abstract Algebra

Let $x$ and $y$ be distinct 3-cycles in $S_5$ with $x \neq y^{-1} $.
  Prove that if $x$ and $y$ do not fix a common element of $\{1,2,3,4,5\}$ then $\langle x,y\rangle =A_5$.

I know that using brute force one can find $31$ distinct permutations in $\langle x,y \rangle$, and then Lagrange's Theorem finishes the proof. 
However, is there a more elegant way to go about this? (Preferably without using the fact that $A_5$ is simple, since it wasn't proven up to this point in the text.)
Thank you!

Yet another answer. We can assume $x=(1,2,3)$, $y=(3,4,5)$. Let $H=\langle x,y \rangle$. Now $z := x^y = (1,2,4) \in H$. Clearly $H$ is transitive on $\{1,2,3,4,5\}$, $\langle x,z \rangle \le H_5$ is transitive on $\{1,2,3,4\}$ and $\langle x \rangle \le H_{4,5}$ is transitive on $\{1,2,3\}$, so by the Orbit-Stabilizer Theorem, $|H| \ge 5 \times 4 \times 3 = 60$.
