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!

Let $H = \langle x,y \rangle \leq A_{5}.$ Then $H$ has more than one Sylow $3$-subgroup, so has either $4$ or $10$ Sylow $3$-subgroups. If $H$ has $10$ Sylow $3$-subgroups, then $|H|$ is divisible by $30.$ If $|H| = 60,$ we are done. If $|H| = 30,$ then $H \lhd A_{5},$ and furthermore, since $H$ already contains $20$-elements of order $3,$ it can't have $6$ Sylow $5$-subgroups. Hence $H$ has a unique Sylow $5$-subgroup $S$ so $S^{g} = S$ for each $g \in A_{5}$ (since $S^{g} \leq H).$  But then $S \lhd A_{5},$ and $A_{5}$ has a unique Sylow $5$-subgroup, a contradiction. 
Hence we may now suppose that $H$ has $4$ Sylow $3$-subgroups. Hence $12$ divides $|H|.$ Suppose that $H \neq A_{5}.$ Then $|H| = 12.$ Now $H$ contains $8$ elements of order $3,$ so $H$ has a normal Sylow $2$-subgroup. Now the action of $A_{5}$ on $5$-points may be realised by (or, if you prefer, identified with)  its conjugation action on its five Sylow $2$-subgroups. Now $H$ is the stabilizer of one of these Sylow $2$-subgroups in that action, and $x$ and $y$ have a common fixed point.
