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!

Hint: $y(1,2,3)y^{-1} = (y(1),y(2),y(3))
$
solution:
You can assume that $x=(1,2,3)$. As $x$ and $y$ do not fix either 4 or 5, you can assume that 
$y=(3,4,5)$.
Now
$$
y(1,2,3)y^{-1} = (y(1),y(2),y(3)) = (1,2,4)
$$
If you replace $y$ with $y^2$, you get that $\langle x,y\rangle$ contains
$(1,2,3),(1,2,4),(1,2,5)$.
Do the same switching $x$ and $y$ you get $(3,4,5),(1,4,5),(2,4,5)$.
To get the remaining 3 cycles:
$$
x(1,2,4)x^ {-1} = (2,3,4)\\
x(2,3,4)x^ {-1} = (3,1,4)\\
x(1,2,5)x^ {-1} = (2,3,5)\\
x(2,3,5)x^ {-1} = (3,1,5)\\
$$
You have proved that 
$\langle x,y\rangle$ contains
$\{(1,2,3),(1,2,4),(1,2,5),(3,4,5),(1,4,5),(2,4,5),
(2,3,4),(3,1,4),(2,3,5),(3,1,5)\}$ and the powers of these elements are all the 3 cycles. Then $\langle x,y\rangle\supseteq A_5$ and eventually $\langle x,y\rangle=A_5$. 

To fit OP's definition of alternate permutations: as
$
(1,2)(2,3) = (1,2,3)$ every product of an odd number of transpositions is a product of 3-cycles.
