Order of group from its presentation Say we want to determine the order of a group generated by $x$ and $y$ who satisfy $x^2y = xy^3 = 1$.
Ok so it would be nice to know the order of $x$ and $y$ respectively. We can readily conclude from the equations above that $x^2 = y^{-1}$ and $x^{-1} = y^3$. 
Next we can derive the relationship $x = y^2$ and plugging this into the given equations we get $y^5 = 1$, hence $y$ is of order $5$. 
Now since $x = y^2 = y \cdot y$, we have $y^{-1} x = y$ and we knew $y^{-1} = x^2$ so we get $x^3 = y$. But since $y^5 = 1$ we get $x^{15} = 1$. Hence the $x$ is of order $15$.
$x$ and $y$ generate the group so every element can be written as $x^a y^b$ where $a$ and $b$ range from 0 to 14 and 0 to 5 respectively. But the group is not necessarily abelian, so are we not missing out elements of the form $y^kx^j$?
Not sure how to conclude the argument, help much appreciated!
 A: There are some subtleties missing from the other answer (or perhaps brushed under the carpet for ease), so I thought I would answer this question formally, and in a way which can be generalised. The question asks,

Say we want to determine the order of a group generated by $x$ and $y$ who satisfy $x^2y = xy^3 = 1$.

Okay. But then the trivial group clearly satisfies these properties! So, the "correct" way to think about this problem is to note that any group which satisfies these relations is a homomorphic image of the group with the following presentation.
$$
G=\langle x, y; x^2y=1, xy^3=1\rangle
$$
As $y=x^{-2}$, we can remove the generator $y$ and the relation $x^2y=1$ and replace every instance of $y$ with $x^{-2}$ in the remaining relations to obtain an isomorphic group. This is called a Tietze transformation. We therefore have the following.
$$
\begin{align*}
G&\cong\langle x; x(x^{-2})^3=1\rangle\\
&\cong\langle x; x^5=1\rangle
\end{align*}
$$
Hence, $G$ is cyclic of order 5.
We therefore conclude:

If a group $H$ is generated by $x$ and $y$ and satisfies $x^2y = xy^3 = 1$ then $H$ is a homomorphic image of the cyclic group of order 5. Hence, $H\cong \mathbb{Z}_5$ or $H$ is trivial.

To give a non-trivial example of what I am getting at, consider the following, alternative question.

Say we want to determine the order of a group generated by $x$ and $y$ who satisfy $x^{100}y = xy^3 = 1$.

By the same process as above, every such group is a homomorphic image of the cyclic group of order $299$. Hence, we conclude,

If a group $H$ is generated by $x$ and $y$ and satisfies $x^{100}y = xy^3 = 1$ then $H$ is a homomorphic image of the cyclic group of order 299. Hence, $H\cong \mathbb{Z}_{299}$, $H\cong \mathbb{Z}_{13}$, $H\cong \mathbb{Z}_{23}$ or $H$ is trivial.

A: $x^2 = y^{-1}, x = y^{-3} \Rightarrow x^6 = y^{-3} = x \Rightarrow x^5 = 1 \Rightarrow |x| = 5 \Rightarrow |x^2| = 5 \Rightarrow |y^{-1}| = 5 \Rightarrow |y| = 5.$ Now $y \in <x> \Rightarrow$ the group is the cyclic group of order $5$.
