Finding formula of reduced scandium oxide using its mass and mass of obtained metal An oxide of scandium, with a mass of $\pu{1.423 g},$ is chemically reduced with $\ce{H2}$ to give $\ce{H2O}$ and $\pu{0.929 g}$ of $\ce{Sc}$ metal. What is the formula of scandium oxide? How much water is formed?

I know that the reaction is of the form:
$$\ce{Sc_xO_y + H2 -&gt; H2O + Sc},$$
where $x$ and $y$ are the units to define scandium oxide molecular formula (not empirical).
I don’t know how to get the constants using the info given, and what the info given tells me. I typically use a linear system of equations to solve for the reactants and products. In this case, we’d have:
$$
\begin{align}
\ce{Sc}&amp;: &amp;\quad  Ax - D &amp;= 0;\\
\ce{O}&amp;: &amp;\quad   Ay - C &amp;= 0;\\
\ce{H}&amp;: &amp;\quad   2B - 2C &amp;= 0.
\end{align}
$$
Usually, we would have an equation where $x$ and $y$ aren’t present and you can solve the system of equations. Here, it prevents you from doing the balancing. Instead, you’re just guessing. Any suggestions when linear algebra doesn’t work?
Since scandium oxide is an ionic compound, the method is exactly the same as obtaining the empirical formula, from the ratio of the amounts of elements $n(\ce{Sc})$ and $n(\ce{O}).$
$$n(\ce{Sc})=\frac{m(\ce{Sc})}{M(\ce{Sc})}=\frac{\pu{0.929 g}}{\pu{44.96 g mol-1}}=\pu{2.07E-2 mol}$$
The mass difference comes from the loss of oxygen atoms:
$$n(\ce{O})=\frac{m(\ce{Sc_xO_y})-m(\ce{Sc})}{M(\ce{O})}=\frac{\pu{1.423 g}-\pu{0.929 g}}{\pu{16.00 g mol-1}}=\pu{3.09E-2 mol}$$
The ratio between the amounts of elements is close to $2:3$, therefore the formula is $\ce{Sc2O3}$. You can check that using oxidation states.