Problem (Algebra):
Solve the system of equations:

$$
\begin{aligned}
& \log _{10}\left(x^{3}\right)+\log _{10}\left(y^{2}\right)=11 \\
& \log _{10}\left(x^{2}\right)-\log _{10}\left(y^{3}\right)=3
\end{aligned}
$$

Solution:
Using the facts that $\log _{10} A+\log _{10} B=\log _{10} A B$ and that $\log _{10} A-\log _{10} B=\log _{10} \frac{A}{B}$, then we can convert the two equations to

$$
\begin{aligned}
\log _{10}\left(x^{3} y^{2}\right) & =11 \\
\log _{10}\left(\frac{x^{2}}{y^{3}}\right) & =3
\end{aligned}
$$

Raising both sides to the power of 10 , we obtain

$$
\begin{aligned}
x^{3} y^{2} & =10^{11} \\
\frac{x^{2}}{y^{3}} & =10^{3}
\end{aligned}
$$

To eliminate the $y$ 's, we raise the first equation to the power 3 and the second to the power 2 to obtain

$$
\begin{aligned}
x^{9} y^{6} & =10^{33} \\
\frac{x^{4}}{y^{6}} & =10^{6}
\end{aligned}
$$

and multiply to obtain $x^{9} x^{4}=x^{13}=10^{39}=10^{33} 10^{6}$.

Therefore, since $x^{13}=10^{39}$, then $x=10^{3}$.



Substituting back into $x^{3} y^{2}=10^{11}$, we get $y^{2}=10^{2}$, and so $y= \pm 10$. However, substituting into $\frac{x^{2}}{y^{3}}=10^{3}$ we see that $y$ must be positive, so $y=10$.

Therefore, the solution to the system of equation is $x=10^{3}$ and $y=10$.

Since the domain of the logarithm is the positive real numbers, then the quantities $\log _{10}\left(x^{3}\right)$ and $\log _{10}\left(y^{3}\right)$ tell us that $x$ and $y$ are positive.

Using the fact that $\log _{10}\left(a^{b}\right)=b \log _{10}(a)$, we rewrite the equations as

$$
\begin{aligned}
& 3 \log _{10} x+2 \log _{10} y=11 \\
& 2 \log _{10} x-3 \log _{10} y=3
\end{aligned}
$$

We solve the system of equations for $\log _{10} x$ and $\log _{10} y$ by multiplying the first equation by 3 and adding two times the second equation in order to eliminate $\log _{10} y$. Thus we obtain $13 \log _{10} x=39$ or $\log _{10} x=3$.

Substituting back into the first equation, we obtain $\log _{10} y=1$.

Therefore, $x=10^{3}$ and $y=10$.

Final answer: $10^{3},10$