Solve the following system of equations:

$$
\begin{aligned}
x^{5} y^{17} & =a, \\
x^{2} y^{7} & =b
\end{aligned}
$$
If $a=0$, then either $x=0$ and $y$ is any number, or $y=0$ and $x$ is any number. In both cases, the system only has a solution if $b$ is also 0, and in this case, there are infinitely many solutions. Similarly, if $b=0$ and $a \neq 0$, there is no definite solution. Hereafter, we assume that neither $a$ nor $b$ is 0, in which case neither $x$ nor $y$ can be 0.

We could eliminate $x$ if its exponent in both equations were the same. This can be achieved by raising our equations to powers, since in this way the exponent of $x$ (1) in successive powers would be $2,4,6,8,10,12, \ldots$, (2) in powers would be $5,10,15, \ldots$, and $c$ has a common element in the two sequences, which is 10.

Raising (1) to the fourth power and (2) to the fifth power, we get

\[
\begin{aligned}
& x^{10} y^{34}=a^{2} \\
& x^{10} y^{35}=b^{5}
\end{aligned}
\]

and dividing the latter by the former,

\[
y=\frac{b^{5}}{a^{2}}
\]

To determine $x$, similarly raising (1) to the 7th power and (2) to the 17th power, and then dividing,

\[
\frac{a^{7}}{b^{17}}=\frac{x^{35} y^{119}}{x^{34} y^{119}}=x
\]

Substituting (3) and (4) back into (1) and (2), we see that we have obtained the solution.

Remarks. 1. Mathematically, it would have been more elegant to calculate $x$ by substituting (3) into either (1) or (2) to form a single-variable equation for $x$. If we substitute into (2):

\[
x^{2}=b\left(\frac{1}{y}\right)^{7}=\frac{a^{14}}{b^{34}}, \quad x= \pm \frac{a^{7}}{b^{17}}
\]

and the verification by substitution shows that only the upper sign is correct.

If signs have already been mentioned, it is good to state that (3) and (4) are valid without any restriction on the signs of $a$ and $b$ (neither can be 0). According to (3), $y$ has the same sign as $b$ - this is easily seen from (2) - and according to (4), $x$ is positive if and only if $a$ and $b$ have the same sign, and negative if they have opposite signs. (The latter can also be stated from (1): $x>0$ if $a$ and $y$ have the same sign.)

2. Several people have pointed out that the system of equations

\[
x^{c} y^{d}=a, \quad x^{e} y^{f}=b
\]

can be solved with the same line of thought, where $a \neq 0, b \neq 0$, and $c, d, e, f$ are natural numbers such that $|c f-d e|=1$. This is correct, but it is not advisable to state it in this form in general, rather immediately for any $c f-d e=g \neq 0$ in the following form:

\[
|x|=\sqrt[g]{\left|\frac{a^{f}}{b^{d}}\right|} \quad|y|=\sqrt[g]{\left|\frac{b^{c}}{a^{e}}\right|}
\]

and - in the case of even $g$ - the signs must be determined separately.

3. Those who are already familiar with the concept of logarithms and their identities can also solve the system of equations using them. Taking the logarithm of (1) and (2), we arrive at the system of equations

\[
\begin{aligned}
5 \lg x+17 \lg y & =\lg a \\
2 \lg x+7 \lg y & =\lg b
\end{aligned}
\]

which is a first-degree system of equations for $\lg x$ and $\lg y$. However, we must now assume that both $a$ and $b$ are positive numbers, from which it follows that $x$ and $y$ are also positive.

(If either $a$ or $b$ is negative, then the absolute value of $a$ and $b$ can be used to calculate the absolute value of $x$ and $y$, and their signs can be determined by the above considerations.)