Express the value of the ratio $(13 x+16 y):(2 x+5 y)$ in terms of $m$ and $n$, given that $(5 x+7 y):(3 x+2 y)=m: n$.
I. Solution: According to the assumption,

$$
(5 x+7 y): m=(3 x+2 y): n
$$

is also a correct pair of ratios. Denoting the common value of these two ratios by $k$,

$$
5 x+7 y=k m, \quad 3 x+2 y=k n .
$$

This is a system of equations for determining $x, y$, and its solution is:

$$
x=k(7 n-2 m) / 11, \quad y=k(3 m-5 n) / 11
$$

Substituting these into the desired ratio,

$$
\begin{gathered}
\frac{13 x+16 y}{2 x+5 y}=\frac{11(13 x+16 y)}{11(2 x+5 y)}=\frac{13 k(7 n-2 m)+16 k(3 m-5 n)}{2 k(7 n-2 m)+5 k(3 m-5 n)}= \\
=\frac{22 m+11 n}{11 m-11 n}=\frac{2 m+n}{m-n}
\end{gathered}
$$

Wáczek Zsuzsa (Budapest, Szilágyi E. lg. II. o. t.)

Remarks. 1. Essentially, we solve the problem this way even if we assume $y \neq 0$ and divide the first two terms of the given ratio pair by $y$, solve the resulting first-degree equation for the ratio $z=x / y$, and substitute $z$ into the ratio obtained from the question by simplifying with $y$.

And this is what we do even if we express $y$ from the given ratio in terms of $m, n$, and $x$, substitute the resulting expression into the desired ratio, and verify that the requirement has been met, as $x$ cancels out.

Molnár Katalin (Körmend, Kölcsey F. g. II. o. t.)

2. The elimination of $x$ and $k$ above hinges on the fact that - and also that the ratio $x / y$ can be expressed only in terms of $m$ and $n$ - in the ratios containing $x$ and $y$, there is no term that is independent of both $x$ and $y$. This is typically expressed by saying that the expressions $5 x+7 y, 3 x+2 y, 13 x+16 y$, and $2 x+5 y$ are all of the same degree in $x$ and $y$, "homogeneous".
3. Of course, $n$ cannot be $0$, because then the known ratio $m: n$ would be meaningless; similarly, the second term of the given ratio pair: $3 x+2 y$ cannot be $0$. Therefore, it cannot be that the values of $x$ and $y$ are both $0$ at the same time, so the desired ratio always makes sense.

II. Solution: Find numbers $\lambda, \mu$ such that for every pair of numbers $x, y$, the following holds: $\lambda(5 x+7 y)+\mu(3 x+2 y)=$ $13 x+16 y$ (except, of course, the pair $x=y=0$, for which the ratio has no meaning). Our requirement is satisfied if the coefficients of $x$ and $y$ on both sides are equal:

$$
\begin{aligned}
& 5 \lambda+3 \mu=13 \\
& 7 \lambda+2 \mu=16
\end{aligned}
$$

From this system of equations, $\lambda=2$ and $\mu=1$. - Similarly, requiring that $2 x+5 y=\nu(5 x+7 y)+\xi(3 x+2 y)$ be an identity, we get: $\nu=1, \xi=-1$.

From these and the assumption (1),

$$
\begin{aligned}
13 x+16 y & =\lambda k m+\mu k n=k(2 m+n) \\
2 x+5 y & =\nu k m+\xi k n=k(m-n)
\end{aligned}
$$

therefore,

$$
(13 x+16 y):(2 x+5 y)=(2 m+n):(m-n)
$$