If $-3 a^{m-1} b^{m+n}$ and $5 a^{n-2 m} b^{3 m+n-4}$ are similar terms, then the solution of the equation set $\left\{\begin{array}{l}m x+(n-2) y=24 \\ 2 m x+n y=46\end{array}\right.$ $\qquad$.
（English-Chinese Mini Dictionary: similar terms 同类项; equation set 方程组）
Translation: If $-3 a^{m-1} b^{m+n}$ and $5 a^{n-2 m} b^{3 m+n-4}$ are similar terms, then the solution of the equation set $\left\{\begin{array}{l}m x+(n-2) y=24 \\ 2 m x+n y=46\end{array}\right.$ is $\qquad$.
Solution: According to the problem, we have
$$
\left\{\begin{array}{l}
m-1=n-2 m \\
m+n=3 m+n-4
\end{array},\right.
$$

Solving, we get
$$
\left\{\begin{array}{l}
m=2 \\
n=5
\end{array} .\right.
$$

Substituting into the system of equations $\left\{\begin{array}{l}m x+(n-2) y=24 \\ 2 m x+n y=46\end{array}\right.$, we get
$$
\left\{\begin{array}{l}
2 x+3 y=24 \\
4 x+5 y=46
\end{array}\right. \text {, }
$$

Solving, we get
$$
\left\{\begin{array}{l}
x=9 \\
y=2
\end{array} .\right.
$$