Let's solve the given system of equations step by step. The system of equations is:

1. \(\frac{1}{x-2} + \frac{1}{y-1} = 2\)
2. \(\frac{2}{x-2} - \frac{3}{y-1} = 1\)

First, let's introduce substitutions to simplify the equations. Let:
\[ u = \frac{1}{x-2} \]
\[ v = \frac{1}{y-1} \]

Now, the system of equations becomes:

1. \( u + v = 2 \)
2. \( 2u - 3v = 1 \)

We can solve this system of linear equations using substitution or elimination. Let's use the substitution method.

From the first equation:
\[ u + v = 2 \]
\[ v = 2 - u \]

Substitute \( v = 2 - u \) into the second equation:
\[ 2u - 3(2 - u) = 1 \]
\[ 2u - 6 + 3u = 1 \]
\[ 5u - 6 = 1 \]
\[ 5u = 7 \]
\[ u = \frac{7}{5} \]

Now substitute \( u = \frac{7}{5} \) back into \( v = 2 - u \):
\[ v = 2 - \frac{7}{5} \]
\[ v = \frac{10}{5} - \frac{7}{5} \]
\[ v = \frac{3}{5} \]

Now we have:
\[ u = \frac{7}{5} \]
\[ v = \frac{3}{5} \]

Recall the substitutions we made:
\[ u = \frac{1}{x-2} \]
\[ v = \frac{1}{y-1} \]

So:
\[ \frac{1}{x-2} = \frac{7}{5} \]
\[ x - 2 = \frac{5}{7} \]
\[ x = 2 + \frac{5}{7} \]
\[ x = \frac{14}{7} + \frac{5}{7} \]
\[ x = \frac{19}{7} \]

And:
\[ \frac{1}{y-1} = \frac{3}{5} \]
\[ y - 1 = \frac{5}{3} \]
\[ y = 1 + \frac{5}{3} \]
\[ y = \frac{3}{3} + \frac{5}{3} \]
\[ y = \frac{8}{3} \]

So the solution to the system of equations is:
\[ x = \frac{19}{7} \]
\[ y = \frac{8}{3} \]