\[ \begin{cases} \frac{3x + 4}{x - 1} + \frac{2y - 3}{y + 2} = 5 \\ \frac{2x - 1}{x - 1} - \frac{4y - 3}{y + 2} = 10 \end{cases} \]

To solve the system of equations:

\[
\begin{cases}
\frac{3x + 4}{x - 1} + \frac{2y - 3}{y + 2} = 5 \quad (1) \\
\frac{2x - 1}{x - 1} - \frac{4y - 3}{y + 2} = 10 \quad (2)
\end{cases}
\]

we'll proceed step-by-step.

### Equation (1):
\[
\frac{3x + 4}{x - 1} + \frac{2y - 3}{y + 2} = 5
\]
Multiply through by \((x - 1)(y + 2)\) to eliminate the fractions:
\[
(3x + 4)(y + 2) + (2y - 3)(x - 1) = 5(x - 1)(y + 2)
\]

Expanding:
1. Left side:
- \(3xy + 6x + 4y + 8 + 2xy - 3x - 2y + 3\)
- Combine like terms: \((3xy + 2xy) + (6x - 3x) + (4y - 2y) + 8 + 3 = 5xy + 3x + 2y + 11\)

2. Right side:
- \(5(xy + 2x - y - 2) = 5xy + 10x - 5y - 10\)

Setting them equal:
\[
5xy + 3x + 2y + 11 = 5xy + 10x - 5y - 10
\]
After canceling \(5xy\) from both sides:
\[
3x + 2y + 11 = 10x - 5y - 10
\]
Rearranging gives:
\[
3x - 10x + 2y + 5y + 11 + 10 = 0 \implies -7x + 7y + 21 = 0
\]
Dividing by -7:
\[
x - y - 3 = 0 \implies y = x + 3 \quad (3)
\]

### Equation (2):
Substituting \(y = x + 3\) into equation (2):
\[
\frac{2x - 1}{x - 1} - \frac{4(x + 3) - 3}{(x + 3) + 2} = 10
\]
This simplifies to:
\[
\frac{2x - 1}{x - 1} - \frac{4x + 12 - 3}{x + 5} = 10
\]
Further simplification gives:
\[
\frac{2x - 1}{x - 1} - \frac{4x + 9}{x + 5} = 10
\]
Multiply through by \((x - 1)(x + 5)\):
\[
(2x - 1)(x + 5) - (4x + 9)(x - 1) = 10(x - 1)(x + 5)
\]

Expanding each term:
1. Left side:
- From \((2x - 1)(x + 5)\): \(2x^2 + 10x - x - 5 = 2x^2 + 9x - 5\)
- From \((4x + 9)(x - 1)\): \(4x^2 - 4x + 9x - 9 = 4x^2 + 5x - 9\)
- Combine: \((2x^2 + 9x - 5) - (4x^2 + 5x - 9) = -2x^2 + 4x + 4\)

2. Right side:
- \(10(x^2 + 4x - 5) = 10x^2 + 40x - 50\)

Setting left to right:
\[
-2x^2 + 4x + 4 = 10x^2 + 40x - 50
\]
Rearranging gives a quadratic:
\[
-12x^2 - 36x + 54 = 0 \implies 2x^2 + 6x - 9 = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 2 \cdot (-9)}}{2 \cdot 2}
\]
Calculating gives:
\[
x = \frac{-6 \pm \sqrt{36 + 72}}{4} = \frac{-6 \pm \sqrt{108}}{4} = \frac{-6 \pm 6\sqrt{3}}{4} = \frac{-3 \pm 3\sqrt{3}}{2}
\]
The two possible solutions for \(x\) are:
\[
x_1 = \frac{-3 + 3\sqrt{3}}{2}, \quad x_2 = \frac{-3 - 3\sqrt{3}}{2}
\]

Using \(y = x + 3\) gives:
\[
y_1 = \frac{-3 + 3\sqrt{3}}{2} + 3 = \frac{3 + 3\sqrt{3}}{2}, \quad y_2 = \frac{-3 - 3\sqrt{3}}{2} + 3 = \frac{3 - 3\sqrt{3}}{2}
\]

### Final Solutions:
\[
\left( \frac{-3 + 3\sqrt{3}}{2}, \frac{3 + 3\sqrt{3}}{2} \right), \quad \left( \frac{-3 - 3\sqrt{3}}{2}, \frac{3 - 3\sqrt{3}}{2} \right)
\]
These are the solutions for the equations.