`{:((1)/(2(x + 2y)) + (5)/(3(3x - 2y))=(-3)/(2)),((5)/(4(x + 2y)) - (3)/(5(3x - 2y)) = (61)/(60)):}`

Answer

To solve the given system of equations:

1. \(\frac{1}{2(x + 2y)} + \frac{5}{3(3x - 2y)} = -\frac{3}{2}\)
2. \(\frac{5}{4(x + 2y)} - \frac{3}{5(3x - 2y)} = \frac{61}{60}\)

we will use substitution to simplify the equations.

### Step 1: Substitute Variables
Let:
- \( a = \frac{1}{x + 2y} \)
- \( b = \frac{1}{3x - 2y} \)

Now, we can rewrite the equations as:
1. \(\frac{a}{2} + \frac{5b}{3} = -\frac{3}{2}\)
2. \(\frac{5a}{4} - \frac{3b}{5} = \frac{61}{60}\)

### Step 2: Clear the Fractions
Multiply the first equation by 6 (the least common multiple of the denominators 2 and 3):
\[ 6 \left(\frac{a}{2}\right) + 6 \left(\frac{5b}{3}\right) = 6 \left(-\frac{3}{2}\right) \]
This simplifies to:
\[ 3a + 10b = -9 \quad \text{(Equation 1)} \]

Now, multiply the second equation by 20 (the least common multiple of the denominators 4 and 5):
\[ 20 \left(\frac{5a}{4}\right) - 20 \left(\frac{3b}{5}\right) = 20 \left(\frac{61}{60}\right) \]
This simplifies to:
\[ 25a - 12b = \frac{61 \times 20}{60} = \frac{1220}{60} = \frac{61}{3} \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have the system:
1. \(3a + 10b = -9\)
2. \(25a - 12b = \frac{61}{3}\)

To eliminate \(b\), we can multiply the first equation by 12:
\[ 12(3a + 10b) = 12(-9) \]
This gives:
\[ 36a + 120b = -108 \quad \text{(Equation 3)} \]

Now, multiply the second equation by 10:
\[ 10(25a - 12b) = 10\left(\frac{61}{3}\right) \]
This gives:
\[ 250a - 120b = \frac{610}{3} \quad \text{(Equation 4)} \]

### Step 4: Add Equations to Eliminate \(b\)
Now, add Equation 3 and Equation 4:
\[ (36a + 120b) + (250a - 120b) = -108 + \frac{610}{3} \]
This simplifies to:
\[ 286a = -108 + \frac{610}{3} \]
Convert -108 to a fraction:
\[ -108 = -\frac{324}{3} \]
So:
\[ 286a = -\frac{324}{3} + \frac{610}{3} = \frac{286}{3} \]
Dividing both sides by 286:
\[ a = \frac{1}{3} \]

### Step 5: Substitute \(a\) Back to Find \(b\)
Substituting \(a = \frac{1}{3}\) back into Equation 1:
\[ 3\left(\frac{1}{3}\right) + 10b = -9 \]
This simplifies to:
\[ 1 + 10b = -9 \]
\[ 10b = -10 \quad \Rightarrow \quad b = -1 \]

### Step 6: Substitute \(a\) and \(b\) Back to Original Variables
Recall:
\[ a = \frac{1}{x + 2y} \quad \Rightarrow \quad x + 2y = 3 \]
\[ b = \frac{1}{3x - 2y} \quad \Rightarrow \quad 3x - 2y = -1 \]

### Step 7: Solve the System of Linear Equations
Now we have:
1. \(x + 2y = 3\)
2. \(3x - 2y = -1\)

Adding these two equations:
\[ x + 2y + 3x - 2y = 3 - 1 \]
This simplifies to:
\[ 4x = 2 \quad \Rightarrow \quad x = \frac{1}{2} \]

Substituting \(x = \frac{1}{2}\) into \(x + 2y = 3\):
\[ \frac{1}{2} + 2y = 3 \]
\[ 2y = 3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2} \]
\[ y = \frac{5}{4} \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = \frac{1}{2}, \quad y = \frac{5}{4} \]