`{:(2x + y = (7xy)/(3)),(x + 3y = (11xy)/(3)):}`

Answer

To solve the given system of equations:

1. **Equations Given:**
\[ 2x + y = \frac{7xy}{3} \quad \text{(1)} \]
\[ x + 3y = \frac{11xy}{3} \quad \text{(2)} \]

2. **Eliminate the fractions:**
Multiply both sides of equation (1) by 3:
\[ 3(2x + y) = 7xy \]
This simplifies to:
\[ 6x + 3y = 7xy \quad \text{(3)} \]

Multiply both sides of equation (2) by 3:
\[ 3(x + 3y) = 11xy \]
This simplifies to:
\[ 3x + 9y = 11xy \quad \text{(4)} \]

3. **Rearranging equations (3) and (4):**
Rearranging equation (3):
\[ 6x + 3y - 7xy = 0 \quad \text{(5)} \]
Rearranging equation (4):
\[ 3x + 9y - 11xy = 0 \quad \text{(6)} \]

4. **Dividing by xy:**
Divide equation (5) by xy:
\[ \frac{6}{y} + \frac{3}{x} = 7 \quad \text{(7)} \]
Divide equation (6) by xy:
\[ \frac{3}{y} + \frac{9}{x} = 11 \quad \text{(8)} \]

5. **Substituting variables:**
Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \). Then, equations (7) and (8) become:
\[ 6b + 3a = 7 \quad \text{(9)} \]
\[ 9a + 3b = 11 \quad \text{(10)} \]

6. **Multiplying equation (9) by 2:**
Multiply equation (9) by 2 to align coefficients of \( b \):
\[ 12b + 6a = 14 \quad \text{(11)} \]

7. **Subtracting equations (10) from (11):**
Subtract equation (10) from equation (11):
\[ (12b + 6a) - (9a + 3b) = 14 - 11 \]
This simplifies to:
\[ 3b - 3a = 3 \]
Thus:
\[ b - a = 1 \quad \text{(12)} \]

8. **Substituting back into equation (9):**
From equation (12), we can express \( b \) in terms of \( a \):
\[ b = a + 1 \]
Substitute \( b \) into equation (9):
\[ 6(a + 1) + 3a = 7 \]
Simplifying gives:
\[ 6a + 6 + 3a = 7 \]
\[ 9a + 6 = 7 \]
\[ 9a = 1 \]
\[ a = \frac{1}{9} \]

9. **Finding \( b \):**
Substitute \( a \) back into equation (12):
\[ b = \frac{1}{9} + 1 = \frac{10}{9} \]

10. **Finding \( x \) and \( y \):**
Recall \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \):
\[ \frac{1}{x} = \frac{1}{9} \implies x = 9 \]
\[ \frac{1}{y} = \frac{10}{9} \implies y = \frac{9}{10} \]

Thus, the solution to the equations is:
\[ x = 9, \quad y = \frac{9}{10} \]

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