Solve: ` 3 (2x+ y ) = 7xy `
` 3(x+ 3y ) = 11xy , x ne 0 , y ne 0 `

Answer

To solve the equations:

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

2. **Rearranging Equation 1:**
\[ 3(2x + y) - 7xy = 0 \]
This can be rewritten as:
\[ 6x + 3y - 7xy = 0 \quad \text{(Equation 3)} \]

3. **Rearranging Equation 2:**
\[ 3(x + 3y) - 11xy = 0 \]
This can be rewritten as:
\[ 3x + 9y - 11xy = 0 \quad \text{(Equation 4)} \]

4. **Equating Right-Hand Sides of Equations 3 and 4:**
Since both equations equal zero, we can set the left-hand sides equal to each other:
\[ 6x + 3y = 7xy \quad \text{and} \quad 3x + 9y = 11xy \]

5. **Cross-Multiplying:**
From the equations:
\[ \frac{6x + 3y}{7} = xy \quad \text{and} \quad \frac{3x + 9y}{11} = xy \]
We can set them equal:
\[ \frac{6x + 3y}{7} = \frac{3x + 9y}{11} \]

6. **Cross-Multiplying Again:**
\[ 11(6x + 3y) = 7(3x + 9y) \]
Expanding both sides:
\[ 66x + 33y = 21x + 63y \]

7. **Rearranging:**
\[ 66x - 21x = 63y - 33y \]
Simplifying gives:
\[ 45x = 30y \]

8. **Finding Relationship Between x and y:**
Dividing both sides by 15:
\[ 3x = 2y \quad \Rightarrow \quad x = \frac{2}{3}y \quad \text{(Equation 5)} \]

9. **Substituting x in Equation 1:**
Substitute \( x = \frac{2}{3}y \) into Equation 1:
\[ 3(2(\frac{2}{3}y) + y) = 7(\frac{2}{3}y)y \]
Simplifying:
\[ 3(\frac{4}{3}y + y) = \frac{14}{3}y^2 \]
\[ 3(\frac{7}{3}y) = \frac{14}{3}y^2 \]
\[ 7y = \frac{14}{3}y^2 \]

10. **Rearranging to Form a Quadratic Equation:**
\[ 14y^2 - 21y = 0 \]
Factoring out \( y \):
\[ y(14y - 21) = 0 \]

11. **Finding Values of y:**
Since \( y \neq 0 \):
\[ 14y - 21 = 0 \quad \Rightarrow \quad 14y = 21 \quad \Rightarrow \quad y = \frac{21}{14} = \frac{3}{2} \]

12. **Finding Value of x:**
Substitute \( y = \frac{3}{2} \) back into Equation 5:
\[ x = \frac{2}{3}(\frac{3}{2}) = 1 \]

**Final Values:**
\[ x = 1, \quad y = \frac{3}{2} \]

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