Solve :
` 3( 2u +v) = 7uv `
` 3(u + 3v) = 11 uv `

Answer

To solve the system of equations given by:

1. \( 3(2u + v) = 7uv \)
2. \( 3(u + 3v) = 11uv \)

we will follow these steps:

### Step 1: Expand both equations

**Equation 1:**
\[ 3(2u + v) = 7uv \implies 6u + 3v = 7uv \]

**Equation 2:**
\[ 3(u + 3v) = 11uv \implies 3u + 9v = 11uv \]

### Step 2: Rearrange both equations

Rearranging both equations to isolate terms involving \(u\) and \(v\):

**From Equation 1:**
\[ 6u + 3v - 7uv = 0 \quad \text{(Equation 1)} \]

**From Equation 2:**
\[ 3u + 9v - 11uv = 0 \quad \text{(Equation 2)} \]

### Step 3: Express one variable in terms of the other

From Equation 1, we can express \(v\) in terms of \(u\):
\[ 3v = 7uv - 6u \implies v = \frac{7uv - 6u}{3} \]

From Equation 2, we can express \(v\) in terms of \(u\):
\[ 9v = 11uv - 3u \implies v = \frac{11uv - 3u}{9} \]

### Step 4: Set the two expressions for \(v\) equal to each other

Equating the two expressions for \(v\):
\[ \frac{7uv - 6u}{3} = \frac{11uv - 3u}{9} \]

### Step 5: Cross-multiply to eliminate the fractions

Cross-multiplying gives:
\[ 9(7uv - 6u) = 3(11uv - 3u) \]

### Step 6: Expand and simplify

Expanding both sides:
\[ 63uv - 54u = 33uv - 9u \]

Rearranging gives:
\[ 63uv - 33uv = 54u - 9u \]
\[ 30uv = 45u \]

### Step 7: Factor out common terms

Factoring out \(u\):
\[ u(30v - 45) = 0 \]

### Step 8: Solve for \(u\) and \(v\)

This gives us two cases:
1. \(u = 0\) (not valid since we are looking for non-zero solutions)
2. \(30v - 45 = 0 \implies v = \frac{45}{30} = \frac{3}{2}\)

### Step 9: Substitute \(v\) back to find \(u\)

Substituting \(v = \frac{3}{2}\) into one of the original equations to find \(u\):
Using \(3(2u + v) = 7uv\):
\[ 3(2u + \frac{3}{2}) = 7u \cdot \frac{3}{2} \]
\[ 6u + \frac{9}{2} = \frac{21u}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 12u + 9 = 21u \]
Rearranging gives:
\[ 21u - 12u = 9 \implies 9u = 9 \implies u = 1 \]

### Final Solution

Thus, the solution is:
\[ u = 1, \quad v = \frac{3}{2} \]

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