Solve :

3(2u + v) = 7 uv

3(u + 3v) = 11 uv

6u + 3v = 7uv

3u + 9v = 11uv

Multiply second equation by 2 and subtract from first equation, we get

(3u + 9v = 11uv) x 2

$\begin{matrix} & 6u & + & 3v & = & 7uv \ & 6u & + & 18v & = & 22uv \ & - & - & & & - \ \hline & & - & 15v & = & 7uv - 22uv\ \Rightarrow & & - & 15v & = & -15uv \end{matrix}$

⇒ u = $\dfrac{-15v}{-15v}$

⇒ u = 1

Substituting the value of u in first equation, we get:

⇒ 6 x 1 + 3v = 7 x 1 x v

⇒ 6 + 3v = 7v

⇒ 7v - 3v = 6

⇒ 4v = 6

⇒ v = $\dfrac{6}{4}$

⇒ v = $\dfrac{3}{2}$

Hence, the value of u = 1 and v = $\dfrac{3}{2}$.