Solve for the given equations:
$3\left( 2u+v \right)=7uv$ and
$3\left( u+3v \right)=11uv$

Hint: Divide both the given equations to get a linear equation in two variables u and v. Now, get the value of u or v in terms of b or u respectively. Now, substitute it in equation (i) or (ii) to get the value of u or v. Hence, get another variable using the equation received on division of both the given equation.

Complete step-by-step answer:
Given equations in the problem are
$\begin{align}
  & 3\left( 2u+v \right)=7uv...............\left( i \right) \\
 & 3\left( u+3v \right)=11uv................\left( ii \right) \\
\end{align}$
On dividing both the equation (i) and (ii), we can get an equation as
\[\begin{align}
  & \dfrac{3\left( 2u+v \right)}{3\left( u+3v \right)}=\dfrac{7uv}{11uv} \\
 & \Rightarrow \dfrac{2u+v}{u+3v}=\dfrac{7}{11} \\
\end{align}\]
On cross-multiplying the above equation, we get
\[\begin{align}
  & \Rightarrow 11\left( 2u+v \right)=7\left( u+3v \right) \\
 & \Rightarrow 22u+11v=7u+21v \\
 & \Rightarrow 22u-7u=21v-11v \\
 & \Rightarrow 15u=10v \\
\end{align}\]
On dividing the above equation by 15, we get
$\Rightarrow u=\dfrac{10}{15}v=\dfrac{2}{3}v...............\left( iii \right)$
Now, we can put the value of u in terms of v from the equation (iii) to any of the equation (i) or (ii) to get a relation in terms of ‘v’ only. So, put $u=\dfrac{2}{3}v$ in equation (i) as
$\begin{align}
  & 3\left( 2\times \dfrac{2}{3}v+v \right)=7\times \dfrac{2}{3}v\times v \\
 & \Rightarrow 3\left( \dfrac{4}{3}v+\dfrac{v}{1} \right)=\dfrac{14}{3}{{v}^{2}} \\
\end{align}$
On taking L.C.M of denominators of fractions written inside brackets of L.H.S. we get
\[\begin{align}
  & \Rightarrow 3\left( \dfrac{4v+3v}{3} \right)=\dfrac{14}{3}{{v}^{2}} \\
 & \Rightarrow 3\left( \dfrac{7v}{3} \right)=\dfrac{14}{3}{{v}^{2}} \\
 & \Rightarrow 7v=\dfrac{14}{3}{{v}^{2}} \\
\end{align}\]
On dividing the whole equation by $7v$ , we get
$\begin{align}
  & \Rightarrow 1=\dfrac{2v}{3} \\
 & \Rightarrow v=\dfrac{3}{2} \\
\end{align}$
Now, we can put the value of u by putting $v=\dfrac{3}{2}$ to the equation (iii). So, we get
$\Rightarrow u=\dfrac{2}{3}\times \dfrac{3}{2}=1$
Hence, values of u and v on solving both the given equations in the problem are
$u=1,v=\dfrac{3}{2}$
Hence, $u=1,v=\dfrac{3}{2}$ is the answer.

Note: Another approach would be that we can eliminate u or v from one equation by using another equation. As an example, value of u from the first equation is given as
\[\begin{align}
  & \Rightarrow 6u+3v=7uv \\
 & \Rightarrow u\left( 6-7v \right)=-3v \\
 & \Rightarrow u=\dfrac{-3v}{6-7v} \\
\end{align}\]
Now, put the value of u to the second equation to get the required answers. It is a little complex approach and would be longer than given in the solution. But answers will be same from both the approaches
Another approach would be that we can rewrite the given equations as
$3\left( 2u+v \right)=7uv$ and $3\left( u+3v \right)=11uv$
Divide them with uv on both sides. We get $3\left( \dfrac{2}{v}+\dfrac{1}{u} \right)=7$ and $3\left( \dfrac{1}{v}+\dfrac{3}{u} \right)=11$
Now, put $\dfrac{1}{u}=x$ and $\dfrac{1}{v}=y$ and solve two simple linear equations in two variables and use $\dfrac{1}{u}=x$ and $\dfrac{1}{v}=y$ to get the required answer.
It means there can be several approaches to solve a question. We need to observe the question and decide the way of solution, so that it will not be complex and flexible as well.