Solve the following equations using elimination method.
$3\left( 2x+y\right) =7xy$
$3\left( x+3y\right) =11xy$

Hint: In this question it is given we have to solve $3\left( 2x+y\right) =7xy$; $3\left( x+3y\right) =11xy$ by using elimination method. So to find the solution we first need to transform the given equation into linear equations of two variables and after that we are going to apply elimination method, i.e, if $$a_{1}x+b_{1}y+c_{1}=0$$ and $$a_{2}x+b_{2}y+c_{2}=0$$ be two linear equations then to eliminate ‘x’ we have to first equate the coefficient of ‘x’ in both equations and then we have to subtract them, in this way we will get third equation by eliminating ‘x' , and from that third equation we can able to find the value of ‘y’.

Complete step-by-step solution:
Given equations,
$3\left( 2x+y\right) =7xy$..............(1)
$3\left( x+3y\right) =11xy$.............(2)
Now dividing both side of the equation(1) and (2) by ‘xy’, we get,
Equation (1) implies,
$$\dfrac{3\left( 2x+y\right) }{xy} =\dfrac{7xy}{xy}$$
$$\Rightarrow 3\left( \dfrac{2}{y} +\dfrac{1}{x} \right) =7$$............(3)
Equation (2) implies,
$$\dfrac{3\left( x+3y\right) }{xy} =\dfrac{11xy}{xy}$$
$$\Rightarrow 3\left( \dfrac{1}{y} +\dfrac{3}{x} \right) =11$$..........(4)
Now let, $$\dfrac{1}{x} =p$$ $$\dfrac{1}{y} =q$$,
Therefore, equation (3) and (4) can be written as,
$$3\left( 2q+p\right) =7$$
$$\Rightarrow 3p+6q=7$$.........(5)
And,
$$3\left( q+3p\right) =11$$
$$\Rightarrow 9p+3q=11$$.........(6)
 Now we are going to eliminate ‘p’, so for this we have to equate the coefficient of ‘p’, so for equating we have to take LCM(Least Common Multiple) of their coefficients.
Therefore, LCM of 3 and 9 is 9.
Now multiplying equation(5) with 3 and equation (6) with 1 , we get,
$3\left( 3p+6q\right) =3\times 7$=$9p+18q=21$
$1\left( 9p+3q\right) =1\times 11$=$9p+3q=11$
Now by subtracting,
$$\left( 9p+18q\right) -\left( 9p+3q\right) =21-11$$
$$\Rightarrow 9p+18q-9p-3q=10$$
$$\Rightarrow 15q=10$$
$$\Rightarrow q=\dfrac{10}{15}$$
$$\Rightarrow q=\dfrac{2}{3}$$

Therefore,
$$\Rightarrow \dfrac{1}{y} =\dfrac{2}{3}$$ [ since, $$\dfrac{1}{y} =q$$]
$$\Rightarrow y=\dfrac{3}{2}$$
Now putting the value of ‘q’ in equation (5) we get,
$$3p+6q=7$$
$$\Rightarrow 3p+6\times \dfrac{2}{3} =7$$
$$\Rightarrow 3p+4=7$$
$$\Rightarrow 3p=7-4$$
$$\Rightarrow 3p=3$$
$$\Rightarrow p=1$$
$$\Rightarrow \dfrac{1}{x} =1$$
$$\Rightarrow x=1$$
Hence the solution is,
$$x=1$$
$$y=\dfrac{3}{2}$$

Note: While applying the elimination method you need to know that the equations should be in the linear form and if they are not linear, then you have to make it by using substitution, just like we did in the during solution. Also in the solution part we initially eliminate the first variable ‘p’, but you can eliminate ‘q’ also by equating their coefficient.