Solve the following simultaneous equations: $ \dfrac{10}{x+y}+\dfrac{2}{x-y}=4 $ and $ \dfrac{5}{x+y}-\dfrac{5}{3\left( x-y \right)}=\dfrac{-2}{3} $ .

Hint:We will assume $ \dfrac{1}{x+y}=A $ and $ \dfrac{1}{x-y}=B $ to make the given equation easy to solve, then we will solve the obtained equation $ 10A+2B=4 $ and $ 15A-5B+2=0 $ to get the values of A and B respectively. Now, from A and B, we will calculate the value of x and y.

Complete step-by-step answer:
It is given in the question that to solve the system of equation $ \dfrac{10}{x+y}+\dfrac{2}{x-y}=4 $ and $ \dfrac{5}{x+y}-\dfrac{5}{3\left( x-y \right)}=\dfrac{-2}{3} $ , we will first assume $ \dfrac{1}{x+y}=A $ and $ \dfrac{1}{x-y}=B $ in both the equations.
Now, using these assumptions, our equation gets modified as follows $ 10A+2B=4 $ and $ 15A-5B+2=0 $ .
We multiply equation $ 10A+2B=4 $ with 3 and equation $ 15A-5B+2=0 $ with 2 and then, subtracting $ 3\left( 10A+2B=4 \right) $ from equation $ 2\left( 15A-5B+2=0 \right) $ , we get $ \left( 30A-10B \right)-\left( 30A+6B \right)=-4-12 $
Simplifying further, we get,
  $ -16B=-16 $
  $ B=1 $
Now, putting the value of $ B=1 $ in equation $ 10A+2B=4 $ , we get
  $ 10A+2\left( 1 \right)=4 $
Solving further, we get,
  $ 10A=4-2=2 $ ,
Hence we get the value of A as
  $ A=\dfrac{2}{10}=\dfrac{1}{5} $
Therefore, we have $ A=\dfrac{1}{5} $ and $ B=1 $ . We have assumed that $ \dfrac{1}{x+y}=A $ and $ \dfrac{1}{x-y}=B $ , therefore, we get \[~\dfrac{1}{x+y}=\dfrac{1}{5}\] and $ \dfrac{1}{x-y}=1 $ .
On cross-multiplying both the equations, we get
  $ x+y=5 $ and $ x-y=1 $
Adding both of the equations, we get,
  $ \left( x+y \right)+\left( x-y \right)=1+5=6 $
  $ 2x=6 $
  $ x=\dfrac{6}{2}=3 $
Now putting the value of $ x=3 $ in equation $ x+y=5 $ , we get value of y as
  $ 3+y=5 $
Solving further, we get,
  $ y=5-3=2 $
Thus, the value of $ x=3 $ and the value $ y=2 $ for the given set of simultaneous equations.

Note: Many times students are stuck initially by seeing the equation in fraction. As a result most of the students leave such questions in examination, thinking that these are time consuming questions. But, we see that with suitable assumptions, we make the equations look simple and easy to solve. It is observed that many students leave the solution incomplete by finding the value of only A and B, but it should be kept in mind that we have assumed these variables, and our final answer could not be in these variables. Thus, we have to calculate the solution in terms of the given variables, that is, x and y.