$${\textbf{Step -1: Simplifying the two equation.}}$$
$${\text{Given that }}\dfrac{{20}}{{{\text{x}} + {\text{y}}}} + \dfrac{3}{{{\text{x}} - {\text{y}}}} = 7$$
$${\text{Let us consider, }}\dfrac{1}{{{\text{x + y}}}} = {\text{u and }}\dfrac{1}{{{\text{x}} - {\text{y}}}} = {\text{v}}$$
$${\text{So, we have, }}$$
$${\text{20u}} + 3{\text{v}} = 7{\text{ - - - - - - - (i)}}$$
$${\text{8v}} - 15{\text{u}} = 5$$
$$\Rightarrow - 15{\text{u}} + 8{\text{v}} = 5{\text{ - - - - - - - - (ii)}}$$
$${\textbf{Step -2: Solving the two equations}}.$$
$${\text{Solving the equation (i) and (ii),}}$$
$$\text{Multiplying (i) by 8 and (ii) by 3 and subtracting, we get}$$
$$\Rightarrow 160{\text{u}} + 24{\text{v}} + 45{\text{u}} - 24{\text{v}} = 56-15$$
$$\Rightarrow 205{\text{u}} = 41$$
$$\Rightarrow {\text{u}} = \dfrac{41}{205}=\dfrac{1}{5}$$
$$\text{Putting value of u in (ii)}$$
$$\Rightarrow -3 + 8{\text{v}} = 5$$
$$\Rightarrow 8{\text{v}} = 8$$
$$\Rightarrow {\text{v}} = 1$$
$$\therefore {\text{u}} = \dfrac{1}{5}{\text{ and v}} = 1$$
$${\textbf{Step -3: Back substituting the value of u and v to find the value of x and y}}.$$
$${\text{As we have,}}$$
$$\dfrac{1}{{{\text{x + y}}}} = {\text{u and }}\dfrac{1}{{{\text{x}} - {\text{y}}}} = {\text{v}}$$
$${\text{So we get, }}$$
$$\dfrac{1}{{{\text{x}} + {\text{y}}}} = \dfrac{1}{5}{\text{ and }}\dfrac{1}{{{\text{x}} - {\text{y}}}} = 1$$
$$ \Rightarrow {\text{x}} + {\text{y}} = 5{\text{ and x}} - {\text{y}} = 1$$
$${\text{adding these equations, we get }}$$
$$\Rightarrow 2{\text{x}} = 6$$
$$\Rightarrow {\text{x}} = 3$$
$${\text{Substituting the value of x in one of the above equation, we get,}}$$
$${\text{x}} + {\text{y}} = 5$$
$$\Rightarrow 3 + {\text{y}} = 5$$
$$\Rightarrow {\text{y}} = 5 - 3$$
$$\Rightarrow {\text{y}} = 2$$
$${\textbf{Hence,The solution of }}\mathbf{\dfrac{{20}}{{{\textbf{x}} + {\textbf{y}}}} + \dfrac{3}{{{\textbf{x}} - {\textbf{y}}}} = 7{\textbf{ and }}\dfrac{8}{{{\textbf{x}} - {\textbf{y}}}} - \dfrac{{15}}{{{\textbf{x}} + {\textbf{y}}}} = 5}{\textbf{ is x}} = \mathbf{3}{\textbf{ and y}} = \mathbf{2}.$$