Solve for x and y. 35/x+y + 14/x-y = 19 14/x+y + 35/x-y = 37

The given system of equations are;
$$\begin{gathered} \frac{35}{x+y}+\frac{14}{x-y} = 19 ...\left(\mathrm{i}\right) \\ \frac{14}{\mathrm{x}+\mathrm{y}}+\frac{35}{\mathrm{x}-\mathrm{y}} = 37 ...\left(\mathrm{ii}\right) \end{gathered}$$
Substituting $u = \frac{1}{x+y} \mathrm{and} v = \frac{1}{x-y}$ in both the above equations we get;
$$\begin{gathered} 35u+14v = 19 ...\left(\mathrm{iii}\right) \\ 14u+35v = 37 ...\left(\mathrm{iv}\right) \end{gathered}$$
Multiplying (iii) by 4 and (iv) by 10 and then subtracting (iv) from (iii) we get;
$$\begin{gathered} 140u+56v-140u-350v = 76-370 \\ \Rightarrow -294v = -294 \\ \Rightarrow v = 1 \end{gathered}$$
Putting v = 1 in (iii) we get;
$$\begin{gathered} 35u+14\times 1 = 19 \\ \Rightarrow 35u = 19-14 \\ \Rightarrow u = \frac{5}{35} = \frac{1}{7} \end{gathered}$$
Substituting back $u = \frac{1}{x+y} \mathrm{and} v = \frac{1}{x-y}$; we get;
$$\begin{gathered} \frac{1}{x+y} = \frac{1}{7} \\ \Rightarrow x+y = 7 ...\left(\mathrm{v}\right) \\ \frac{1}{x-y} = 1 \\ \Rightarrow x-y = 1 ...\left(\mathrm{vi}\right) \end{gathered}$$
Solving (v) and (vi) we get;
$$\begin{gathered} x+y+x-y = 7+1 \\ \Rightarrow 2x = 8 \\ \Rightarrow x = 4 \\ \mathrm{Then}, y = 7-4 = 3 \end{gathered}$$
Therefore x = 4 and y = 3