Question

Solve for x and y:
$$\begin{gathered} \frac{10}{x+y}+\frac{2}{x-y}=4 \\ \frac{15}{x+y}-\frac{9}{x-y}=-2 \end{gathered}$$

Answer 1

The given equations are
$$\begin{gathered} \frac{10}{x+y}+\frac{2}{x-y}=4.....\left(i\right) \\ \frac{15}{x+y}-\frac{9}{x-y}=-2.....\left(ii\right) \end{gathered}$$
Substituting $\frac{1}{x+y}=u\mathrm{and}\frac{1}{x-y}=v$ in (i) and (ii), we get
$$\begin{gathered} 10u+2v=4.....\left(iii\right) \\ 15u-9v=-2.....\left(iv\right) \end{gathered}$$
Multiplying (iii) by 9 and (iv) by 2 and adding, we get
$$\begin{gathered} 90u+30u=36-4 \\ \Rightarrow 120u=32 \\ \Rightarrow u=\frac{32}{120}=\frac{4}{15} \\ \Rightarrow x+y=\frac{15}{4}\left(\because \frac{1}{x+y}=u\right).....\left(v\right) \end{gathered}$$
Now, substituting $u=\frac{4}{15}$ in (iii), we get
$$\begin{gathered} 10\times \frac{4}{15}+2v=4 \\ \frac{8}{3}+2v=4 \\ \Rightarrow 2v=4-\frac{8}{3}=\frac{4}{3} \\ \Rightarrow v=\frac{2}{3} \\ \Rightarrow x-y=\frac{3}{2}\left(\because \frac{1}{x-y}=v\right).....\left(vi\right) \end{gathered}$$
Adding (v) and (vi), we get
$2x=\frac{15}{4}+\frac{3}{2}\Rightarrow 2x=\frac{21}{4}\Rightarrow x=\frac{21}{8}$
Substituting $x=\frac{21}{8}$ in (v), we have
$\frac{21}{8}+y=\frac{15}{4}\Rightarrow y=\frac{15}{4}-\frac{21}{8}=\frac{9}{8}$
Hence, $x=\frac{21}{8}\mathrm{and}y=\frac{9}{8}$.