Question

Solve the following simultaneous equations :
$\left(\mathrm{v}\right)\frac{16}{x+y}+\frac{2}{x-y}=1;\frac{8}{x+y}-\frac{12}{x-y}=7$

Answer 1

$$\begin{gathered} \frac{16}{x+y}+\frac{2}{x-y}=1-------\left(1\right) \\ \frac{8}{x+y}-\frac{12}{x-y}=7-------\left(2\right) \\ \mathrm{Substituting}\frac{1}{x+y}=m\mathrm{and}\frac{1}{x-y}=n\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equations},\mathrm{we}\mathrm{get}: \\ 16m+2n=1--------\left(3\right) \\ 8m-12n=7--------\left(4\right) \\ \mathrm{Multiplying}\mathrm{equation}\left(3\right)\mathrm{by}6,\mathrm{we}\mathrm{get}: \\ 96m+12n=6-------\left(5\right) \\ \mathrm{Adding}\mathrm{equations}\left(4\right)\mathrm{and}\left(5\right),\mathrm{we}\mathrm{get}: \\ 96m+12n=6 \\ \underline{8m-12n=7} \\ 104m=13 \\ \Rightarrow m=\frac{13}{104}=\frac{1}{8} \\ \mathrm{Substituting}m=\frac{1}{8}\mathrm{in}\mathrm{equation}\left(3\right),\mathrm{we}\mathrm{get}: \\ n\mathit{}=-\frac{1}{2} \\ \mathrm{Replacing}\mathrm{the}\mathrm{values}\mathrm{of}m\mathit{}\mathrm{and}n,\mathit{}\mathrm{we}\mathrm{get}: \\ \frac{1}{x+y}\mathit{}=\frac{1}{8}\mathrm{and}\frac{1}{x-y}=-\frac{1}{2} \\ \therefore x+y=8\mathrm{and}x-y=-2 \\ \mathrm{Adding}\mathrm{equations}\left(4\right)\mathrm{and}\left(5\right),\mathrm{we}\mathrm{get}: \\ 2x=6 \\ \therefore x=3\mathrm{and}y=8-3=5 \end{gathered}$$