Question

Solve the following simultaneous equations :
$\left(\mathrm{IV}\right)\frac{27}{x-2}+\frac{31}{y+3}=85;\frac{31}{x-2}+\frac{27}{y+3}=89$

Answer 1

$$\begin{gathered} \frac{27}{x-2}+\frac{31}{y+3}=85-------\left(1\right) \\ \frac{31}{x-2}+\frac{27}{y+3}=89-------\left(2\right) \\ \mathrm{Substituting}\frac{1}{x-2}=m\mathrm{and}\frac{1}{y+3}=n\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equations},\mathrm{we}\mathrm{get}: \\ 27m+31n=85--------\left(3\right) \\ 31m+27n=89--------\left(4\right) \\ \mathrm{Multiplying}\mathrm{equation}\left(3\right)\mathrm{by}27\mathrm{and}\mathrm{equation}\left(4\right)\mathrm{by}31,\mathrm{we}\mathrm{get}: \\ 729m+837n=2295-------\left(5\right) \\ 961m+837n=2759-------\left(6\right) \\ \mathrm{Subtracting}\mathrm{equation}\left(5\right)\mathrm{from}\left(6\right),\mathrm{we}\mathrm{get}: \\ 961m+837n=275 \\ 729m+837n=2295 \\ \underline{---} \\ 232m=464 \\ \Rightarrow m=\frac{464}{232}=2 \\ \mathrm{Substituting}m\mathit{}=2\mathrm{in}\mathrm{equation}\left(3\right),\mathrm{we}\mathrm{get}: \\ n\mathit{}=1 \\ \mathrm{Replacing}\mathrm{the}\mathrm{values}\mathrm{of}m\mathit{}\mathrm{and}n,\mathit{}\mathrm{we}\mathrm{get}: \\ \frac{1}{x-2}\mathit{}=2\mathrm{and}\frac{1}{y+3}=1\Rightarrow 2x-4=1\mathrm{and}y+3=1 \\ \therefore x=5\mathrm{and}y=-2 \end{gathered}$$