Question

Solve the following simultaneous equations :
$\left(\mathrm{iii}\right)\frac{1}{3x}+\frac{1}{5y}=\frac{1}{15};\frac{1}{2x}+\frac{1}{3y}=\frac{1}{12}$

Answer 1

$$\begin{gathered} \frac{1}{3x}+\frac{1}{5y}=\frac{1}{15}-------\left(1\right) \\ \frac{1}{2x}+\frac{1}{3y}=\frac{1}{12}-------\left(2\right) \\ \mathrm{Substituting}\frac{1}{x}=m\mathrm{and}\frac{1}{y}=n\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ \frac{m}{3}+\frac{n}{5}=\frac{1}{15} \\ \mathrm{Multiplying}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{by}15,\mathrm{we}\mathrm{get}: \\ 5m+3n=1--------\left(3\right) \\ \mathrm{Substituting}\frac{1}{x}=m\mathrm{and}\frac{1}{y}=n\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get}: \\ \frac{m}{2}+\frac{n}{3}=\frac{1}{12} \\ \mathrm{Multiplying}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{by}12,\mathrm{we}\mathrm{get}: \\ 6m+4n=1--------\left(4\right) \\ \mathrm{Multiplying}\mathrm{equation}\left(3\right)\mathrm{by}4,\mathrm{we}\mathrm{get}: \\ 20m+12n=4-------\left(5\right) \\ \mathrm{Multiplying}\mathrm{equation}\left(4\right)\mathrm{by}3,\mathrm{we}\mathrm{get}: \\ 18m+12n=3-------\left(6\right) \\ \mathrm{Subtracting}\left(6\right)\mathrm{from}\left(5\right),\mathrm{we}\mathrm{get}: \\ 20m+12n=4 \\ 18m+12n=3 \\ \underline{\underline{---}} \\ 2m=1 \\ \Rightarrow m=\frac{1}{2} \\ \mathrm{Substituting}m\mathit{}=\frac{1}{2}\mathrm{in}\mathrm{equation}\left(4\right),\mathrm{we}\mathrm{get}: \\ n\mathit{}=-\frac{1}{2} \\ \mathrm{Resubstituting}\mathrm{the}\mathrm{values}\mathrm{of}m\mathit{}\mathrm{and}n,\mathit{}\mathrm{we}\mathrm{get}: \\ \frac{1}{x}\mathit{}=\frac{1}{2}\mathrm{and}\frac{1}{y}=-\frac{1}{2} \\ \therefore x=2\mathrm{and}y=-2 \end{gathered}$$