Question

(i) $\frac{x}{3}+\frac{y}{4}=4,\frac{5x}{6}-\frac{y}{8}=4$

Answer 1

$$\begin{gathered} \frac{x}{3}+\frac{y}{4}=4 \\ \Rightarrow 4x+3y=48...\left(\mathrm{i}\right) \\ \frac{5x}{6}-\frac{y}{8}=4 \\ 20x-3y=96...\left(\mathrm{ii}\right) \\ \mathrm{Multiplying}\mathrm{eq}\left(\mathrm{i}\right)\mathrm{by}5,\mathrm{we}\mathrm{get}: \\ 20x+15y=240...\left(\mathrm{iii}\right) \\ \mathrm{Subtracting}\mathrm{eq}\left(\mathrm{ii}\right)\mathrm{from}\mathrm{eq}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}: \\ 20x+15y=240 \\ 20x-3y=96 \\ \underline{\underline{-+-}} \\ 18y=144 \\ \therefore y=8 \\ \mathrm{Substituting}y=8\mathrm{in}\text{eq.}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ \Rightarrow 4x+3\times 8=48 \\ \Rightarrow 4x=24 \\ \therefore x=6 \\ \mathrm{Hence},x=6,y=8\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equations}. \end{gathered}$$