Question

(iii) $\frac{x}{3}+\frac{y}{4}=4;\frac{x}{2}-\frac{y}{4}=1$

Answer 1

$$\begin{gathered} \frac{x}{3}+\frac{y}{4}=4 \\ \Rightarrow 4x+3y=48...\left(\mathrm{i}\right) \\ \frac{x}{2}-\frac{y}{4}=1 \\ \Rightarrow 2x-y=4...\left(\mathrm{ii}\right) \\ \mathrm{Multiplying}\mathrm{eq}\left(\mathrm{ii}\right)\mathrm{by}2,\text{we get:} \\ 4x-2y=8...\left(\mathrm{iii}\right) \\ \mathrm{Subtracting}\mathrm{eq}\left(\mathrm{iii}\right)\mathrm{from}\mathrm{eq}\left(\mathrm{i}\right),\text{we get:} \\ 4x+3y=48 \\ 4x-2y=8 \\ \underline{-+-} \\ 5y=40 \\ \therefore y=8 \\ \mathrm{Substituting}y=8\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ 4x+3\times 8=48 \\ \Rightarrow 4x=24 \\ \therefore x=6 \\ \mathrm{Hence},x=6,y=8\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equations}. \end{gathered}$$