Question

(vi) $\frac{2x}{3}+\frac{3y}{2}=8\frac{1}{3};\frac{3x}{12}+\frac{2y}{3}=13\frac{1}{3}$

Answer 1

$$\begin{gathered} \frac{2x}{3}+\frac{3y}{2}=8\frac{1}{3} \\ \Rightarrow 4x+9y=50 \\ \therefore x=\frac{50-9y}{4}...\left(\mathrm{i}\right) \\ \frac{3x}{12}+\frac{2y}{3}=13\frac{1}{3} \\ 3x+8y=160...\left(\mathrm{ii}\right) \\ \mathrm{Substituting}\mathrm{eq}\left(\mathrm{i}\right)\mathrm{in}\mathrm{eq}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get}: \\ 3\times \frac{50-9y}{4}+8y=160 \\ \Rightarrow 150-27\mathrm{y}+32y=640 \\ \Rightarrow 5y=490 \\ \therefore y=98 \\ \mathrm{Substituting}y=98\mathrm{in}\mathrm{eq}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ x=\frac{50-9\times 98}{4} \\ \Rightarrow x=-208 \\ \mathrm{Hence},x=-208\mathit{,}\mathit{}y=98\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equations}. \end{gathered}$$