Question

$\frac{3}{2}x+\frac{2}{3}y=13\frac{1}{3};\frac{2}{3}x+\frac{3}{2}y=8\frac{1}{3}$

Answer 1

$$\begin{gathered} \frac{3}{2}x+\frac{2}{3}y=13\frac{1}{3} \\ \Rightarrow \frac{9x+4y}{6}=\frac{40}{3} \\ \Rightarrow 9x+4y=80...\left(\mathrm{i}\right) \\ \frac{2}{3}x+\frac{3}{2}y=8\frac{1}{3} \\ \Rightarrow \frac{4x+9y}{6}=\frac{25}{3} \\ \Rightarrow 4x+9y=50...\left(\mathrm{ii}\right) \\ \mathrm{Here},\mathrm{we}\mathrm{observe}\mathrm{that}\mathrm{the}\mathrm{coefficients}\mathrm{of}x\mathrm{and}y\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{equation}\mathrm{are}\mathrm{interchanged}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{equation}. \\ \mathrm{Adding}\mathrm{these}\mathrm{two}\mathrm{equations},\text{we get}: \\ 9x+4y=80 \\ \underline{4x+9y=50} \\ 13x+13y=130 \\ \therefore x+y=10...\left(\mathrm{iii}\right) \\ \mathrm{Subtracting}\mathrm{eq}\left(\text{ii}\right)\mathrm{from}\left(\text{i}\right),\text{we get:} \\ 9x+4y=80 \\ 4x+9y=50 \\ \underline{\underline{---}} \\ 5x-5y=30...\left(\mathrm{iv}\right) \\ \therefore x-y=6 \\ \mathrm{From}\left(\mathrm{iii}\right)\&\left(\mathrm{iv}\right),\mathrm{we}\text{get}: \\ x+y=10\mathrm{and}x-y=6 \\ \mathrm{Adding}\left(\mathrm{iii}\right)\mathrm{and}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}: \\ x+y=10 \\ \underline{x-y=6} \\ 2x=16 \\ \therefore x=8 \\ \mathrm{Substituting}x=8\mathrm{in}\mathrm{eq}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}: \\ 8+y=10 \\ \therefore y=2 \\ \mathrm{Hence},x=8\mathrm{and}y=2\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equations}. \end{gathered}$$