Question

(v) $\frac{2x-y}{3}+3x-2y=3;4x-3y+\frac{5x-4y}{4}=1$

Answer 1

$$\begin{gathered} \frac{2x-y}{3}+3x-2y=3 \\ \Rightarrow 2x-y+9x-6y=9 \\ \Rightarrow 11x-7y=9...\left(\mathrm{i}\right) \\ 4x-3y+\frac{5x-4y}{4}=1 \\ \Rightarrow 16x-12y+5x-4y=4 \\ \Rightarrow 21x-16y=4...\left(\mathrm{ii}\right) \\ \mathrm{Multiplying}\mathrm{eq}\left(\mathrm{i}\right)\mathrm{by}16\text{and}\mathrm{eq}\left(\mathrm{ii}\right)\mathrm{by}7,\text{we get}: \\ 176x-112y=144...\left(\mathrm{iii}\right) \\ 147x-112y=28...\left(\mathrm{iv}\right) \\ \mathrm{Subtracting}\mathrm{eq}\left(\mathrm{iv}\right)\mathrm{from}\left(\mathrm{iii}\right),\text{we get:} \\ 176x-112y=144 \\ 147x-112y=28 \\ \underline{-+-} \\ 29x=116 \\ \therefore x=4 \\ \mathrm{Substituting}x=4\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ 11\times 4-7y=9 \\ \Rightarrow 44-7y=9 \\ \therefore y=5 \\ \mathrm{Hence},x=4,y=5\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equations}. \end{gathered}$$