The graphs of the equations $$4x + \frac{1}{3}y = \frac{8}{3}$$   and $$\frac{1}{2}x + \frac{3}{4}y + \frac{5}{2} = 0$$    intersect at a point P. The point P also lies on the graph of the equation:

Solution (By Examveda Team)

$$\eqalign{ & 4x + \frac{1}{3}y = \frac{8}{3} \cr & 12x + y = 8{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \frac{1}{2}x + \frac{3}{4}y + \frac{5}{2} = 0 \cr & 2x + 3y = - 10{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{Solve equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 12x + y = 8 \cr & \underline {2x + 3y = - 10} \,\,\,\,\, * 6 \cr & 12x + y = 8 \cr & 12x + 18y = - 60 \cr & \underline { - \,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,} \cr & - 17y = 68 \cr & y = - 4 \cr & x = 1 \cr & P = \left( {1,\, - 4} \right) \cr & {\text{Only option }}\left( {\text{D}} \right){\text{satisfy in this point }}P \cr & 3x - y - 7 = 0 \cr & 3 \times 1 - \left( { - 4} \right) - 7 = 0 \cr & 0 = 0\,\,\,\left[ {{\text{satisfy}}} \right] \cr} $$