Let the equation of the line having equal intercepts on the axes be ax+ay=1
x+y=a…..(1)
On solving equations 4x+7y −3=0 and 2x −3y+1=0, we obtain
x=131 and y=135
∴(131,135) is the point of intersection of two given lines.
Since equation (1) passes through point (131,135)
131+135=a
a=136
∴ Equation (1) becomes
x+y=136
13x+13y=6
Thus, the required equation of the line is 13x+13y=6.