The point (2, 3) lies on the graph of the linear equation 3x - (a -1) y = 2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a) y = 3b, then find the value of b.

Given, point (2, 3) lies on the line, 3x-(a-1)y = 2a-1

So, the point (2, 3) is the solution of

3x - (a - 1)y = 2a -1

∴3×2-(a-1)×3 = 2a-1 [put x=2,y = 3]

=> 6 - 3a+ 3 = 2a - 1

=> -3a -2a = -1-3-6

=> -5a = -10

=>a=2

Also, (2, 3) is the solution of 5x + (1 - 2a) y = 3b.

∴ 5×2 + (1 - 2a) × 3 = 3b [put x=2,y = 3]

On putting the value of a = 2, we get