For the equations to have an unique solution "a" and "b" can have any value except for the values which will make the equations have infinite solutions. So, for the equation to have infinite solutions,

Finding valid values of a and b

To find the valid values of a and b, we need to use the concept of determinant. For a system of linear equations to have a unique solution, the determinant of the coefficient matrix should not be equal to zero.

Using determinant to find valid values of a and b

The coefficient matrix for the given system of equations is:

| 2a-1 -3 |
| 3 (b-2)|

The determinant of this matrix is:

(2a-1)(b-2) - (-3)(3)
= 2ab - 4a - b + 6

For the system to have a unique solution, the determinant should not be equal to zero. Therefore, we have:

2ab - 4a - b + 6 ≠ 0

Simplifying this expression, we get:

2a(b-2) - (b-2) ≠ 0
(b-2)(2a-1) ≠ 0

Therefore, the valid values of a and b are:

a ≠ 1/2
b ≠ 2

Finding values of p and q

To find the values of p and q, we need to solve the given system of equations. We can use the method of substitution to do this.

From the first equation, we have:

(2a-1)x - 3y = 5
=> 2a-1 = (5+3y)/x

Substituting this value of 2a-1 in the second equation, we get:

3x(b-2)(px+q) = 3
=> x(b-2)(px+q) = 1
=> px^2 + qx - 1/(b-2) = 0

This is a quadratic equation in x, and we can solve for x using the quadratic formula:

x = (-q ± sqrt(q^2 + 4p/(b-2)))/(2p)

Since the system has a unique solution, the discriminant of the quadratic must be greater than zero. Therefore,

q^2 + 4p/(b-2) > 0

Simplifying this expression, we get:

q^2(b-2) + 4p > 0

Therefore, we have:

p > 0
q ≠ 0

To find the values of p and q, we need to solve the quadratic equation for x and substitute the values of p and q in the given equation for y. This would give us the required solution.