To solve this system of equations, you can use either the substitution method or the elimination method.

Let's use the elimination method to solve this system of equations:

Multiply the second equation by 5 to make the coefficients of y the same in both equations:
5(2x + 3y) = 5(9)
10x + 15y = 45

Now, we have two equations:
3x - 5y = 23
10x + 15y = 45

Add the two equations together:
(3x - 5y) + (10x + 15y) = 23 + 45
13x + 10y = 68

To solve for x, we can isolate x in the equation above:
13x = 68 - 10y
13x = 68 - 10y
x = (68 - 10y) / 13

Now, substitute this value of x into either of the original equations. Let's use the first equation:
3(68 - 10y) / 13 - 5y = 23
204 - 30y / 13 - 5y = 23
204 - 30y - 65y = 299
204 - 95y = 299
-95y = 95
y = -1

Now that we have found the value of y, we can substitute it back into one of the original equations to solve for x. Let's use the first equation:
3x - 5(-1) = 23
3x + 5 = 23
3x = 18
x = 6

Therefore, the solution to the system of equations 3x - 5y = 23 and 2x + 3y = 9 is x = 6 and y = -1.