When the line 2y-3x+5 = 0 meets the line 3y-4x-8 = 0, the x and y coordinates are the same. We can solve the system of equations

2y-3x+5 = 0 ...(1)

3y-4x-8 = 0 ...(2)

Using substitution, from (1)

2y-3x+5 = 0

=> 2y = 3x - 5

=> y = (3/2)x - 5/2

Substitute this value for y in (2)

3*((3/2)x - 5/2) - 4x - 8 = 0

=> 9/2x - 15/2 - 4x - 8 = 0

=> x(9/2 - 8/2) = 16/2 + 15/2

=> x = 31

y = (3/2)x - 5/2

=> 3*31/2 - 5/2

=> 44

**The point of intersection is (31, 44)**

Given the lines:

2y-3x + 5 = 0...........(1)

3y-4x -8= 0..............(2)

Both line meets at a point, This point is the solution to the system.

We will use the elimination method to find the points.

We will multiply (1) by 3 and (2) by -2

3*(1) ==> 6y -9x +15 = 0

-2*(2)==> -6y +8x +16 = 0

Now we will add both equations.

==> -x + 31 = 0

==> -x = -31

==> x = 31

==> y= (3x-5)/2 = 3*31-5 /2 = 93-5 /2 = 88/2 = 44

**Then the lines meets at the point (31, 44)**