To start, let's use algebra to solve this equation for x:
2ax-15=3(x+5)+5(x-1)
2ax - 15 = 3x + 15 + 5x - 5
2ax - 15 = 8x +10
(2a - 8)x = 25
x = 25/(2a-8)
The only way that this equation does not have a solution is if 2a - 8 = 0, as that would mean x = 25/0, and we can't divide by 0. If 2a - 8 = 0, then a = 4. And so the original equation, 2ax-15=3(x+5)+5(x-1) will not have a solution when a = 4.