I can easily solve this problem by finding A and B, and then A+B. My question is where there is a way to obtain A+B without finding A and B first. The problem is supposed to be challenging, but it looks too easy. That is why I think there must be a way to find A+B directly.
If $$\frac{7x-14}{2x^2-9x+4}=\frac{A}{x-4}+\frac{B}{2x-1}$$ find $A+B$.
On
Observe that:
$$(2x-4)(x-1) = 2x^2-9x+1 $$
Therefore:
$$ \frac{A}{x-4} + \frac{B}{2x-1} = \frac{(2x-1)A+ (x-4)B}{2x^2-9x+1}$$
But we have that
$$ \frac{A}{x-4} + \frac{B}{2x-1} = \frac{7x-14}{2x^2-9x+1}$$
From before.
So we conclude that:
$${(2x-1)A+ (x-4)B} = 7x-14$$
At this point we deduce by common terms (the X terms and the constant terms that)
$$ 2A +B = 7$$ $$ -A -4B = -14$$
Now I don't see a fast trick to get A+B out of here that isn't less work than solving the system of equations for A,B so to answe your question: my gut is that it is the fastest way. Perhaps other answers can shed light.
Make $x-4=2x-1$ by choosing $x=-3$. The R.H.S is then $(A+B)(-1/7)$.