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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 $A+B$.

Kushal Bhuyan
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learning
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2 Answers2

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Make $x-4=2x-1$ by choosing $x=-3$. The R.H.S is then $(A+B)(-1/7)$.

Aravind
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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.

Sidharth Ghoshal
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