Write a formula as a sum of fractions with constant numerators I'm supposed to write this formula:
$$\frac {9a + 43}{a^2 + 9a + 20}$$
As a sum of fractions with constant numerators as:
$$\frac {7}{a+5} + \frac {2}{a+4}$$
The first step is of course:
$$\frac {9a + 43}{(a + 5)(a + 4)}$$
Now it is possible to write it as a sum using the following method:
$$\frac {u}{a+5} + \frac {v}{a + 4}$$
$$u + v = 9a + 43$$
Which gives me:
$$u = 9a + 43 - v$$
$$v = 9a + 43 - u$$
$$u = 9a + 43 - (9a + 43 - u)$$
$$u = u$$
But that doesn't help much knowing that $u = u$. Therefor my question: how can I get $u = 7$ and $v = 2$? Any hints are appreciated.

You made a mistake when you wrote this:

Now it is possible to write it as a sum using the following method:
  $$\frac {u}{a+5} + \frac {v}{a + 4}$$
  $$u + v = 9a + 43$$

This should have been: $$u(a + 4) + v(a + 5) = 9a + 43$$ or, 
$$(u+v)a+(4u+5v)=9a+43$$
Since this is an identity in $a$, we have $$u+v=9$$ and $$4u+5v=43$$ Equate this to get $u$ and $v$.
