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 want to write $\frac{9a+ 43}{(a+ 5)(a+ 4)}$ as fractions with constant numerators which clearly must be of the form $\frac{p}{a+5}+ \frac{q}{a+4}$.  Getting the "common denominator", $\frac{p(a+ 4)}{(a+ 5)(a+ 4)}+ \frac{q(a+ 5)}{(a+5)(a+ 4)}= \frac{pa+ 4p+ qa+ 5q}{(a+ 5)(a+ 4)}$ and you want $pa+ 4p+ qa+ 5q= (p+ q)a+ (4p+ 5q)= 9a+ 43$ for all a.  That means that you want p+q= 9 and 4p+ 5q= 43.  Solve those equations for p and q.
