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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.

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    $\begingroup$ The correct answer is $\cfrac{2}{a+5} + \cfrac{7}{a+4}$. Which is what I have in my answer below. $\endgroup$ – BLAZE Oct 24 '15 at 14:19
  • $\begingroup$ For more information about this type of transformation search for "partial fraction decomposition" on google. $\endgroup$ – Thomas Oct 24 '15 at 22:16
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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.

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  • $\begingroup$ This answer made the most sense to me. Thanks to you all. $\endgroup$ – Tim Oct 24 '15 at 12:08
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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 $v$.

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You have \begin{equation*} \frac{9a+43}{(a+5)(a+4)} = \frac{u}{a+5} + \frac{v}{a+4} \end{equation*}

Multiply both sides by $(a+5)(a+4)$ and equate the parts with the variable $a$ and without $a$.

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You use the Partial Fractions method which involves the following identity:

$$\frac {9a + 43}{(a + 5)(a + 4)}\equiv \frac {A}{a+5} + \frac {B}{a + 4}$$

$$\implies {9a + 43}\equiv {A}{(a+4)} + {B}{(a + 5)}$$

Since this is an identity (true for all real $a$) we can substitute any values of $a$ into it:

$$a=-5 \implies -2= -A \implies A=2$$ $$a=-4 \implies B=7 $$

So $$\color{blue}{\frac {9a + 43}{(a + 5)(a + 4)} = \frac {2}{a+5} + \frac {7}{a + 4}}$$

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