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The complete method for decomposing fractions is obvious . For example $y = \frac{2x+1}{(x-1)(x+3)} = \frac{a}{x-1} + \frac{b}{x+3} = \frac{a(x+3) + b(x-1)}{(x-1)(x+3)} \Rightarrow $$ \left\{ \begin{array}{c} a+b= 2 \\ 3a - b = 1 \end{array} \right. $$ \Rightarrow a = 3/4 , b= 5/4 $

But I was interested in finding a quick way . So I did another operation : $y = \frac{2x+1}{(x-1)(x+3)} = \frac{a}{x-1} + \frac{b}{x+3} = \frac{a(x+3) + b(x-1)}{(x-1)(x+3)} \Rightarrow 2x+1 = a(x+3) +b(x-1) \ \ \star$

Put $x= 1$ then $a = 3/4$ and $x = -3$ then $b =5/4 $ . I was wondering why this method works correctly because the relation $\star$ works for all values except $1$ and $-3$ . I've tried many examples and strangely this method was correct all the time .

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  • $\begingroup$ Your method uses 'partial fractions': tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx $\endgroup$ – TheSimpliFire Nov 29 '17 at 19:36
  • $\begingroup$ @TheSimpliFire Thank you but that article hasn't explained why it works . $\endgroup$ – S.H.W Nov 29 '17 at 20:01
  • $\begingroup$ It works because on expanding $\star$ we get $$2x+1=(a+b)x+(3a-b)$$ and equating like terms we have two equations which are the same as those obtained in your complete method. $\endgroup$ – TheSimpliFire Nov 30 '17 at 7:59

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