# A quick way for decomposing fractions

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 .

• Your method uses 'partial fractions': tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx – TheSimpliFire Nov 29 '17 at 19:36
• @TheSimpliFire Thank you but that article hasn't explained why it works . – S.H.W Nov 29 '17 at 20:01
• 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. – TheSimpliFire Nov 30 '17 at 7:59