SOLUTION: Hello, I found out that a fraction of : (7x+1)/(x+1)(x-2) simplifies as follows : 2/(x+1) + 5/(x-2) Could you please guide me as to how I can simplify similar fractions? Is it

Question 326720: Hello, I found out that a fraction of : (7x+1)/(x+1)(x-2) simplifies as follows :
2/(x+1) + 5/(x-2)
Could you please guide me as to how I can simplify similar fractions? Is it based solely on experience or is there some kind of methodology? I would be really grateful if you'd help me on this!
Sincerely yours,
Kollias Panos
Found 2 solutions by AAfter Search, J2R2R:
Answer by AAfter Search(61) (Show Source): You can put this solution on YOUR website!
Let (7x+1)/(x+1)(x-2) = m/(x + 1) + n/(n - 2), m and n being integers.
Multiplying by (x+1)(x-2) on both sides of equality, we get
(7x+1) = m(x - 2) + n(x + 1)
=> 7x + 1 = (m + n)x + n - 2m
Equating the coefficients of x and the constant terms on both sides of equality, we get
m + n = 7
-2m + n = 1
------------
(subtracting) 3m = 6
m = 2
Hence, n = 5
Hence,
(7x+1)/(x+1)(x-2) = 2/(x+1) + 5/(x-1)
Answer by J2R2R(94) (Show Source): You can put this solution on YOUR website!
Using partial fractions we have
(7x + 1)/[(x + 1)(x - 2)] = A/(x + 1) + B/(x - 2)

Multiply throughout by (x + 1)(x - 2) gives
7x + 1 = A(x - 2) + B(x + 1)

By putting x=2 we can eliminate A and get
7(2) + 1 = A(2 - 2) + B(2 + 1)
15 = 3B
B = 5

Similarly by putting x=-1 we can eliminate B and get
7(-1) + 1 = A(-1 - 2) + B(-1 + 1)
-6 = -3A
A = 2

Therefore (7x + 1)/[(x + 1)(x - 2)] = 2/(x + 1) + 5/(x - 2)

Check:
(7x + 1)/[(x + 1)(x - 2)] = [2(x - 2) + 5/(x + 1)]/[(x + 1)(x - 2)]
(2x - 4 + 5x + 5)/[ (x + 1)(x - 2)] = (7x + 1)/[(x + 1)(x - 2)]

So using this example should enable you to work out other fractions like this using this method.
It is a bit more difficult for repeated factors and factors of higher orders but for the moment we will stick to non-repeated linear factors.

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