Write the fraction (3x-2)/(x-3)(x+1) as partial fractions.

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We have to write (3x-2)/(x-3)(x+1) as the sum of fractions.

Let us write (3x-2)/(x-3)(x+1) = A / (x - 3) + B/(x +1)

=> [A(x +1) + B(x - 3)]/(x-3)(x+1) = (3x-2)/(x-3)(x+1)

=> Ax + A + Bx - 3B = 3x - 2

equate the numeric terms and those with x

=> Ax + Bx = 3x and A - 3B = -2

Now we have A + B = 3 and A - 3B  = -2

=> A + B - A + 3B = 3 + 2

=> 4B = 5

=> B = 5/4

A = 3 - B = 3 - 5/4 = 7/4

Therefore (3x-2)/(x-3)(x+1) = 7/ 4*(x - 3) + 5/ 4*(x +1)

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