You need to remember the technique called partial fraction decomposition such that:

(3x-2)/(x-3)(x+1) = a/(x-3) + b/(x+1)

Notice that the numerators are not known yet, hence two variables are assigned.

You may get rid of all denominator multiplying all terms by the common denominator (x-3)(x+1) such that:

3x-2 = a(x+1) + b(x-3)

Opening the brackets yields: 3x-2=ax+a+bx-3b

Group the x terms and the constant terms such that: 3x-2=(a+b)x + (a-3b)

Equating the coefficients of like powers yields:

a+b = 3 => a = 3-b

a - 3b = -2 => 3 - b - 3b = -2 => -4b = -5 => b = 5/4

a = 3 - 5/4 = 7/4

**Hence, the original fractions were the following: ****(3x-2)/(x-3)(x+1) = 7/(4x-12) + 5/(4x+14)**

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