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Algebra Difficulty 3.1 AMC 10/12 Find the answer

Let 35x−29x2−3x+2=N1x−1+N2x−2\frac {35x - 29}{x^2 - 3x + 2} = \frac {N_1}{x - 1} + \frac {N_2}{x - 2} be an identity in xx. The numerical value of N1N2N_1N_2 is:

Pick one

Solution

N1x−1+N2x−2=N1(x−2)+N2(x−1)(x−1)(x−2)=(N1+N2)x−(2N1+N2)x2−3x+2\frac {N_1}{x - 1} + \frac {N_2}{x - 2} = \frac{N_1(x-2) + N_2(x-1)}{(x-1)(x-2)} = \frac{(N_1 + N_2)x - (2N_1 + N_2)}{x^2 - 3x + 2}
Comparing coefficients, we have the system of two equations
N1+N2=35N1+2N2=29\begin{align*} N_1 + N_2 &= 35\\ N_1 + 2N_2 &= 29 \end{align*}
Subtracting the first equation from the second yields N2=−6N_2 = -6. Substituting yields N1=41N_1 = 41, and their product is −246⇒(A)-246 \Rightarrow \mathrm{(A)}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.