1966 AHSME Problems/Problem 7

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Problem

Let $\frac {35x - 29}{x^2 - 3x + 2} = \frac {N_1}{x - 1} + \frac {N_2}{x - 2}$ be an identity in $x$ . The numerical value of $N_1N_2$ is:

$\text{(A)} \ - 246 \qquad \text{(B)} \ - 210 \qquad \text{(C)} \ - 29 \qquad \text{(D)} \ 210 \qquad \text{(E)} \ 246$

Solution

$\frac {N_1}{x - 1} + \frac {N_2}{x - 2} = \frac{N_1(x-2) + N_2(x-1)}{(x-1)(x-2)} = \frac{(N_1 + N_2)x - (2N_1 + N_2)}{x^2 - 3x + 2}$

Comparing coefficients, we have the system of two equations

$\begin{align*}N_1 + N_2 &= 35\\N_1 + 2N_2 &= 29\end{align*}$

Subtracting the first equation from the second yields $N_2 = -6$ . Substituting yields $N_1 = 41$ , and their product is $-246 \Rightarrow \mathrm{(A)}$ .

1966 AHSME (Problems)
Preceded by Problem 6	Followed by Problem 8
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1966 AHSME Problems/Problem 7

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Problem

Solution

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