1956 AHSME Problems/Problem 17

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Problem 17

The fraction $\frac {5x - 11}{2x^2 + x - 6}$ was obtained by adding the two fractions $\frac {A}{x + 2}$ and $\frac {B}{2x - 3}$. The values of $A$ and $B$ must be, respectively:

$\textbf{(A)}\ 5x,-11\qquad\textbf{(B)}\ -11,5x\qquad\textbf{(C)}\ -1,3\qquad\textbf{(D)}\ 3,-1\qquad\textbf{(E)}\ 5,-11$

Solution

This is essentially asking for the partial fraction decomposition of \[\frac{5x-11}{2x^2 + x - 6}\] Looking at $A$ and $B$, we can write \[A(2x-3) + B(x+2) = 5x-11\]. Substituting $x= -2$, we get \[A(-4-3) + B(0) = -21 \Rrightarrow A = 3\] Substituting $x = \frac{3}{2}$, we get \[A(0) + B(\frac{7}{2}) = \frac{15}{2} - \frac{22}{2} \Rrightarrow B = -1\] Thus, our answer is $\boxed{D}$

~JustinLee2017

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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