# 1956 AHSME Problems/Problem 17

## 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 (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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