# 1959 AHSME Problems/Problem 17

## Problem 17

If $y=a+\frac{b}{x}$, where $a$ and $b$ are constants, and if $y=1$ when $x=-1$, and $y=5$ when $x=-5$, then $a+b$ equals: $\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

## Solution

Plugging in the x and y values, we obtain the following system of equations: $$\begin {cases} 1 = a - b \\ 5 = a - \frac{b}{5} \end {cases}$$ We can then subtract the equations to obtain the equation $4 = 0.8b$, which works out to $b = 5$.

Plugging that in to the original system of equations: $$\begin {cases} 1 = a - 5 \\ 5 = a - \frac{5}{5} \end {cases}$$ It is easy to see that $a = 6$, and that $a+b$ is (E) 11.

## See also

 1959 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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