2016-II-05

Ans: A
$\left\{ \begin{array}{ll}
4\alpha + \beta = 5 & \ldots \unicode{x2460} \\
7\alpha + 3\beta = 5 & \ldots \unicode{x2461}
\end{array} \right.$

$\unicode{x2460} \times 3 – \unicode{x2461}$, we have

$\begin{array}{rcl}
5 \alpha & = & 10 \\
\alpha & = & 2
\end{array}$

Sub. $\alpha=2$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
4(2) + \beta & = & 5 \\
\beta & = & -3
\end{array}$