Ans: C
According to the given system of equations, we have
According to the given system of equations, we have
$\left\{ \begin{array}{ll}
p = 2q-7 & \ldots \unicode{x2460} \\
q^2-12q+6=2q-7 & \ldots \unicode{x2461}
\end{array} \right.$
From $\unicode{x2461}$, we have
$\begin{array}{rcl}
q^2 -12q + 6 & = & 2q – 7 \\
q^2 -12q + 6 -2q +7 & = & 0 \\
q^2 -14q +13 & = & 0 \\
(q-1)(q-13) & = & 0
\end{array}$
Therefore, $q=1$ or $q=13$.
Sub. $q=1$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
p & = & 2(1) – 7 \\
& = & -5
\end{array}$
Sub. $q=13$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
p & = & 2(13) – 7 \\
& = & 19
\end{array}$
Therefore, $p=-5$ or $p=19$.
