Ans: B
Let $y = k_1 + k_2 x$, where $k_1$ and $k_2$ are constants.
Let $y = k_1 + k_2 x$, where $k_1$ and $k_2$ are constants.
When $x=2$, $y=17$, we have
$\begin{array}{rcl}
k_1 + k_2(2) & = & 17 \\
k_1 + 2k_2 & = & 17 \ \ldots \unicode{x2460}
\end{array}$
When $x=4$, $y=11$, we have
$\begin{array}{rcl}
k_1 + k_2(4) & = & 11 \\
k_1 + 4k_2 & = & 11 \ \ldots \unicode{x2461}
\end{array}$
$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
2k_2 & = & -6 \\
k_2 & = & -3
\end{array}$
Sub. $k_2$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
k_1 + 2(-3) & = & 17 \\
k_1 & = & 23
\end{array}$
$\therefore y= 23 -3x$.
For $y=5$, we have
$\begin{array}{rcl}
5 & = & 23 – 3x \\
3x & = & 18 \\
x & = & 6
\end{array}$
