Consider $\left\{ \begin{array}{ll} y = 9 & \ldots \unicode{x2460} \\ x – y – 9 = 0 & \ldots \unicode{x2461} \\ x + y – 9 = 0 & \ldots \unicode{x2462} \end{array}\right.$
Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
x – 9 – 9 & = & 0 \\
x & = & 18
\end{array}$
Therefore, the intersection point is $(18,9)$.
Sub. $\unicode{x2460}$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
x + 9 – 9 & = & 0 \\
x & = & 0
\end{array}$
Therefore, the intersection point is $(0,9)$.
$\unicode{x2461} + \unicode{x2462}$, we have
$\begin{array}{rcl}
2x – 18 & = & 0 \\
x & = & 9
\end{array}$
Sub. $x=9$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
9 + y – 9 & = & 0 \\
y & = & 0
\end{array}$
Therefore, the intersection point is $(9,0)$.
At point $(18,9)$, the value of $x – 2y + 43$
$\begin{array}{cl}
= & 18 – 2(9) + 43 \\
= & 43
\end{array}$
At point $(0,9)$, the value of $x – 2y + 43$
$\begin{array}{cl}
= & 0 – 2(9) + 43 \\
= & 25
\end{array}$
At point $(9,0)$, the value of $x – 2y + 43$
$\begin{array}{cl}
= & 9 – 2(0) + 43 \\
= & 52
\end{array}$
Hence, the greatest value of $x – 2y + 43$ is $52$.
