$\left\{ \begin{array}{ll}
x + 2y = 20 & \ldots \unicode{x2460} \\
7x – 6y = 20 & \ldots \unicode{x2461} \\
13x + 6y = 20 & \ldots \unicode{x2462}
\end{array} \right.$
$3 \times \unicode{x2460} + \unicode{x2461}$, we have
$\begin{array}{rcl}
10x & = & 80 \\
x & = & 8
\end{array}$
Sub. $x = 8$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
8 + 2y & = & 20 \\
y & = & 6
\end{array}$
Therefore, the intersection point of $\unicode{x2460}$ and $\unicode{x2461}$ is $(8, 6)$.
$\unicode{x2462} – 3 \times \unicode{x2460}$, we have
$\begin{array}{rcl}
10x & = & – 40 \\
x & = & -4
\end{array}$
Sub. $x = -4$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
-4 + 2y & = & 20 \\
y & = & 12
\end{array}$
Therefore, the intersection point of $\unicode{x2460}$ and $\unicode{x2462}$ is $(-4, 12)$.
$\unicode{x2461} + \unicode{x2462}$, we have
$\begin{array}{rcl}
20 x & = & 40 \\
x & = & 2
\end{array}$
Sub. $x=2$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
7(2) – 6y & = & 20 \\
-6y & = & 6 \\
y & = & -1
\end{array}$
Therefore, the intersection point of $\unicode{x2461}$ and $\unicode{x2462}$ is $(2, -1)$.
At the point $(8,6)$, the value of $7x + 8y + 9$
$\begin{array}{cl}
= & 7(8) + 8(6) + 9 \\
= & 113
\end{array}$
At the point $(-4,12)$, the value of $7x + 8y + 9$
$\begin{array}{cl}
= & 7(-4) + 8(12) + 9 \\
= & 77
\end{array}$
At the point $(2,-1)$, the value of $7x + 8y + 9$
$\begin{array}{cl}
= & 7(2) + 8(-1) + 9 \\
= & 15
\end{array}$
Therefore, the greatest value of $7x + 8y + 9$ is $113$.
