Because both functions are linear, each has a constant rate of change. For $f$, when $x$ increases by $2$, the output decreases by $6$, so the slope is $-3$. Since $f(0)=8$, the equation is $f(x)=8-3x$. For $g$, when $x$ increases by $2$, the output increases by $6$, so the slope is $3$. Since $g(0)=-1$, the equation is $g(x)=3x-1$. At the intersection, the outputs are equal:

So the graphs intersect when $x=\frac{3}{2}$.

Because the solution has $x=2$, the two expressions for $y$ must be equal when $x=2$:

So

which simplifies to

The condition that the system has exactly one solution means the lines are not the same line and not parallel, but to find what must be true, equating the $y$-values at $x=2$ is enough. Therefore, the statement that must be true is $k=2m-11$.

Let $s$ be the number of student tickets and $a$ be the number of adult tickets. The total number of tickets gives

The total money collected gives

Substitute $s=120-a$ into the revenue equation:

Then

so

Subtract $720$ from both sides:

and divide by $4$:

Therefore, $48$ adult tickets were sold.

Because $(2,b)$ is a solution, substitute $x=2$ into both equations. From $2x+3y=13$, we get $2(2)+3b=13$, so $4+3b=13$ and $3b=9$, which gives $b=3$. Then use $x=2$ and $y=3$ in the second equation: $4(2)-k(3)=2$, so $8-3k=2$. This means $-3k=-6$, so $k=2$. Therefore,

First find the slope of line $k$ using its two points:

A line perpendicular to $k$ has slope $\frac{2}{3}$. Since line $m$ passes through $(2,1)$, its equation is

Line $k$ has $y$-intercept $6$ and slope $-\frac{3}{2}$, so its equation is

Set the two expressions for $y$ equal:

Simplify the right side:

Then solve:

Multiply by $6$:

So

and