First find the equation of the line through $(12,0)$ and $(4,6)$. Its slope is

Using point-slope form with $(12,0)$,

Simplify:

The signs are placed where the $y$-coordinate is at least $9$, so solve

Subtract $9$ from both sides:

Divide by $-\frac{3}{4}$, which reverses the inequality:

Let the smaller number be $s$ and the larger number be $l$. The first condition gives $s+l=18$. The second condition gives $3s+2l=41$. Substitute $l=18-s$ into the second equation:

Then

Check the third condition: $13$ is at least $4$ greater than $5$ because $13-5=8$. So all three conditions are satisfied. Therefore, the pair must be $5$ and $13$.

To make

equivalent to

both inequalities must have exactly the same solution set. Rewrite the absolute value inequality as a compound inequality:

Add $5$ throughout:

Now divide by $2$:

Since this must match

the endpoints must be equal:

From the first equation, $5-k=2$, so $k=3$. Checking the second gives $5+3=8$, and $8/2=4$, which works. Therefore, the statement that must be true is $k=3$.

Because the walkway has width $x$ on both sides of each dimension, the outer length is $30+2x$ and the outer width is $18+2x$. The total area is therefore $(18+2x)(30+2x)$. Since this area is at least $900$ square feet, the correct inequality is

Let the setup fee be $s$ dollars and the hourly rate be $r$ dollars per hour. Then the two given charges satisfy

and

Subtracting gives $3r=120$, so $r=40$. Then

So the pricing model is

where $h$ is the number of hours.

If a third customer was charged more than \$250,then$

Subtract $50$ from both sides:

Divide by $40$:

So the third customer must have used more than $5$ hours of service.