First find the equation of the line through $(1,7)$ and $(4,13)$. Its slope is

Since one point on the line is $(1,7)$, use point-slope or slope-intercept form:

which simplifies to

This is the same as the first equation, so both equations represent the same line. Therefore, every point on the line satisfies both equations, and the system has infinitely many solutions.

Let the width be $w$ and the length be $l$. The perimeter gives the equation

so

The length is $7$ feet greater than the width, so

Substituting into $l+w=23$ gives

so

and

Then

The area is

Because $(2,4)$ is a solution to the system, it must satisfy both equations. Substitute $x=2$ and $y=4$ into the second equation:

which simplifies to

So the value that must be equal to $2a+4b$ is $23$.

Let the total monthly costs be $A=12+3m$ and $B=6+5m$. Since the plans cost the same for $m$ movies, set the expressions equal: $12+3m=6+5m$. Solving gives $6=2m$, so $m=\frac{12-6}{5-3}$. The total cost of either plan at that value is found by substituting this expression for $m$ into one of the cost formulas. Using Plan A gives $12+3\left(\frac{12-6}{5-3}\right)$, which matches the correct answer.

Because $(4,1)$ is the solution to the system, it must satisfy both equations. First use $y=ax+b$:

Next, since the point also lies on $3x-2y=10$, substitute $(4,1)$ to confirm it works:

So the point is on the second line as expected. For the system to have exactly one solution at $(4,1)$, the line $y=ax+b$ must intersect $3x-2y=10$ at that point and not be the same line.

Rewrite the second equation in slope-intercept form:

This line has slope $\frac{3}{2}$. Since there is exactly one solution, the line $y=ax+b$ cannot have the same slope, so $a\ne \frac{3}{2}$.

Using $1=4a+b$, we get

So we need a value consistent with the answer choices and with $a\ne \frac{3}{2}$. Test the choices by solving $1-3a=$ each value.

If $a+b=-\frac{1}{2}$, then

Then

So the line is

which passes through $(4,1)$ and has slope different from $\frac{3}{2}$, so it intersects the other line exactly once. Thus,