If both equations represent the same line, you only need to work with one equation. Try solving $12x - 10y = 24$ for $y$ in terms of $x$.

Once you have $y$ as a linear expression in $x$, think of $x$ as the parameter $t$. Which choice keeps $t$ as the $x$-coordinate and uses the same expression for $y$?

In Desmos, enter the two equations 2(6x - 5y) = 24 and 18x - 15y = 36. You should see that the two graphs lie exactly on top of each other, confirming they represent the same line.

Take the simpler equation 12x - 10y = 24 (the expanded form of the first equation) and rearrange it algebraically into the form y = (something in x) / 5. You can type this rearranged equation into Desmos to check that it still produces the same line.

Treat $x$ as the parameter $t$. Replace $x$ by $t$ in the equation for $y$ that Desmos (or your algebra) gives you, then compare that $(x,y)$ pattern to the answer choices and pick the one whose $y$-coordinate has exactly the same expression in terms of $t$. You should also notice in Desmos that plugging that expression into the coordinates always lands points on the graphed line as $t$ varies.

First expand the first equation:

Now compare this with the second equation:

Notice that every coefficient in the second equation is $\tfrac{3}{2}$ times the corresponding coefficient in $12x - 10y = 24$ (since $18 = \tfrac{3}{2}\cdot 12$, $15 = \tfrac{3}{2}\cdot 10$, and $36 = \tfrac{3}{2}\cdot 24$). This means the two equations represent the same line, so any solution of one is a solution of the other.

Because both equations describe the same line, it is enough to work with just one of them. Use the simpler form:

We will solve this equation for $y$ in terms of $x$ to describe every point on the line.

Start with

Move $12x$ to the right side:

Divide both sides by $-10$:

Simplify the fraction by factoring out $12$ in the numerator and $-10$ in the denominator:

So every solution on this line has coordinates $(x, \tfrac{6x - 12}{5})$ for some real number $x$.

The question uses a parameter $t$ to represent "any real number." If we let $x = t$, then from the relationship we just found,

So all points that lie on both graphs can be written as

This matches answer choice C.