To write $y=3x-5$ in standard form, move $y$ to the left side: $-3x+y=-5$. Multiplying both sides by $-1$ gives the equivalent equation $3x-y=5$. So the same line is represented by $3x-y=5$.

A point lies on the line if its coordinates satisfy the equation $y=2x-3$. For $(0,-3)$, substitute $x=0$ into the equation: $y=2(0)-3=-3$. This matches the point's $y$-coordinate, so $(0,-3)$ lies on the line.

For a line in slope-intercept form, $y=mx+b$. Substitute each given point into the equation. Using $(0,4)$ gives $4=0m+b$, and using $(2,0)$ gives $0=2m+b$. That matches the system shown in $\text{A}$.

Use the slope formula with the two points $(0,4)$ and $(2,0)$:

So the statement that must be true is that the slope of the line is $-2$.

Substitute $x=3$ into the equation of the line: $y=2(3)+1=6+1=7$. So the point on the path has $y$-coordinate $7$.