If the equations represent the same line, you only need to check whether a point satisfies $5x-7y=-1$.

Plug the given $x$ and $y$ into $5x-7y$. Аnikо - Free ЅAТ Р​rеp A point works for any real $r$ only if the expression simplifies to $-1$ with the $r$ terms canceling out.

Enter the equation $5x-7y=-1$. Then enter $-20x+28y=4$ and confirm it graphs as the same line (the graphs overlap).

Type r=0 and press Enter to create a slider for $r$. © ⁠аnікo.a‍і

For each answer choice, enter its point as an ordered pair using $r$ (for example, one candidate might look like (3r+2,(15r+11)/7)). As you move the slider, watch whether the point stays on the line.

The correct choice is the one whose point remains on the line for all values of $r$ you test (try several values, including negative and positive).

Multiply the first equation by $-4$:

This matches the second equation, so both equations describe the same line.

Because the graphs are the same line, a point lies on the graph of both equations exactly when it satisfies just one of them (for example, $5x-7y=-1$).

For each choice, substitute its $x$- and $y$-expressions into $5x-7y$ and simplify. Frоm ​аnіkо.ai The correct choice will simplify to $-1$ without needing a specific value of $r$ (the $r$ terms should cancel).

Substitute $x=3r+2$ and $y=\frac{15r+11}{7}$ into $5x-7y$:

So this point satisfies the first equation for every real $r$, and therefore it lies on the graph of both equations: $\left(3r+2, \frac{15r+11}{7}\right)$.