SAT Question 9b7a1b67: Systems of Linear Equations answer and explanation with Desmos solution

Choice A is correct. It's given that if line $k$ is graphed in the xy-plane with line $h$, the result is the graph of a system of two linear equations that has no solution. It follows that line $h$ and line $k$ don't intersect. This means that line $k$ and line $h$ have the same slope and different y-intercepts. The graph shows that the y-intercept of line $h$ is $\left(0,-9\right)$. The y-coordinate of the y-intercept of line $k$ is the value of $y$ when $x=0$ in the equation defining line $k$. Substituting $0$ for $x$ in the equation $sx+48y=t$ yields $s\left(0\right)+48y=t$, or $48y=t$. Dividing both sides of this equation by $48$ yields $y=\frac{t}{48}$. Therefore, the y-coordinate of the y-intercept of line $k$ is $\frac{t}{48}$, so the y-intercept of line $k$ is $\left(0,\frac{t}{48}\right)$. Since line $k$ and line $h$ have different y-intercepts, $\frac{t}{48}$ can't be equal to $-9$. It follows that $t$ can't be equal to $-9\left(48\right)$, or $-432$. Thus, $-432$ isn't a possible value of $t$.