SAT Question 4dc5c6f9: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice D is correct. A point $\left(x,y\right)$ is a solution to a system of equations if it lies on the graphs of both equations in the xy-plane. In other words, a solution to a system of equations is a point $\left(x,y\right)$ at which the graphs intersect. It's given that the first equation is $y=18$. Substituting $18$ for $y$ in the second equation yields $18=-3{\left(x-18\right)}^2+15$. Subtracting $15$ from each side of this equation yields $3=-3{\left(x-18\right)}^2$. Dividing each side of this equation by $-3$ yields $-1={\left(x-18\right)}^2$. Since the square of a real number is at least $0$, this equation can't have any real solutions. Therefore, the graphs of the equations intersect at zero points.

Alternate approach: The graph of the second equation is a parabola that opens downward and has a vertex at $\left(18,15\right)$. Therefore, the maximum value of this parabola occurs when $y=15$. The graph of the first equation is a horizontal line at $18$ on the y-axis, or $y=18$. Since $18$ is greater than $15$, or the horizontal line is above the vertex of the parabola, the graphs of these equations intersect at zero points.