SAT Question 4becad44: Systems of Linear Equations answer and explanation with Desmos solution

Choice B is correct. A system of two linear equations in two variables, $x$ and $y$, has no solution when the lines in the xy-plane representing the equations are parallel and distinct. Two lines are parallel and distinct if their slopes are the same and their y-intercepts are different. The slope of the graph of the given equation, $3x=36y-45$, in the xy-plane can be found by rewriting the equation in the form $y=mx+b$, where $m$ is the slope of the graph and $\left(0,b\right)$ is the y-intercept. Adding $45$ to each side of the given equation yields $3x+45=36y$. Dividing each side of this equation by $36$ yields $\frac{1}{12}x+\frac{5}{4}=y$, or $y=\frac{1}{12}x+\frac{5}{4}$. It follows that the slope of the graph of the given equation is $\frac{1}{12}$ and the y-intercept is $\left(0,\frac{5}{4}\right)$. Therefore, the graph of the second equation in the system must also have a slope of $\frac{1}{12}$, but must not have a y-intercept of $\left(0,\frac{5}{4}\right)$. Multiplying each side of the equation given in choice B by $\frac{1}{4}$ yields $\frac{1}{12}x=y$, or $y=\frac{1}{12}x$. It follows that the graph representing the equation in choice B has a slope of $\frac{1}{12}$ and a y-intercept of $\left(0,0\right)$. Since the slopes of the graphs of the two equations are equal and the y-intercepts of the graphs of the two equations are different, the equation in choice B could be the second equation in the system.