SAT Question 080e75c2: Systems of Linear Equations answer and explanation with Desmos solution

Choice C is correct. A line in the xy-plane that passes through the points $(x_1,y_1)$ and $(x_2,y_2)$ has slope $m$, where $m=\frac{y_2-y_1}{x_2-x_1}$. The equation of the line can be written in the form $y=mx+b$, where $b$ is the y-intercept represented by $(0,b)$. One of the lines shown in the graph passes through the points $(0,7)$ and $(11,11)$. Substituting $0$ for $x_1$, $7$ for $y_1$, $11$ for $x_2$, and $11$ for $y_2$ in the equation $m=\frac{y_2-y_1}{x_2-x_1}$ yields $m=\frac{11-7}{11-0}$, or $m=\frac{4}{11}$. Substituting $\frac{4}{11}$ for $m$ and $7$ for $b$ in the equation $y=mx+b$ yields $y=\frac{4}{11}x+7$. Therefore, an equation of this line is $y=\frac{4}{11}x+7$. Similarly, the other line shown in the graph passes through the points $(0,13)$ and $(11,11)$. Substituting $0$ for $x_1$, $13$ for $y_1$, $11$ for $x_2$, and $11$ for $y_2$ in the equation $m=\frac{y_2-y_1}{x_2-x_1}$ yields $m=\frac{11-13}{11-0}$, or $m=\frac{-2}{11}$. Substituting $\frac{-2}{11}$ for $m$ and $13$ for $b$ in the equation $y=mx+b$ yields $y=\frac{-2}{11}x+13$. Therefore, an equation of this line is $y=\frac{-2}{11}x+13$. So, the functions $f$ and $g$ could be represented by $f(x)=\frac{-2}{11}x+13$ and $g(x)=\frac{4}{11}x+7$.