SAT Question 3f5a3602: Systems of Linear Equations answer and explanation with Desmos solution

Choice D is correct. A line in the xy-plane that passes through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ has slope $m$, where $m=\frac{y_2-y_1}{x_2-x_1}$, and can be defined by an equation of the form $y-y_1=m\left(x-x_1\right)$. One of the lines shown in the graph passes through the points $\left(8,0\right)$ and $\left(3,4\right)$. Substituting $8$ for $x_1$, $0$ for $y_1$, $3$ for $x_2$, and $4$ for $y_2$ in the equation $m=\frac{y_2-y_1}{x_2-x_1}$ yields $m=\frac{4-0}{3-8}$, or $m=-\frac{4}{5}$. Substituting $-\frac{4}{5}$ for $m$, $8$ for $x_1$ and $0$ for $y_1$ in the equation $y-y_1=m\left(x-x_1\right)$ yields $y-0=-\frac{4}{5}\left(x-8\right)$, which is equivalent to $y=-\frac{4}{5}x+\frac{32}{5}$. Adding $\frac{4}{5}x$ to both sides of this equation yields $\frac{4}{5}x+y=\frac{32}{5}$. Multiplying both sides of this equation by $-10$ yields $-8x-10y=-64$. Therefore, an equation of this line is $-8x-10y=-64$. Similarly, the other line shown in the graph passes through the points $\left(8,0\right)$ and $\left(3,2\right)$. Substituting $8$ for $x_1$, $0$ for $y_1$, $3$ for $x_2$, and $2$ for $y_2$ in the equation $m=\frac{y_2-y_1}{x_2-x_1}$ yields $m=\frac{2-0}{3-8}$, or $m=-\frac{2}{5}$. Substituting $-\frac{2}{5}$ for $m$, $8$ for $x_1$, and $0$ for $y_1$ in the equation $y-y_1=m\left(x-x_1\right)$ yields $y-0=-\frac{2}{5}\left(x-8\right)$, which is equivalent to $y=-\frac{2}{5}x+\frac{16}{5}$. Adding $\frac{2}{5}x$ to both sides of this equation yields $\frac{2}{5}x+y=\frac{16}{5}$. Multiplying both sides of this equation by $10$ yields $4x+10y=32$. Therefore, an equation of this line is $4x+10y=32$. So, the system of linear equations represented by the lines shown is $4x+10y=32$ and $-8x-10y=-64$.