SAT Question d0e614a6: Linear Equations In Two Variables answer and explanation with Desmos solution

Choice D is correct. Each of the tables gives the same three values of $x$: $1$, $2$, and $4$. Substituting $1$ for $x$ in the given equation yields $\left(\frac{3}{5}\right)\left(1\right)+\frac{3}{4}y=7$, or $\frac{3}{5}+\frac{3}{4}y=\frac{35}{5}$. Subtracting $\frac{3}{5}$ from both sides of this equation yields $\frac{3}{4}y=\frac{32}{5}$. Multiplying both sides of this equation by $\frac{4}{3}$ yields $y=\frac{128}{15}$. Therefore, when $x=1$, the corresponding value of $y$ for the given equation is $\frac{128}{15}$. Substituting $2$ for $x$ in the given equation yields $\left(\frac{3}{5}\right)\left(2\right)+\frac{3}{4}y=7$, or $\frac{6}{5}+\frac{3}{4}y=\frac{35}{5}$. Subtracting $\frac{6}{5}$ from both sides of this equation yields $\frac{3}{4}y=\frac{29}{5}$. Multiplying both sides of this equation by $\frac{4}{3}$ yields $y=\frac{116}{15}$. Therefore, when $x=2$, the corresponding value of $y$ for the given equation is $\frac{116}{15}$. Substituting $4$ for $x$ in the given equation yields $\left(\frac{3}{5}\right)\left(4\right)+\frac{3}{4}y=7$, or $\frac{12}{5}+\frac{3}{4}y=\frac{35}{5}$. Subtracting $\frac{12}{5}$ from both sides of this equation yields $\frac{3}{4}y=\frac{23}{5}$. Multiplying both sides of this equation by $\frac{4}{3}$ yields $y=\frac{92}{15}$. Therefore, when $x=4$, the corresponding value of $y$ for the given equation is $\frac{92}{15}$. The table in choice D gives x-values of $1$, $2$, and $4$ and corresponding y-values of $\frac{128}{15}$, $\frac{116}{15}$, and $\frac{92}{15}$, respectively. Therefore, the table in choice D gives three values of $x$ and their corresponding values of $y$ for the given equation.