SAT Question 153ee763: Linear Equations In One Variable answer and explanation with Desmos solution

Choice B is correct. A linear equation in one variable has no solution if and only if the equation is false; that is, when there is no value of $x$ that produces a true statement. It's given that in the equation $-3x+21px=84$, $p$ is a constant and the equation has no solution for $x$. Therefore, the value of the constant $p$ is one that results in a false equation. Factoring out the common factor of $-3x$ on the left-hand side of the given equation yields $-3x\left(1-7p\right)=84$. Dividing both sides of this equation by $-3$ yields $x\left(1-7p\right)=-28$. Dividing both sides of this equation by $\left(1-7p\right)$ yields $x=\frac{-28}{1-7p}$. This equation is false if and only if $1-7p=0$. Adding $7p$ to both sides of $1-7p=0$ yields $1=7p$. Dividing both sides of this equation by $7$ yields $\frac{1}{7}=p$. It follows that the equation $x=\frac{-28}{1-7p}$ is false if and only if $p=\frac{1}{7}$. Therefore, the given equation has no solution if and only if the value of $p$ is $\frac{1}{7}$.