SAT Question 7a5a74a6: Linear Equations In One Variable answer and explanation with Desmos solution

Choice B is correct. Because $2$ is a factor of both $2x$ and $6$, the expression $2x-6$ can be rewritten as $2(x-3)$. Substituting $2(x-3)$ for $2x-6$ on the left-hand side of the given equation yields $3\times 2(x-3)-11=4(x-3)+6$, or $6(x-3)-11=4(x-3)+6$. Subtracting $4(x-3)$ from both sides of this equation yields $2(x-3)-11=6$. Adding $11$ to both sides of this equation yields $2(x-3)=17$. Dividing both sides of this equation by $2$ yields $x-3=\frac{17}{2}$. Alternate approach: Distributing $3$ to the quantity $2x-6$ on the left-hand side of the given equation and distributing $4$ to the quantity $x-3$ on the right-hand side yields $6x-18-11=4x-12+6$, or $6x-29=4x-6$. Subtracting $4x$ from both sides of this equation yields $2x-29=-6$. Adding $29$ to both sides of this equation yields $2x=23$. Dividing both sides of this equation by $2$ yields $x=\frac{23}{2}$. Therefore, the value of $x-3$ is $\frac{23}{2}-3$, or $\frac{17}{2}$.