SAT Question fc3dfa26: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice C is correct. Each fraction in the given equation can be expressed with the common denominator $x^2-9$. Multiplying $\frac{2x}{x+3}$ by $\frac{x-3}{x-3}$ yields $\frac{2x^2-6x}{x^2-9}$, and multiplying $\frac{1}{x-3}$ by $\frac{x+3}{x+3}$ yields $\frac{x+3}{x^2-9}$. Therefore, the given equation can be written as $\frac{4x^2}{x^2-9}-\frac{2x^2-6x}{x^2-9}=\frac{x+3}{x^2-9}$. Multiplying each fraction by the denominator results in the equation $4x^2-(2x^2-6x)=x+3$, or $2x^2+6x=x+3$. This equation can be solved by setting a quadratic expression equal to $0$, then solving for $x$. Subtracting $x+3$ from both sides of this equation yields $2x^2+5x-3=0$. The expression $2x^2+5x-3$ can be factored, resulting in the equation $(2x-1)\times (x+3)=0$. By the zero product property, $2x-1=0$ or $x+3=0$. To solve for $x$ in $2x-1=0$, $1$ can be added to both sides of the equation, resulting in $2x=1$. Dividing both sides of this equation by $2$ results in $x=\frac{1}{2}$. Solving for $x$ in $x+3=0$ yields $x=-3$. However, this value of $x$ would result in the second fraction of the original equation having a denominator of $0$. Therefore, $x=-3$ is an extraneous solution. Thus, the only value of $x$ that satisfies the given equation is $x=\frac{1}{2}$.