what is true about the solution of $\frac{x^{2}}{2x - 6}=\frac{9}{6x - 18}$?\n$x=pmsqrt{3}$, and they are…

what is true about the solution of $\frac{x^{2}}{2x - 6}=\frac{9}{6x - 18}$?\n$x=pmsqrt{3}$, and they are actual solutions.\n$x=pmsqrt{3}$, but they are extraneous solutions.\n$x = 3$, and it is an actual solution.\n$x=-3$, but it is an extraneous solution.

what is true about the solution of $\frac{x^{2}}{2x - 6}=\frac{9}{6x - 18}$?\n$x=pmsqrt{3}$, and they are actual solutions.\n$x=pmsqrt{3}$, but they are extraneous solutions.\n$x = 3$, and it is an actual solution.\n$x=-3$, but it is an extraneous solution.

Answer

Answer:

A. $x = \pm\sqrt{3}$, and they are actual solutions.

Explanation:

Step1: Simplify the right - hand side denominator

Factor out 3 from $6x - 18$: $6x-18 = 3(2x - 6)$. The equation $\frac{x^{2}}{2x - 6}=\frac{9}{6x - 18}$ becomes $\frac{x^{2}}{2x - 6}=\frac{9}{3(2x - 6)}$.

Step2: Cross - multiply (assuming $x\neq3$)

Since $2x - 6\neq0$ (we'll check this later), we can cross - multiply: $3x^{2}(2x - 6)=9(2x - 6)$. Divide both sides by $(2x - 6)$ (temporarily, $x\neq3$) to get $3x^{2}=9$.

Step3: Solve for $x$

Divide both sides of $3x^{2}=9$ by 3: $x^{2}=3$. Take the square root of both sides: $x=\pm\sqrt{3}$.

Step4: Check for extraneous solutions

The original equation has denominators $2x - 6$ and $6x - 18$. When $x=\pm\sqrt{3}$, $2x-6=2(\pm\sqrt{3})-6\neq0$ and $6x - 18=6(\pm\sqrt{3})-18\neq0$. So $x = \pm\sqrt{3}$ are actual solutions.