Question

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

Explanation:

Step1: Simplify denominators

Factor denominators:
$2x-6=2(x-3)$, $6x-18=6(x-3)$
Equation becomes: $\frac{x^2}{2(x-3)}=\frac{9}{6(x-3)}$

Step2: Eliminate common factors

Multiply both sides by $6(x-3)$ (where $x
eq3$):
$3x^2=9$

Step3: Solve for $x$

Divide by 3: $x^2=3$
Take square roots: $x=\pm\sqrt{3}$

Step4: Check for extraneous solutions

Substitute $x=\sqrt{3}$ and $x=-\sqrt{3}$ into original equation:
Denominators $2(\sqrt{3})-6
eq0$, $6(\sqrt{3})-18
eq0$; $2(-\sqrt{3})-6
eq0$, $6(-\sqrt{3})-18
eq0$. Both satisfy the equation.

Answer:

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