What is true about the solution of \( \frac{x^{2}}{2 x-6}=\frac{9}{6 x-18} ? \)

To solve the equation \( \frac{x^{2}}{2 x-6}=\frac{9}{6 x-18} \), you'll first notice that the denominator in the second fraction, \( 6x - 18 \), can be factored: \( 6x - 18 = 6(x - 3) \). Similarly, the first denominator \( 2x - 6 \) also factors: \( 2x - 6 = 2(x - 3) \). This means both sides have a common factor of \( x - 3 \), which leads us to potential solutions, but we must also check for extraneous solutions that may arise from cancelling these factors. Next, by cross-multiplying to eliminate the fractions, you'll see the equation turns into \( x^2 \cdot 6(x - 3) = 9 \cdot 2(x - 3) \). Here, we can safely divide both sides by \( x - 3 \) (keeping in mind that \( x \neq 3 \) to avoid division by zero). Simplifying reveals that \( x^2 = 3 \), leading to two potential solutions: \( x = \sqrt{3} \) and \( x = -\sqrt{3} \). However, our original constraints mean we need to discard \( x = 3 \) from the solution set, leaving us with the two valid solutions.