What is true about the solution of (x^2)/(2x-6)=(9)/(6x-18) x=pm sqrt (3) , and they are actual solutions. x=pm sqrt (3) , but they are extraneous solutions. x=3, and it is an actual solution. x=-3 but it is an extraneous solution.

To determine which statement is true about the solution of the equation $\frac{x^2}{2x-6}=\frac{9}{6x-18}$, we need to solve the equation step by step.<br /><br />First, let's simplify the equation:<br /><br />$\frac{x^2}{2x-6}=\frac{9}{6x-18}$<br /><br />We can factor out a common factor of 2 from the denominator of the left side:<br /><br />$\frac{x^2}{2(x-3)}=\frac{9}{6(x-3)}$<br /><br />Next, we can simplify the right side by dividing both the numerator and denominator by 3:<br /><br />$\frac{x^2}{2(x-3)}=\frac{3}{2(x-3)}$<br /><br />Now, we can cross-multiply to eliminate the fractions:<br /><br />$x^2 \cdot 2(x-3) = 3 \cdot 2(x-3)$<br /><br />Simplifying further, we get:<br /><br />$2x^3 - 6x^2 = 6x - 9$<br /><br />Rearranging the terms, we have:<br /><br />$2x^3 - 6x^2 - 6x + 9 = 0$<br /><br />This is a cubic equation, and solving it may require numerical methods or factoring techniques. However, we can observe that x = 3 is a solution by substituting it into the equation:<br /><br />$2(3)^3 - 6(3)^2 - 6(3) + 9 = 54 - 54 - 18 + 9 = 0$<br /><br />Therefore, x = 3 is an actual solution to the equation.<br /><br />So, the correct statement is: "x=3, and it is an actual solution."