What is true about the solution of $$ \frac{x^{2}}{2 x-6}=\frac{9}{6 x-18} ? $$ $$ x= \pm \sqrt{3} $$, 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.

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**Step 1. Domain Restriction**

The given equation is

$$
\frac{x^{2}}{2x-6}=\frac{9}{6x-18}\,.
$$

Notice that the denominators cannot be zero. Thus, we require:

- $$2x-6 \neq 0 \implies x \neq 3$$,
- $$6x-18 \neq 0 \implies x \neq 3$$.

Thus, $$x=3$$ is excluded.

**Step 2. Simplify the Equation**

Observe that

$$
6x-18 = 3(2x-6)\,.
$$

Thus, the equation can be written as:

$$
\frac{x^2}{2x-6}=\frac{9}{3(2x-6)}\,.
$$

Multiply both sides of the equation by $$3(2x-6)$$ (which is allowed as $$2x-6 \neq 0$$):

$$
3(2x-6)\cdot\frac{x^2}{2x-6} = 3(2x-6)\cdot\frac{9}{3(2x-6)}\,.
$$

Simplifying, we cancel the common factor $$2x-6$$:

$$
3x^2 = 9\,.
$$

**Step 3. Solve for $$x$$**

Divide both sides by 3:

$$
x^2 = 3\,.
$$

Taking the square root:

$$
x = \pm \sqrt{3}\,.
$$

**Step 4. Verify the Solutions**

The solutions $$x=\sqrt{3}$$ and $$x=-\sqrt{3}$$ do not violate the domain restriction $$x \neq 3$$.

**Conclusion**

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

What is true about the solution of \( \frac{x^{2}}{2 x-6}=\frac{9}{6 x-18} ? \) \( x= \pm \sqrt{3} \), 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.

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