Question: User: $\frac{11x-33}{x-3}$ =x
1. **State the problem:** Solve the equation $$\frac{11x-33}{x-3} = x$$ for $x$.
2. **Recall the formula and rules:** To solve rational equations, multiply both sides by the denominator to eliminate the fraction, but remember $x \neq 3$ because the denominator cannot be zero.
3. **Multiply both sides by $x-3$:**
$$\frac{11x-33}{x-3} \times (x-3) = x \times (x-3)$$
which simplifies to
$$11x - 33 = x(x - 3)$$
4. **Expand the right side:**
$$11x - 33 = x^2 - 3x$$
5. **Bring all terms to one side to form a quadratic equation:**
$$0 = x^2 - 3x - 11x + 33$$
$$0 = x^2 - 14x + 33$$
6. **Factor the quadratic:**
We look for two numbers that multiply to $33$ and add to $-14$. These are $-11$ and $-3$.
$$0 = (x - 11)(x - 3)$$
7. **Solve for $x$:**
$$x - 11 = 0 \Rightarrow x = 11$$
$$x - 3 = 0 \Rightarrow x = 3$$
8. **Check for restrictions:**
Since $x = 3$ makes the denominator zero in the original equation, it is not a valid solution.
9. **Final answer:**
$$\boxed{11}$$