Question: User: $\frac{11x-33}{x-3} = x$ $x=11$
1. **State the problem:** Solve the equation $$\frac{11x-33}{x-3} = x$$ and verify the solution $x=11$.
2. **Recall the formula and rules:** The equation involves a rational expression. To solve, 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:** $$0 = x^2 - 3x - 11x + 33$$ which simplifies to $$0 = x^2 - 14x + 33$$
6. **Factor the quadratic:** Find two numbers that multiply to 33 and add to -14. These are -11 and -3.
$$0 = (x - 11)(x - 3)$$
7. **Solve for $x$:** Set each factor equal to zero:
$$x - 11 = 0 \Rightarrow x = 11$$
$$x - 3 = 0 \Rightarrow x = 3$$
8. **Check for restrictions:** $x=3$ is not allowed because it makes the denominator zero.
9. **Verify the solution $x=11$ in the original equation:**
$$\frac{11(11) - 33}{11 - 3} = \frac{121 - 33}{8} = \frac{88}{8} = 11$$ which equals the right side $x=11$.
**Final answer:** $$x = 11$$