Question: Determine whether the equation below has one solution, no solutions, or an infinite number of solutions. Afterwards, determine two values of $x$ that support your conclusion.
$$\frac{33}{x} = -3$$
1. **State the problem:** We need to find whether the equation $$\frac{33}{x} = -3$$ has one solution, no solutions, or infinitely many solutions.
2. **Recall the formula and rules:** The equation is a rational equation. To solve it, multiply both sides by $x$ (assuming $x \neq 0$) to clear the denominator:
$$33 = -3x$$
3. **Solve for $x$:**
Divide both sides by $-3$:
$$x = \frac{33}{-3}$$
Show cancellation:
$$x = \frac{\cancel{33}}{\cancel{-3}} = -11$$
4. **Interpretation:** There is exactly one solution, $x = -11$.
5. **Check values:**
- Substitute $x = -11$ into the original equation:
$$\frac{33}{-11} = -3$$
which is true.
- Substitute a different value, for example $x = 1$:
$$\frac{33}{1} = 33 \neq -3$$
which is false.
6. **Conclusion:** The equation has one solution, $x = -11$.
Two values of $x$ to support this conclusion are:
- $x = -11$ (makes the equation true)
- $x = 1$ (makes the equation false)