Question: User: $\frac{x^2 - 9}{x - 1} < 0$ show the union and the number line representation and solution
1. **State the problem:**
Solve the inequality $$\frac{x^2 - 9}{x - 1} < 0$$ and find the solution set with its union and number line representation.
2. **Rewrite the expression:**
Factor the numerator:
$$x^2 - 9 = (x - 3)(x + 3)$$
So the inequality becomes:
$$\frac{(x - 3)(x + 3)}{x - 1} < 0$$
3. **Identify critical points:**
The critical points are where numerator or denominator is zero:
- $x = 3$ (numerator zero)
- $x = -3$ (numerator zero)
- $x = 1$ (denominator zero, undefined)
4. **Determine intervals:**
The critical points divide the real line into four intervals:
- $(-\infty, -3)$
- $(-3, 1)$
- $(1, 3)$
- $(3, \infty)$
5. **Test each interval:**
Choose test points in each interval and evaluate the sign of the expression:
- For $x = -4$ in $(-\infty, -3)$:
$$\frac{(-4 - 3)(-4 + 3)}{-4 - 1} = \frac{(-7)(-1)}{-5} = \frac{7}{-5} = -\text{negative}$$
Expression is negative.
- For $x = 0$ in $(-3, 1)$:
$$\frac{(0 - 3)(0 + 3)}{0 - 1} = \frac{(-3)(3)}{-1} = \frac{-9}{-1} = 9$$
Expression is positive.
- For $x = 2$ in $(1, 3)$:
$$\frac{(2 - 3)(2 + 3)}{2 - 1} = \frac{(-1)(5)}{1} = -5$$
Expression is negative.
- For $x = 4$ in $(3, \infty)$:
$$\frac{(4 - 3)(4 + 3)}{4 - 1} = \frac{(1)(7)}{3} = \frac{7}{3}$$
Expression is positive.
6. **Determine solution intervals:**
We want where the expression is less than zero (negative):
- $(-\infty, -3)$
- $(1, 3)$
7. **Check critical points:**
- At $x = -3$ and $x = 3$, numerator is zero, so expression equals zero, not less than zero, exclude these points.
- At $x = 1$, denominator zero, expression undefined, exclude.
8. **Final solution:**
$$(-\infty, -3) \cup (1, 3)$$
9. **Number line representation:**
- Mark points $-3$, $1$, and $3$.
- Open circles at these points (excluded).
- Shade intervals $(-\infty, -3)$ and $(1, 3)$.
**Answer:**
The solution to $$\frac{x^2 - 9}{x - 1} < 0$$ is $$(-\infty, -3) \cup (1, 3)$$.