Question

Solve the inequality and graph the solution set on a real number line. Express the solution set in interval notation.
StartFraction x squared minus 3 x minus 4 Over x squared minus 10 x plus 21 EndFraction greater than 0
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Part 1
The solution set is
  
enter your response here.
​(Type your answer in interval notation. Use integers or fractions for any numbers in the​ expression.)

Answer 1

To solve the inequality

\[
\frac{x^2 - 3x - 4}{x^2 - 10x + 21} > 0,
\]

we first factor both the numerator and the denominator.

1. **Factor the numerator:**

The numerator \(x^2 - 3x - 4\) can be factored as:
\[
x^2 - 3x - 4 = (x - 4)(x + 1).
\]

2. **Factor the denominator:**

The denominator \(x^2 - 10x + 21\) can be factored as:
\[
x^2 - 10x + 21 = (x - 7)(x - 3).
\]

Now we rewrite the inequality as:

\[
\frac{(x - 4)(x + 1)}{(x - 7)(x - 3)} > 0.
\]

Next, we identify the critical points by setting the numerator and denominator equal to zero:

- For \( (x - 4)(x + 1) = 0 \):
- \(x - 4 = 0 \Rightarrow x = 4\)
- \(x + 1 = 0 \Rightarrow x = -1\)

- For \( (x - 7)(x - 3) = 0 \):
- \(x - 7 = 0 \Rightarrow x = 7\)
- \(x - 3 = 0 \Rightarrow x = 3\)

The critical points are \(x = -1\), \(x = 3\), \(x = 4\), and \(x = 7\).

Now, we will test the intervals determined by these critical points:

1. **Intervals to test**: \((- \infty, -1)\), \((-1, 3)\), \((3, 4)\), \((4, 7)\), \((7, \infty)\).

2. **Choose test points for each interval**:

- For \(x < -1\) (let's take \(x = -2\)):
\[
\frac{(-2 - 4)(-2 + 1)}{(-2 - 7)(-2 - 3)} = \frac{(-6)(-1)}{(-9)(-5)} = \frac{6}{45} > 0.
\]

- For \(-1 < x < 3\) (let's take \(x = 0\)):
\[
\frac{(0 - 4)(0 + 1)}{(0 - 7)(0 - 3)} = \frac{(-4)(1)}{(-7)(-3)} = \frac{-4}{21} < 0.
\]

- For \(3 < x < 4\) (let's take \(x = 3.5\)):
\[
\frac{(3.5 - 4)(3.5 + 1)}{(3.5 - 7)(3.5 - 3)} = \frac{(-0.5)(4.5)}{(-3.5)(0.5)} = \frac{-2.25}{-1.75} > 0.
\]

- For \(4 < x < 7\) (let's take \(x = 5\)):
\[
\frac{(5 - 4)(5 + 1)}{(5 - 7)(5 - 3)} = \frac{(1)(6)}{(-2)(2)} = \frac{6}{-4} < 0.
\]

- For \(x > 7\) (let's take \(x = 8\)):
\[
\frac{(8 - 4)(8 + 1)}{(8 - 7)(8 - 3)} = \frac{(4)(9)}{(1)(5)} = \frac{36}{5} > 0.
\]

3. **Summary of the test results**:
- The inequality is satisfied in the intervals: \((- \infty, -1)\), \((3, 4)\), and \((7, \infty)\).

4. **Final solution in interval notation**:
Now we combine the intervals:

\[
(-\infty, -1) \cup (3, 4) \cup (7, \infty).
\]

So the solution set is:

\[
\text{The solution set is } (-\infty, -1) \cup (3, 4) \cup (7, \infty).
\]