Solving rational inequalities homework assignment with eight problems involving algebraic expressions and inequalities.

To solve the given rational inequalities, we will follow a systematic approach for each problem. The general steps involve:

1. Factorize the numerator and denominator (if possible).
2. Find the critical points (where the numerator or denominator is zero).
3. Determine the intervals defined by these critical points.
4. Test the sign of the expression in each interval.
5. Consider the inequality condition to determine the solution set.

Let's solve each problem step by step.

---

Problem 1:

\[
\frac{(x+2)(x-1)}{(x+4)(x-7)} > 0
\]

#### Step 1: Identify critical points
The critical points are where the numerator or denominator is zero:
- Numerator: \( (x+2)(x-1) = 0 \) → \( x = -2, 1 \)
- Denominator: \( (x+4)(x-7) = 0 \) → \( x = -4, 7 \)

#### Step 2: Determine intervals
The critical points divide the real line into intervals:
\[
(-\infty, -4), (-4, -2), (-2, 1), (1, 7), (7, \infty)
\]

#### Step 3: Test the sign in each interval
We test a point in each interval to determine the sign of the expression.

- Interval \( (-\infty, -4) \): Choose \( x = -5 \)
\[
\frac{(-5+2)(-5-1)}{(-5+4)(-5-7)} = \frac{(-3)(-6)}{(-1)(-12)} = \frac{18}{12} > 0
\]

- Interval \( (-4, -2) \): Choose \( x = -3 \)
\[
\frac{(-3+2)(-3-1)}{(-3+4)(-3-7)} = \frac{(-1)(-4)}{(1)(-10)} = \frac{4}{-10} < 0
\]

- Interval \( (-2, 1) \): Choose \( x = 0 \)
\[
\frac{(0+2)(0-1)}{(0+4)(0-7)} = \frac{(2)(-1)}{(4)(-7)} = \frac{-2}{-28} > 0
\]

- Interval \( (1, 7) \): Choose \( x = 2 \)
\[
\frac{(2+2)(2-1)}{(2+4)(2-7)} = \frac{(4)(1)}{(6)(-5)} = \frac{4}{-30} < 0
\]

- Interval \( (7, \infty) \): Choose \( x = 8 \)
\[
\frac{(8+2)(8-1)}{(8+4)(8-7)} = \frac{(10)(7)}{(12)(1)} = \frac{70}{12} > 0
\]

#### Step 4: Consider the inequality
The expression is positive in the intervals \( (-\infty, -4) \), \( (-2, 1) \), and \( (7, \infty) \). Since the inequality is strict (\( > 0 \)), we exclude the critical points where the expression is zero or undefined.

#### Solution:
\[
(-\infty, -4) \cup (-2, 1) \cup (7, \infty)
\]

---

Problem 2:

\[
\frac{3x-12}{x^2-4x-5} \leq 0
\]

#### Step 1: Factorize
- Numerator: \( 3x - 12 = 3(x - 4) \)
- Denominator: \( x^2 - 4x - 5 = (x - 5)(x + 1) \)

So the expression becomes:
\[
\frac{3(x-4)}{(x-5)(x+1)}
\]

#### Step 2: Identify critical points
- Numerator: \( 3(x-4) = 0 \) → \( x = 4 \)
- Denominator: \( (x-5)(x+1) = 0 \) → \( x = -1, 5 \)

#### Step 3: Determine intervals
The critical points divide the real line into intervals:
\[
(-\infty, -1), (-1, 4), (4, 5), (5, \infty)
\]

#### Step 4: Test the sign in each interval
We test a point in each interval to determine the sign of the expression.

- Interval \( (-\infty, -1) \): Choose \( x = -2 \)
\[
\frac{3(-2-4)}{(-2-5)(-2+1)} = \frac{3(-6)}{(-7)(-1)} = \frac{-18}{7} < 0
\]

- Interval \( (-1, 4) \): Choose \( x = 0 \)
\[
\frac{3(0-4)}{(0-5)(0+1)} = \frac{3(-4)}{(-5)(1)} = \frac{-12}{-5} > 0
\]

- Interval \( (4, 5) \): Choose \( x = 4.5 \)
\[
\frac{3(4.5-4)}{(4.5-5)(4.5+1)} = \frac{3(0.5)}{(-0.5)(5.5)} = \frac{1.5}{-2.75} < 0
\]

- Interval \( (5, \infty) \): Choose \( x = 6 \)
\[
\frac{3(6-4)}{(6-5)(6+1)} = \frac{3(2)}{(1)(7)} = \frac{6}{7} > 0
\]

#### Step 5: Consider the inequality
The expression is less than or equal to zero in the intervals \( (-\infty, -1) \) and \( (4, 5) \). We include the points where the expression is zero (\( x = 4 \)) but exclude the points where it is undefined (\( x = -1, 5 \)).

#### Solution:
\[
(-\infty, -1) \cup [4, 5)
\]

---

Problem 3:

\[
\frac{2x^2 + 2x}{x^2 - 3x - 18} \geq 0
\]

#### Step 1: Factorize
- Numerator: \( 2x^2 + 2x = 2x(x + 1) \)
- Denominator: \( x^2 - 3x - 18 = (x - 6)(x + 3) \)

So the expression becomes:
\[
\frac{2x(x+1)}{(x-6)(x+3)}
\]

#### Step 2: Identify critical points
- Numerator: \( 2x(x+1) = 0 \) → \( x = 0, -1 \)
- Denominator: \( (x-6)(x+3) = 0 \) → \( x = -3, 6 \)

#### Step 3: Determine intervals
The critical points divide the real line into intervals:
\[
(-\infty, -3), (-3, -1), (-1, 0), (0, 6), (6, \infty)
\]

#### Step 4: Test the sign in each interval
We test a point in each interval to determine the sign of the expression.

- Interval \( (-\infty, -3) \): Choose \( x = -4 \)
\[
\frac{2(-4)(-4+1)}{(-4-6)(-4+3)} = \frac{2(-4)(-3)}{(-10)(-1)} = \frac{24}{10} > 0
\]

- Interval \( (-3, -1) \): Choose \( x = -2 \)
\[
\frac{2(-2)(-2+1)}{(-2-6)(-2+3)} = \frac{2(-2)(-1)}{(-8)(1)} = \frac{4}{-8} < 0
\]

- Interval \( (-1, 0) \): Choose \( x = -0.5 \)
\[
\frac{2(-0.5)(-0.5+1)}{(-0.5-6)(-0.5+3)} = \frac{2(-0.5)(0.5)}{(-6.5)(2.5)} = \frac{-0.5}{-16.25} > 0
\]

- Interval \( (0, 6) \): Choose \( x = 1 \)
\[
\frac{2(1)(1+1)}{(1-6)(1+3)} = \frac{2(1)(2)}{(-5)(4)} = \frac{4}{-20} < 0
\]

- Interval \( (6, \infty) \): Choose \( x = 7 \)
\[
\frac{2(7)(7+1)}{(7-6)(7+3)} = \frac{2(7)(8)}{(1)(10)} = \frac{112}{10} > 0
\]

#### Step 5: Consider the inequality
The expression is greater than or equal to zero in the intervals \( (-\infty, -3) \), \( (-1, 0] \), and \( (6, \infty) \). We include the points where the expression is zero (\( x = 0, -1 \)) but exclude the points where it is undefined (\( x = -3, 6 \)).

#### Solution:
\[
(-\infty, -3) \cup (-1, 0] \cup (6, \infty)
\]

---

Final Answer:

\[
\boxed{
\begin{aligned}
1. & \quad (-\infty, -4) \cup (-2, 1) \cup (7, \infty) \\
2. & \quad (-\infty, -1) \cup [4, 5) \\
3. & \quad (-\infty, -3) \cup (-1, 0] \cup (6, \infty)
\end{aligned}
}
\]