Worksheet with 20 rational inequality problems for practice.

To solve rational inequalities, we need to follow a systematic approach. Here's the general method:

1. Rewrite the inequality: Ensure one side is zero.
2. Factorize the numerator and denominator if possible.
3. Find critical points: These are the values of \( x \) that make the numerator or denominator zero.
4. Create a sign chart: Use the critical points to divide the number line into intervals. Test a point in each interval to determine the sign of the expression in that interval.
5. Determine the solution set: Based on the inequality sign (e.g., \( <, >, \leq, \geq \)), identify the intervals where the expression satisfies the inequality.

Let's solve a few of these problems step by step as examples.

---

Problem 17: Solve \( \frac{x-1}{x-10} < 0 \)

#### Step 1: Identify critical points
The critical points are where the numerator or denominator is zero:
- Numerator: \( x - 1 = 0 \implies x = 1 \)
- Denominator: \( x - 10 = 0 \implies x = 10 \)

#### Step 2: Create a sign chart
The critical points divide the number line into three intervals: \( (-\infty, 1) \), \( (1, 10) \), and \( (10, \infty) \).

| Interval | Test Point | Sign of \( x-1 \) | Sign of \( x-10 \) | Sign of \( \frac{x-1}{x-10} \) |
|----------------|------------|--------------------|---------------------|----------------------------------|
| \( (-\infty, 1) \) | \( x = 0 \) | \( - \) | \( - \) | \( + \) |
| \( (1, 10) \) | \( x = 5 \) | \( + \) | \( - \) | \( - \) |
| \( (10, \infty) \) | \( x = 11 \) | \( + \) | \( + \) | \( + \) |

#### Step 3: Determine the solution
The inequality \( \frac{x-1}{x-10} < 0 \) is satisfied where the expression is negative. From the sign chart, this occurs in the interval \( (1, 10) \).

#### Final Answer:
\[ \boxed{(1, 10)} \]

---

Problem 19: Solve \( \frac{2x+5}{x^2+2x-35} \geq 0 \)

#### Step 1: Factorize the denominator
The denominator can be factored as:
\[ x^2 + 2x - 35 = (x + 7)(x - 5) \]

#### Step 2: Identify critical points
The critical points are:
- Numerator: \( 2x + 5 = 0 \implies x = -\frac{5}{2} \)
- Denominator: \( x + 7 = 0 \implies x = -7 \)
- Denominator: \( x - 5 = 0 \implies x = 5 \)

#### Step 3: Create a sign chart
The critical points divide the number line into four intervals: \( (-\infty, -7) \), \( (-7, -\frac{5}{2}) \), \( (-\frac{5}{2}, 5) \), and \( (5, \infty) \).

| Interval | Test Point | Sign of \( 2x+5 \) | Sign of \( x+7 \) | Sign of \( x-5 \) | Sign of \( \frac{2x+5}{(x+7)(x-5)} \) |
|--------------------|------------|---------------------|--------------------|--------------------|----------------------------------------|
| \( (-\infty, -7) \) | \( x = -8 \) | \( - \) | \( - \) | \( - \) | \( + \) |
| \( (-7, -\frac{5}{2}) \) | \( x = -6 \) | \( - \) | \( + \) | \( - \) | \( - \) |
| \( (-\frac{5}{2}, 5) \) | \( x = 0 \) | \( + \) | \( + \) | \( - \) | \( - \) |
| \( (5, \infty) \) | \( x = 6 \) | \( + \) | \( + \) | \( + \) | \( + \) |

#### Step 4: Include critical points where the expression is zero or undefined
- The expression is zero at \( x = -\frac{5}{2} \).
- The expression is undefined at \( x = -7 \) and \( x = 5 \).

#### Step 5: Determine the solution
The inequality \( \frac{2x+5}{x^2+2x-35} \geq 0 \) is satisfied where the expression is non-negative. From the sign chart, this occurs in the intervals \( (-\infty, -7) \) and \( (5, \infty) \), including \( x = -\frac{5}{2} \).

#### Final Answer:
\[ \boxed{(-\infty, -7) \cup \left[-\frac{5}{2}, 5\right)} \]

---

Problem 27: Solve \( \frac{x-3}{2x+5} \geq 1 \)

#### Step 1: Rewrite the inequality
\[ \frac{x-3}{2x+5} - 1 \geq 0 \]
\[ \frac{x-3 - (2x+5)}{2x+5} \geq 0 \]
\[ \frac{x-3-2x-5}{2x+5} \geq 0 \]
\[ \frac{-x-8}{2x+5} \geq 0 \]
\[ \frac{x+8}{2x+5} \leq 0 \] (multiplying by -1 reverses the inequality)

#### Step 2: Identify critical points
The critical points are:
- Numerator: \( x + 8 = 0 \implies x = -8 \)
- Denominator: \( 2x + 5 = 0 \implies x = -\frac{5}{2} \)

#### Step 3: Create a sign chart
The critical points divide the number line into three intervals: \( (-\infty, -8) \), \( (-8, -\frac{5}{2}) \), and \( (-\frac{5}{2}, \infty) \).

| Interval | Test Point | Sign of \( x+8 \) | Sign of \( 2x+5 \) | Sign of \( \frac{x+8}{2x+5} \) |
|--------------------|------------|--------------------|---------------------|----------------------------------|
| \( (-\infty, -8) \) | \( x = -9 \) | \( - \) | \( - \) | \( + \) |
| \( (-8, -\frac{5}{2}) \) | \( x = -7 \) | \( + \) | \( - \) | \( - \) |
| \( (-\frac{5}{2}, \infty) \) | \( x = 0 \) | \( + \) | \( + \) | \( + \) |

#### Step 4: Include critical points where the expression is zero or undefined
- The expression is zero at \( x = -8 \).
- The expression is undefined at \( x = -\frac{5}{2} \).

#### Step 5: Determine the solution
The inequality \( \frac{x+8}{2x+5} \leq 0 \) is satisfied where the expression is non-positive. From the sign chart, this occurs in the interval \( [-8, -\frac{5}{2}) \).

#### Final Answer:
\[ \boxed{[-8, -\frac{5}{2})} \]

---

Problem 34: Solve \( \frac{x}{2} \geq \frac{5}{x+1} + 4 \)

#### Step 1: Rewrite the inequality
\[ \frac{x}{2} - \frac{5}{x+1} - 4 \geq 0 \]
\[ \frac{x(x+1) - 10 - 4(x+1)}{2(x+1)} \geq 0 \]
\[ \frac{x^2 + x - 10 - 4x - 4}{2(x+1)} \geq 0 \]
\[ \frac{x^2 - 3x - 14}{2(x+1)} \geq 0 \]
\[ \frac{(x-7)(x+2)}{2(x+1)} \geq 0 \] (factorizing the quadratic)

#### Step 2: Identify critical points
The critical points are:
- Numerator: \( x - 7 = 0 \implies x = 7 \)
- Numerator: \( x + 2 = 0 \implies x = -2 \)
- Denominator: \( x + 1 = 0 \implies x = -1 \)

#### Step 3: Create a sign chart
The critical points divide the number line into four intervals: \( (-\infty, -2) \), \( (-2, -1) \), \( (-1, 7) \), and \( (7, \infty) \).

| Interval | Test Point | Sign of \( x-7 \) | Sign of \( x+2 \) | Sign of \( x+1 \) | Sign of \( \frac{(x-7)(x+2)}{2(x+1)} \) |
|----------------|------------|--------------------|--------------------|--------------------|----------------------------------------|
| \( (-\infty, -2) \) | \( x = -3 \) | \( - \) | \( - \) | \( - \) | \( + \) |
| \( (-2, -1) \) | \( x = -1.5 \) | \( - \) | \( + \) | \( - \) | \( - \) |
| \( (-1, 7) \) | \( x = 0 \) | \( - \) | \( + \) | \( + \) | \( - \) |
| \( (7, \infty) \) | \( x = 8 \) | \( + \) | \( + \) | \( + \) | \( + \) |

#### Step 4: Include critical points where the expression is zero or undefined
- The expression is zero at \( x = -2 \) and \( x = 7 \).
- The expression is undefined at \( x = -1 \).

#### Step 5: Determine the solution
The inequality \( \frac{(x-7)(x+2)}{2(x+1)} \geq 0 \) is satisfied where the expression is non-negative. From the sign chart, this occurs in the intervals \( (-\infty, -2] \) and \( [7, \infty) \).

#### Final Answer:
\[ \boxed{(-\infty, -2] \cup [7, \infty)} \]

---

These are the solutions for a few of the problems. You can apply the same method to solve the remaining problems. If you need solutions for specific problems, let me know!