Solving Rational Equations and Inequalities Part 3 - Free Printable

Problem: Solve the inequality

\[
\frac{x+3}{x+1} \geq \frac{x-2}{x-3}
\]
with the restrictions \( x \neq -1 \) and \( x \neq 3 \).

Solution:

#### Step 1: Understand the problem and restrictions
The inequality involves rational expressions, so we need to be careful about the values of \( x \) that make the denominators zero. The denominators are \( x+1 \) and \( x-3 \), so the restrictions are:
\[
x \neq -1 \quad \text{and} \quad x \neq 3.
\]

#### Step 2: Combine the fractions
To solve the inequality, we first combine the fractions on one side:
\[
\frac{x+3}{x+1} - \frac{x-2}{x-3} \geq 0.
\]

Find a common denominator, which is \( (x+1)(x-3) \):
\[
\frac{(x+3)(x-3) - (x-2)(x+1)}{(x+1)(x-3)} \geq 0.
\]

#### Step 3: Simplify the numerator
Expand the terms in the numerator:
\[
(x+3)(x-3) = x^2 - 9,
\]
\[
(x-2)(x+1) = x^2 - x - 2.
\]

Substitute these back into the numerator:
\[
(x^2 - 9) - (x^2 - x - 2) = x^2 - 9 - x^2 + x + 2 = x - 7.
\]

So the inequality becomes:
\[
\frac{x - 7}{(x+1)(x-3)} \geq 0.
\]

#### Step 4: Determine the critical points
The critical points are the values of \( x \) that make the numerator or the denominator zero. These are:
\[
x = 7, \quad x = -1, \quad x = 3.
\]

These points divide the real number line into intervals:
\[
(-\infty, -1), \quad (-1, 3), \quad (3, 7), \quad (7, \infty).
\]

#### Step 5: Test each interval
We need to determine the sign of the expression \( \frac{x-7}{(x+1)(x-3)} \) in each interval. Choose a test point from each interval and evaluate the sign of the expression.

1. Interval \( (-\infty, -1) \):
Choose \( x = -2 \):
\[
\frac{-2-7}{(-2+1)(-2-3)} = \frac{-9}{(-1)(-5)} = \frac{-9}{5} < 0.
\]

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

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

4. Interval \( (7, \infty) \):
Choose \( x = 8 \):
\[
\frac{8-7}{(8+1)(8-3)} = \frac{1}{(9)(5)} = \frac{1}{45} > 0.
\]

#### Step 6: Include the critical points
We need to check the behavior at the critical points \( x = -1 \), \( x = 3 \), and \( x = 7 \):
- At \( x = -1 \) and \( x = 3 \), the expression is undefined because the denominator is zero.
- At \( x = 7 \), the expression is zero, which satisfies the inequality \( \geq 0 \).

#### Step 7: Combine the results
From the sign analysis, the expression \( \frac{x-7}{(x+1)(x-3)} \geq 0 \) is satisfied in the intervals:
\[
(-1, 3) \quad \text{and} \quad [7, \infty).
\]

#### Final Answer:
\[
\boxed{(-1, 3) \cup [7, \infty)}
\]