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The problem asks to express:

\[
\frac{4}{x - 2} - \frac{3}{y + 1}
\]

as a single fraction and to simplify it.

### Step-by-step solution:

1. **Find a common denominator:**
   The denominators of the two fractions are \(x - 2\) and \(y + 1\). The least common denominator (LCD) will be the product of these two terms: 

   \[
   \text{LCD} = (x - 2)(y + 1)
   \]

2. **Rewrite each fraction with the common denominator:**

   \[
   \frac{4}{x - 2} = \frac{4(y + 1)}{(x - 2)(y + 1)} = \frac{4y + 4}{(x - 2)(y + 1)}
   \]

   \[
   \frac{3}{y + 1} = \frac{3(x - 2)}{(x - 2)(y + 1)} = \frac{3x - 6}{(x - 2)(y + 1)}
   \]

3. **Subtract the two fractions:**

   \[
   \frac{4y + 4}{(x - 2)(y + 1)} - \frac{3x - 6}{(x - 2)(y + 1)} = \frac{(4y + 4) - (3x - 6)}{(x - 2)(y + 1)}
   \]

4. **Simplify the numerator:**

   \[
   (4y + 4) - (3x - 6) = 4y + 4 - 3x + 6 = 4y - 3x + 10
   \]

5. **Final expression:**

   \[
   \frac{4y - 3x + 10}{(x - 2)(y + 1)}
   \]

This is the simplified single fraction.

---

Would you like any further details on the process or a breakdown of any specific step?

Here are 5 questions related to this process:

1. How do you determine the least common denominator (LCD) of two fractions?
2. Why is it necessary to multiply both the numerator and the denominator by the same term when rewriting fractions with a common denominator?
3. What steps would change if the operation were addition instead of subtraction?
4. How can you simplify complex rational expressions further if necessary?
5. Can this process be applied to fractions with more complex denominators, such as those involving exponents?

**Tip:** Always double-check your simplification by expanding the final result and comparing it with the original expressions.

Express 4/(x - 2) - 3/(y + 1) as a single fraction. Give your answer in its simplest form.

Learn how to express the algebraic fractions 4/(x - 2) - 3/(y + 1) as a single simplified fraction. This solution involves finding the least common denominator and simplifying the result. Ideal for students in Grades 9-11.

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