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The problem asks to find the sum of the two fractions \(\frac{3}{x-2}\) and \(\frac{1}{x+3}\).

To add these fractions, we need to find a common denominator:

1. The denominators are \(x-2\) and \(x+3\).
2. The least common denominator (LCD) is \((x-2)(x+3)\).

Now, express each fraction with this common denominator:

\[
\frac{3}{x-2} = \frac{3(x+3)}{(x-2)(x+3)} = \frac{3x + 9}{(x-2)(x+3)}
\]

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

Now, add the two fractions:

\[
\frac{3x + 9}{(x-2)(x+3)} + \frac{x - 2}{(x-2)(x+3)} = \frac{(3x + 9) + (x - 2)}{(x-2)(x+3)}
\]

Simplify the numerator:

\[
\frac{3x + 9 + x - 2}{(x-2)(x+3)} = \frac{4x + 7}{(x-2)(x+3)}
\]

So, the sum is:

\[
\frac{4x + 7}{(x-2)(x+3)}
\]

Would you like more details on this solution, or do you have any questions?

Here are some related questions to consider:
1. What is the least common denominator of two other rational expressions?
2. How do you simplify complex fractions involving multiple variables?
3. What happens to the sum when one of the terms in the denominator is zero?
4. How do you subtract similar rational expressions?
5. What are some applications of adding rational expressions in real-world problems?

**Tip:** Always factor the denominators first when working with rational expressions to simplify finding the least common denominator.

Find the sum of \(\frac{3}{x-2}\) and \(\frac{1}{x+3}\).

Learn how to find the sum of the rational expressions \(\frac{3}{x-2}\) and \(\frac{1}{x+3}\) by determining the least common denominator and simplifying the resulting expression. This problem is suitable for students in Grades 9-10, covering key concepts in algebra.

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