We are given the equation:

\[
\frac{2}{x-3} - \frac{3}{x+2} = \frac{x+7}{x^2 - x - 6}
\]

### Step 1: Factor the denominator on the right-hand side
We need to factor \( x^2 - x - 6 \). We look for two numbers that multiply to \(-6\) and add to \(-1\), which are \(-3\) and \(2\). Thus, we factor the quadratic as:

\[
x^2 - x - 6 = (x - 3)(x + 2)
\]

So, the equation becomes:

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

### Step 2: Combine the left-hand side
To combine the fractions on the left-hand side, we need a common denominator. The common denominator is \((x-3)(x+2)\). We rewrite the terms with this common denominator:

\[
\frac{2}{x-3} = \frac{2(x+2)}{(x-3)(x+2)} \quad \text{and} \quad \frac{3}{x+2} = \frac{3(x-3)}{(x-3)(x+2)}
\]

Thus, the left-hand side becomes:

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

Simplifying the numerator:

\[
2(x+2) = 2x + 4 \quad \text{and} \quad 3(x-3) = 3x - 9
\]

So:

\[
2(x+2) - 3(x-3) = (2x + 4) - (3x - 9) = 2x + 4 - 3x + 9 = -x + 13
\]

Thus, the left-hand side becomes:

\[
\frac{-x + 13}{(x-3)(x+2)}
\]

### Step 3: Set the two sides equal
Now, we set the two sides equal:

\[
\frac{-x + 13}{(x-3)(x+2)} = \frac{x + 7}{(x-3)(x+2)}
\]

Since the denominators are the same, we can equate the numerators:

\[
-x + 13 = x + 7
\]

### Step 4: Solve for \(x\)
Now, solve the equation \( -x + 13 = x + 7 \):

\[
13 - 7 = x + x
\]

\[
6 = 2x
\]

\[
x = 3
\]

### Step 5: Check for extraneous solutions
We must check whether \(x = 3\) is a valid solution by plugging it into the original equation. If \(x = 3\), the terms \(\frac{2}{x-3}\) and \(\frac{3}{x+2}\) would involve division by zero, which is undefined.

Thus, \(x = 3\) is not a valid solution.

### Final Answer:
There is **no solution** to this equation because \(x = 3\) makes the original equation undefined.

---

Would you like further details or have any other questions?

Here are 5 related problems you might find helpful:
1. Solve \( \frac{3}{x-2} - \frac{4}{x+1} = \frac{7}{x^2 - x - 2} \).
2. Solve \( \frac{1}{x-1} + \frac{2}{x+2} = \frac{3}{x^2 + x - 2} \).
3. Factor the expression \(x^2 - 7x + 10\).
4. Solve \( \frac{5}{x-5} - \frac{6}{x+3} = 0 \).
5. Solve \( \frac{1}{x-4} + \frac{2}{x+5} = \frac{3}{x^2+x-20} \).

**Tip**: Always check for extraneous solutions, especially when dealing with rational equations, as division by zero can occur!

Learn how to solve the rational equation 2/(x-3) - 3/(x+2) = (x+7)/(x^2 - x - 6) by factoring the quadratic denominator and combining terms. Includes step-by-step solution, factoring techniques, and a check for extraneous solutions. Suitable for high school students in Grades 9-12.

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