To solve the equation:

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

### Step 1: Eliminate the common denominator
Since \((x + 1)\) is a common factor in the denominators on both sides, we can multiply both sides by \((x + 1)\) to eliminate it:

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

### Step 2: Cross-multiply
Now, cross-multiply to eliminate the remaining fraction:

\[
(x - 3)(x + 2) = 4x - 6
\]

### Step 3: Expand the left side
Expand the expression on the left-hand side:

\[
x^2 + 2x - 3x - 6 = 4x - 6
\]

Simplify the left-hand side:

\[
x^2 - x - 6 = 4x - 6
\]

### Step 4: Bring all terms to one side
Subtract \(4x\) from both sides:

\[
x^2 - 5x - 6 = -6
\]

Add 6 to both sides to simplify:

\[
x^2 - 5x = 0
\]

### Step 5: Factor the quadratic equation
Factor the left-hand side:

\[
x(x - 5) = 0
\]

### Step 6: Solve for \(x\)
Set each factor equal to zero:

\[
x = 0 \quad \text{or} \quad x = 5
\]

### Step 7: Check for extraneous solutions
We need to verify that these solutions do not make any denominators equal to zero. 

For \(x = 0\):
\[
x + 1 = 1 \quad \text{and} \quad x + 2 = 2 \quad \text{(no issues)}
\]

For \(x = 5\):
\[
x + 1 = 6 \quad \text{and} \quad x + 2 = 7 \quad \text{(no issues)}
\]

Both solutions are valid.

### Final Answer:
The solutions to the equation are \(x = 0\) and \(x = 5\).

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

Here are some related questions:
1. How do you solve quadratic equations by factoring?
2. What are extraneous solutions and how do they arise?
3. How do you check if a solution makes a denominator zero?
4. Can quadratic equations have complex roots?
5. How does the distributive property help in solving equations?

**Tip:** Always check that your solutions do not make any denominators in the original equation equal to zero, as this would make the solution invalid.

This step-by-step guide shows how to solve the rational equation ((x-3)/(x+1)) = ((4x-6)/((x+1)(x+2))). Learn algebraic techniques like cross-multiplication and factoring, and explore key concepts such as the zero-product property. Suitable for students in Grades 9-10.

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