To solve the equation 

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

we can cross-multiply to eliminate the fractions. This gives us:

\[
(x - 2)(x + 1) = (x - 1)(x - 3).
\]

Now, we'll expand both sides:

### Expanding the left side:
\[
(x - 2)(x + 1) = x(x + 1) - 2(x + 1) = x^2 + x - 2x - 2 = x^2 - x - 2.
\]

### Expanding the right side:
\[
(x - 1)(x - 3) = x(x - 3) - 1(x - 3) = x^2 - 3x - x + 3 = x^2 - 4x + 3.
\]

Now we equate both sides:

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

Next, subtract \(x^2\) from both sides:

\[
-x - 2 = -4x + 3.
\]

To solve for \(x\), first add \(4x\) to both sides:

\[
3x - 2 = 3.
\]

Now, add 2 to both sides:

\[
3x = 5.
\]

Finally, divide by 3:

\[
x = \frac{5}{3}.
\]

### Check for extraneous solutions:
We should check if \(x = \frac{5}{3}\) makes any denominator in the original equation equal to zero. The denominators are \(x - 3\) and \(x + 1\). For \(x = \frac{5}{3}\), neither denominator equals zero, so this is a valid solution.

Thus, the solution is:

\[
x = \frac{5}{3}.
\]

Would you like any more details or have any questions?

Here are 5 related questions:
1. How would you solve \(\frac{x+2}{x+4} = \frac{x+1}{x+2}\)?
2. What happens if a solution makes the denominator zero?
3. Can cross-multiplication be used in all rational equations?
4. How can you verify the solution of a rational equation?
5. What are extraneous solutions and how do they occur?

**Tip:** Always check if your solution makes any denominator zero, as that would make the solution invalid!

Learn how to solve the rational equation (x-2)/(x-3) = (x-1)/(x+1) step-by-step with detailed explanations and example calculations. This problem involves rational equations and cross-multiplication techniques. Suitable for high school students and anyone preparing for algebra exams.

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