We are given the equation:

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

To solve this, let's follow these steps:

### Step 1: Cross-multiply
We eliminate the denominators by cross-multiplying:

\[
(2x + 3)(x + 2) = (4x - 1)(x - 1)
\]

### Step 2: Expand both sides
First, expand the left-hand side:

\[
(2x + 3)(x + 2) = 2x(x + 2) + 3(x + 2) = 2x^2 + 4x + 3x + 6 = 2x^2 + 7x + 6
\]

Now, expand the right-hand side:

\[
(4x - 1)(x - 1) = 4x(x - 1) - 1(x - 1) = 4x^2 - 4x - x + 1 = 4x^2 - 5x + 1
\]

### Step 3: Set up the equation
Now that we have both sides expanded, set them equal to each other:

\[
2x^2 + 7x + 6 = 4x^2 - 5x + 1
\]

### Step 4: Move all terms to one side
We want to set the equation to 0, so move all terms from the right-hand side to the left:

\[
2x^2 + 7x + 6 - 4x^2 + 5x - 1 = 0
\]

Simplify this:

\[
-2x^2 + 12x + 5 = 0
\]

### Step 5: Solve the quadratic equation
Now, solve the quadratic equation \(-2x^2 + 12x + 5 = 0\). To do this, we can either factor it or use the quadratic formula. The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \(-2x^2 + 12x + 5 = 0\), the coefficients are:

- \(a = -2\)
- \(b = 12\)
- \(c = 5\)

Substitute these into the quadratic formula:

\[
x = \frac{-12 \pm \sqrt{12^2 - 4(-2)(5)}}{2(-2)}
\]
\[
x = \frac{-12 \pm \sqrt{144 + 40}}{-4}
\]
\[
x = \frac{-12 \pm \sqrt{184}}{-4}
\]
\[
x = \frac{-12 \pm 2\sqrt{46}}{-4}
\]
\[
x = \frac{12 \pm \sqrt{46}}{4}
\]

So, the solutions are:

\[
x = \frac{12 + \sqrt{46}}{4} \quad \text{or} \quad x = \frac{12 - \sqrt{46}}{4}
\]

These are the two possible solutions for \(x\).

### Final Step: Check for extraneous solutions
We should check whether these values of \(x\) result in any denominators being zero. Since the original denominators are \(x - 1\) and \(x + 2\), neither solution should make the denominators zero, so both are valid.

Would you like to go into more details about any step?

Here are 5 related questions to extend this topic:
1. How can you factor a quadratic equation instead of using the quadratic formula?
2. What are extraneous solutions, and how do they occur in rational equations?
3. How does cross-multiplication work in solving equations with fractions?
4. Can all quadratic equations be solved using factoring?
5. What is the geometric meaning of the quadratic formula solutions?

**Tip:** Always check for extraneous solutions when dealing with rational equations!

This guide explains how to solve the rational equation 2x + 3/(x - 1) = 4x - 1/(x + 2) by using cross-multiplication and applying the quadratic formula. Key steps include expanding and simplifying the equation, followed by solving the quadratic equation. Suitable for Grades 9-11.

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