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Let's solve the given equation:

\[
\frac{4x + 1}{x + 1} = \frac{12}{x^2 - 1} + 3
\]

### Step 1: Simplify the equation
First, recognize that \(x^2 - 1\) can be factored as:

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

Now, rewrite the equation:

\[
\frac{4x + 1}{x + 1} = \frac{12}{(x - 1)(x + 1)} + 3
\]

### Step 2: Eliminate the denominator by multiplying through by \((x - 1)(x + 1)\)

Multiply both sides of the equation by \((x - 1)(x + 1)\):

\[
(4x + 1)(x - 1) = 12 + 3(x^2 - 1)
\]

### Step 3: Expand both sides

Expand the left-hand side:

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

Expand the right-hand side:

\[
12 + 3(x^2 - 1) = 12 + 3x^2 - 3 = 3x^2 + 9
\]

### Step 4: Set the equation to zero

Now set both sides equal:

\[
4x^2 - 3x - 1 = 3x^2 + 9
\]

Subtract \(3x^2 + 9\) from both sides:

\[
4x^2 - 3x - 1 - 3x^2 - 9 = 0
\]

Simplify:

\[
x^2 - 3x - 10 = 0
\]

### Step 5: Solve the quadratic equation

Factor the quadratic equation:

\[
x^2 - 3x - 10 = (x - 5)(x + 2) = 0
\]

This gives the solutions:

\[
x = 5 \quad \text{or} \quad x = -2
\]

### Step 6: Check for extraneous solutions

We need to check if either of these values makes any denominator in the original equation zero.

- For \(x = 5\), the denominator \(x + 1\) is not zero, and \(x^2 - 1 = 24\), so it's valid.
- For \(x = -2\), the denominator \(x + 1 = -1\), so it's valid.

Both \(x = 5\) and \(x = -2\) are solutions.

### Final Answer:
The correct response is **"both -2 and 5"**.

---

Would you like further details or have any questions?

Here are some related questions:
1. How do you solve quadratic equations using the quadratic formula?
2. What is the significance of checking for extraneous solutions in rational equations?
3. How do you factor quadratic expressions effectively?
4. Why is factoring useful when solving equations?
5. Can the same method be used for equations with higher degrees?

**Tip:** Always check for undefined values in the denominator when solving rational equations!

What will make the equation true? \( \frac{4x + 1}{x + 1} = \frac{12}{x^2 - 1} + 3 \)

Learn how to solve the rational equation \( \frac{4x + 1}{x + 1} = \frac{12}{x^2 - 1} + 3 \) by simplifying, factoring, and solving for x. This solution involves factoring quadratics and checking for extraneous solutions. Suitable for high school students in Grades 9-12.

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