The equation in the image is:

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

Step 1: Find the Least Common Denominator (LCD)

The denominators in the equation are $x + 1$, 1, and $x^2 - 1$. Notice that $x^2 - 1$ can be factored as $(x - 1)(x + 1)$.

The LCD is $(x + 1)(x - 1)$.

Step 2: Multiply Both Sides by the LCD

Multiplying the entire equation by the LCD $(x + 1)(x - 1)$ gives:

$\frac{4x + 1}{x + 1} \cdot (x + 1)(x - 1) - \frac{3}{1} \cdot (x + 1)(x - 1) = \frac{12}{x^2 - 1} \cdot (x + 1)(x - 1)$

Simplifying each term:

1. The first term: $\frac{4x + 1}{x + 1} \cdot (x + 1)(x - 1) = (4x + 1)(x - 1)$

2. The second term: $-3(x + 1)(x - 1)$

3. The third term: $\frac{12}{(x + 1)(x - 1)} \cdot (x + 1)(x - 1) = 12$

So, the equation becomes:

$(4x + 1)(x - 1) - 3(x + 1)(x - 1) = 12$

Step 3: Apply the Distributive Property and Simplify

Now, expand and simplify each part:

1. Expand $(4x + 1)(x - 1)$: $4x^2 - 4x + x - 1 = 4x^2 - 3x - 1$

2. Expand $3(x + 1)(x - 1)$: $3(x^2 - 1) = 3x^2 - 3$

Substitute these back into the equation:

$4x^2 - 3x - 1 - (3x^2 - 3) = 12$

Simplify:

$4x^2 - 3x - 1 - 3x^2 + 3 = 12$

Combine like terms:

$x^2 - 3x + 2 = 12$

Step 4: Solve for $x$

Now, bring all terms to one side of the equation:

$x^2 - 3x + 2 - 12 = 0$

Simplify further:

$x^2 - 3x - 10 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

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

Substitute $a = 1$, $b = -3$, and $c = -10$:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}$

$x = \frac{3 \pm \sqrt{9 + 40}}{2}$

$x = \frac{3 \pm \sqrt{49}}{2}$

$x = \frac{3 \pm 7}{2}$

Thus, the two possible values for $x$ are:

$x = \frac{10}{2} = 5$ $x = \frac{-4}{2} = -2$

Step 5: Check for Extraneous Roots

We substitute $x = 5$ and $x = -2$ back into the original equation to see if they satisfy the equation.

For $x = 5$:

$\frac{4(5) + 1}{5 + 1} - 3 = \frac{12}{(5 - 1)(5 + 1)}$

Simplifying:

$\frac{21}{6} - 3 = \frac{12}{24}$

$\frac{7}{2} - 3 = \frac{1}{2}$

$\frac{7}{2} - \frac{6}{2} = \frac{1}{2}$

$\frac{1}{2} = \frac{1}{2}$

This is true, so $x = 5$ is a solution.

For $x = -2$:

$\frac{4(-2) + 1}{-2 + 1} - 3 = \frac{12}{((-2) - 1)((-2) + 1)}$

Simplifying:

$\frac{-8 + 1}{-1} - 3 = \frac{12}{-3}$

$\frac{-7}{-1} - 3 = -4$

$7 - 3 = -4$

This is not true, so $x = -2$ is an extraneous root.

Final Answer

The only valid solution is $x = 5$.

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Would you like more details on any step or have any questions?

Here are some related questions to explore:

1. How do you factorize a quadratic equation like $x^2 - 3x - 10$?
2. What is the quadratic formula, and how is it derived?
3. How do you determine if a root is extraneous?
4. What are common mistakes to avoid when solving rational equations?
5. How do you handle equations where the denominator is a quadratic expression?

Tip: Always check your potential solutions by substituting them back into the original equation to avoid extraneous roots.