https://file.aiquickdraw.com/mathbot/file/1725360347864anyjfa0e.jpg

Find the least common denominator, multiple both sides of the equation by its the LCD, apply the distributive propoerty and then simplify, find all the possible values of x,check it values by substituting into original equation and reject any extraneous roots

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 \( x = 5 \).

---

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.

Learn how to solve rational equations involving least common denominators (LCD) and quadratic equations. Understand how to find and apply LCD, use the quadratic formula to solve resulting equations, and check for extraneous roots. Suitable for high school students in Grades 10-12.

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