https://file.aiquickdraw.com/mathbot/file/1729004210545sy1sv8p3.jpg

We are given the equation:

\[
\frac{1}{x+1} = \frac{1}{6} + \frac{x+4}{x^2 + 6x + 8}
\]

### Step 1: Factor the quadratic expression in the denominator.
The quadratic expression \(x^2 + 6x + 8\) can be factored as:
\[
x^2 + 6x + 8 = (x + 2)(x + 4)
\]
Thus, the equation becomes:

\[
\frac{1}{x+1} = \frac{1}{6} + \frac{x+4}{(x+2)(x+4)}
\]

### Step 2: Simplify the second term.
Notice that in the term \(\frac{x+4}{(x+2)(x+4)}\), the \(x+4\) cancels out, leaving us with:

\[
\frac{x+4}{(x+2)(x+4)} = \frac{1}{x+2}
\]

Now the equation is:
\[
\frac{1}{x+1} = \frac{1}{6} + \frac{1}{x+2}
\]

### Step 3: Find a common denominator.
We need to combine the terms on the right-hand side. The least common denominator (LCD) of 6 and \(x+2\) is \(6(x+2)\).

Rewriting the right-hand side:
\[
\frac{1}{6} + \frac{1}{x+2} = \frac{x+2}{6(x+2)} + \frac{6}{6(x+2)} = \frac{x+2+6}{6(x+2)} = \frac{x+8}{6(x+2)}
\]

So the equation becomes:
\[
\frac{1}{x+1} = \frac{x+8}{6(x+2)}
\]

### Step 4: Cross-multiply.
Now, we can cross-multiply to eliminate the denominators:
\[
6(x+2) = (x+1)(x+8)
\]

### Step 5: Expand both sides.
Expanding both sides:
\[
6(x+2) = 6x + 12
\]
\[
(x+1)(x+8) = x^2 + 8x + x + 8 = x^2 + 9x + 8
\]

So the equation becomes:
\[
6x + 12 = x^2 + 9x + 8
\]

### Step 6: Rearrange the equation.
Rearrange all terms to one side:
\[
0 = x^2 + 9x + 8 - 6x - 12
\]
\[
0 = x^2 + 3x - 4
\]

### Step 7: Solve the quadratic equation.
Factor the quadratic equation:
\[
x^2 + 3x - 4 = (x - 1)(x + 4) = 0
\]

Thus, the solutions are:
\[
x = 1 \quad \text{or} \quad x = -4
\]

### Step 8: Check for extraneous solutions.
We need to check if either solution makes any denominator in the original equation zero. If we plug in \(x = -4\), the denominator \(x+4\) becomes zero, which makes the second term undefined. Therefore, \(x = -4\) is not a valid solution.

Thus, the only valid solution is \(x = 1\).

### Final Answer:
The correct answer is **D) 1**.

Would you like more details on any part of this solution?

---

Here are some related questions to explore further:
1. How do you factor quadratic expressions like \(x^2 + 6x + 8\)?
2. What is the least common denominator, and why is it important when adding fractions?
3. How can cross-multiplication help in solving rational equations?
4. What are extraneous solutions, and how do they arise in rational equations?
5. How can you check if a solution makes the denominator of a rational equation undefined?

**Tip**: Always check for extraneous solutions when solving equations involving rational expressions, as some solutions may make the equation undefined.

Solve \(\frac{1}{x+1} = \frac{1}{6} + \frac{x+4}{x^2 + 6x + 8}\).

Learn how to solve the rational equation \(\frac{1}{x+1} = \frac{1}{6} + \frac{x+4}{x^2 + 6x + 8}\) with detailed steps. This problem covers algebraic concepts such as factoring, rational expression simplification, and cross-multiplication. Suitable for high school students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation