1 2/(3x) + 3/4 = 1/x

4 9/x - 1 = 3/x + 2

2 2/x = 9/(x - 3)

5 1/x + 2/(x ^ 2) = (x + 9)/(2x ^ 2)

3 (n + 5)/(n + 8) = 1 + 6/(n + 1)

Let's solve each of these equations one by one.

### 1. \( \frac{2}{3x} + \frac{3}{4} = \frac{1}{x} \)
To solve for \(x\), let's first eliminate the fractions by multiplying every term by the least common denominator (LCD) which is \(12x\).

\[
12x \cdot \left(\frac{2}{3x}\right) + 12x \cdot \left(\frac{3}{4}\right) = 12x \cdot \left(\frac{1}{x}\right)
\]

Simplifying:

\[
4 \cdot 2 + 3 \cdot 3x = 12
\]

\[
8 + 9x = 12
\]

Subtract 8 from both sides:

\[
9x = 4
\]

So,

\[
x = \frac{4}{9}
\]

### 2. \( \frac{9}{x} - 1 = \frac{3}{x} + 2 \)
First, get rid of the fractions by multiplying every term by \(x\):

\[
9 - x = 3 + 2x
\]

Next, bring the \(x\) terms to one side:

\[
9 - 3 = 2x + x
\]

Simplifying:

\[
6 = 3x
\]

So,

\[
x = 2
\]

### 3. \( \frac{2}{x} = \frac{9}{x - 3} \)
Cross-multiply to solve for \(x\):

\[
2(x - 3) = 9x
\]

Expanding and simplifying:

\[
2x - 6 = 9x
\]

Subtract \(2x\) from both sides:

\[
-6 = 7x
\]

So,

\[
x = -\frac{6}{7}
\]

### 4. \( \frac{1}{x} + \frac{2}{x^2} = \frac{x + 9}{2x^2} \)
First, let's multiply every term by \(2x^2\) to eliminate the fractions:

\[
2x^2 \cdot \frac{1}{x} + 2x^2 \cdot \frac{2}{x^2} = 2x^2 \cdot \frac{x + 9}{2x^2}
\]

Simplifying:

\[
2x + 4 = x + 9
\]

Subtract \(x\) from both sides:

\[
x + 4 = 9
\]

So,

\[
x = 5
\]

### 5. \( \frac{n + 5}{n + 8} = 1 + \frac{6}{n + 8} \)
Let's first subtract 1 from both sides to combine the fractions:

\[
\frac{n + 5}{n + 8} - 1 = \frac{6}{n + 8}
\]

Combine the fractions on the left:

\[
\frac{n + 5 - (n + 8)}{n + 8} = \frac{6}{n + 8}
\]

Simplifying:

\[
\frac{-3}{n + 8} = \frac{6}{n + 8}
\]

Since the denominators are the same, we equate the numerators:

\[
-3 = 6
\]

This is a contradiction, meaning there is **no solution** for \(n\) that satisfies this equation.

---

Would you like to go over any of these steps in more detail, or do you have any other questions?

### Related Questions:
1. How do you solve equations involving absolute values?
2. What are some methods for solving quadratic equations?
3. Can you explain how to solve systems of linear equations?
4. What is the significance of the discriminant in quadratic equations?
5. How can you solve equations with exponents or logarithms?

**Tip:** When solving equations, always look for opportunities to simplify or factor to make the solving process easier.

Explore how to solve complex equations involving fractions with detailed step-by-step solutions. Covering equations like 2/(3x) + 3/4 = 1/x, 9/x - 1 = 3/x + 2, and more. Suitable for advanced high school students and above.

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