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Let's solve these rational equations one by one.

### 1) \( \frac{x}{6} + \frac{3x}{5} = 2 \)

We will solve this by finding the least common denominator (LCD), which is 30.

Multiply both sides by 30:
\[
30 \left( \frac{x}{6} + \frac{3x}{5} \right) = 30 \cdot 2
\]
Simplifying:
\[
5x + 18x = 60
\]
\[
23x = 60
\]
\[
x = \frac{60}{23}
\]

Thus, the solution is \( x = \frac{60}{23} \).

---

### 2) \( \frac{x+1}{x-5} = \frac{5}{3} \)

Cross-multiply to eliminate the denominators:
\[
3(x+1) = 5(x-5)
\]
Expand both sides:
\[
3x + 3 = 5x - 25
\]
Move all terms involving \(x\) to one side:
\[
3 + 25 = 5x - 3x
\]
\[
28 = 2x
\]
\[
x = \frac{28}{2} = 14
\]

Thus, the solution is \( x = 14 \).

---

### 3) \( \frac{6}{5} = \frac{3a + 2}{2a + 3} \)

Cross-multiply:
\[
6(2a + 3) = 5(3a + 2)
\]
Expand both sides:
\[
12a + 18 = 15a + 10
\]
Move all terms involving \(a\) to one side:
\[
18 - 10 = 15a - 12a
\]
\[
8 = 3a
\]
\[
a = \frac{8}{3}
\]

Thus, the solution is \( a = \frac{8}{3} \).

---

### 4) \( \frac{9}{a + 3} = 1 - \frac{a}{a + 3} \)

First, rewrite the equation:
\[
\frac{9}{a + 3} = \frac{(a + 3) - a}{a + 3}
\]
Simplify the right-hand side:
\[
\frac{9}{a + 3} = \frac{3}{a + 3}
\]
Since the denominators are the same, equate the numerators:
\[
9 = 3
\]
This is a contradiction, meaning there is **no solution** to this equation.

---

### 5) \( \frac{x}{x - 2} = \frac{x + 3}{x + 2} - \frac{x}{x^2 - 4} \)

Notice that \( x^2 - 4 = (x - 2)(x + 2) \), so the second fraction can be rewritten as:
\[
\frac{x}{x^2 - 4} = \frac{x}{(x - 2)(x + 2)}
\]
The least common denominator (LCD) of all fractions is \( (x - 2)(x + 2) \). Multiply every term by \( (x - 2)(x + 2) \) to eliminate the denominators:
\[
(x)(x + 2) = (x + 3)(x - 2) - x
\]
Expand both sides:
\[
x^2 + 2x = (x + 3)(x - 2) - x
\]
Expanding \( (x + 3)(x - 2) \):
\[
x^2 + 2x = x^2 - 2x + 3x - 6 - x
\]
Simplify:
\[
x^2 + 2x = x^2 + x - 6
\]
Cancel out \( x^2 \) on both sides:
\[
2x = x - 6
\]
Move all terms involving \(x\) to one side:
\[
2x - x = -6
\]
\[
x = -6
\]

Thus, the solution is \( x = -6 \).

---

Let me know if you need further details on any of these solutions. Here are some related questions for practice:

1. How do you find the least common denominator for rational equations?
2. How do you solve rational equations by cross-multiplying?
3. Why might an equation have no solution?
4. How do you simplify complex fractions with common factors in the denominator?
5. What happens if you have a contradiction in the middle of solving an equation?

**Tip:** Always check for restrictions in rational equations by ensuring the denominator is never zero for your solutions.

Solve the following rational equations: 1) (x/6) + (3x/5) = 2, 2) (x + 1)/(x - 5) = 5/3, 3) 6/5 = (3a + 2)/(2a + 3), 4) 9/(a + 3) = 1 - a/(a + 3), 5) x/(x - 2) = (x + 3)/(x + 2) - x/(x^2 - 4)

This guide walks through solving rational equations like (x/6) + (3x/5) = 2 and more using cross-multiplication and algebraic techniques. Perfect for Grades 9-11 students.

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