We are given the inequality:

\[
\frac{x + 3}{3x - 6} > 0
\]

Let's solve this step by step.

### Step 1: Factor the denominator
Notice that the denominator \(3x - 6\) can be factored as:

\[
3x - 6 = 3(x - 2)
\]

So the inequality becomes:

\[
\frac{x + 3}{3(x - 2)} > 0
\]

### Step 2: Determine critical points
The expression will change its sign at the points where either the numerator or denominator is zero. So, we need to solve:

1. \(x + 3 = 0\) → \(x = -3\)
2. \(3(x - 2) = 0\) → \(x = 2\)

These points divide the real number line into three intervals: \( (-\infty, -3) \), \( (-3, 2) \), and \( (2, \infty) \).

### Step 3: Test the intervals
Now, we check the sign of the expression in each interval.

- For \(x \in (-\infty, -3)\): Let's test with \(x = -4\):
  \[
  \frac{-4 + 3}{3(-4 - 2)} = \frac{-1}{3(-6)} = \frac{-1}{-18} = \frac{1}{18} > 0
  \]
  So, the expression is positive.

- For \(x \in (-3, 2)\): Let's test with \(x = 0\):
  \[
  \frac{0 + 3}{3(0 - 2)} = \frac{3}{3(-2)} = \frac{3}{-6} = -\frac{1}{2} < 0
  \]
  So, the expression is negative.

- For \(x \in (2, \infty)\): Let's test with \(x = 3\):
  \[
  \frac{3 + 3}{3(3 - 2)} = \frac{6}{3(1)} = 2 > 0
  \]
  So, the expression is positive.

### Step 4: Determine the solution set
We want the expression to be greater than zero, so we choose the intervals where the expression is positive. This occurs in the intervals \( (-\infty, -3) \) and \( (2, \infty) \).

### Step 5: Exclude points where the expression is undefined or zero
The point \(x = 2\) makes the denominator zero, so we must exclude it from the solution. The point \(x = -3\) makes the numerator zero, but since we want strict inequality (\(> 0\)), we exclude \(x = -3\) as well.

### Final solution:
The solution to the inequality is:

\[
x \in (-\infty, -3) \cup (2, \infty)
\]

Would you like further details or have any questions?

Here are some related questions for practice:
1. Solve \(\frac{x - 5}{2x + 4} \geq 0\).
2. Solve \(\frac{2x + 1}{x^2 - 1} > 0\).
3. Solve \(\frac{4 - x}{x - 6} \leq 0\).
4. Solve \(\frac{5x - 3}{x^2 - 9} < 0\).
5. Solve \(\frac{x + 1}{x - 4} \geq 0\).

**Tip:** When solving rational inequalities, always test the intervals determined by critical points and exclude values that make the denominator zero.

Explore how to solve the rational inequality (x+3)/(3x-6) > 0. Learn key algebraic concepts like factorization, sign testing, and solving rational inequalities with critical points. Ideal for students in Grades 9-11.

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