Solve the inequalities for real x and represent solution on number line
`(x+3)/(x-2)>0``xinR`

Answer

To solve the inequality \(\frac{x+3}{x-2} > 0\), we will follow these steps:

### Step 1: Identify the critical points
The critical points occur when the numerator and denominator are equal to zero.

1. **Numerator:** \(x + 3 = 0 \Rightarrow x = -3\)
2. **Denominator:** \(x - 2 = 0 \Rightarrow x = 2\)

These points divide the number line into intervals.

### Step 2: Determine the intervals
The critical points divide the real number line into three intervals:
1. \((-∞, -3)\)
2. \((-3, 2)\)
3. \((2, ∞)\)

### Step 3: Test each interval
We will test a point from each interval to see if the inequality holds.

1. **Interval \((-∞, -3)\)**: Choose \(x = -4\)
\[ \frac{-4 + 3}{-4 - 2} = \frac{-1}{-6} = \frac{1}{6} > 0 \quad \text{(True)} \]

2. **Interval \((-3, 2)\)**: Choose \(x = 0\)
\[ \frac{0 + 3}{0 - 2} = \frac{3}{-2} = -\frac{3}{2} < 0 \quad \text{(False)} \]

3. **Interval \((2, ∞)\)**: Choose \(x = 3\)
\[ \frac{3 + 3}{3 - 2} = \frac{6}{1} = 6 > 0 \quad \text{(True)} \]

### Step 4: Combine the results
From our tests, we find that the inequality \(\frac{x+3}{x-2} > 0\) holds true for the intervals:
- \((-∞, -3)\)
- \((2, ∞)\)

### Step 5: Write the solution
The solution in interval notation is:
\[ x \in (-∞, -3) \cup (2, ∞) \]

### Step 6: Represent the solution on a number line
On a number line, we represent the solution as follows:
- Draw an open circle at \(-3\) and shade to the left (indicating all numbers less than \(-3\)).
- Draw an open circle at \(2\) and shade to the right (indicating all numbers greater than \(2\)).

### Final Answer
The solution to the inequality \(\frac{x+3}{x-2} > 0\) is:
\[ x \in (-∞, -3) \cup (2, ∞) \]

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