Solve : ` (x +2)/(x+1)lt 0`

Explanation

To solve the inequality \(\frac{x + 2}{x + 1} < 0\), we will follow these steps:

### Step 1: Identify the points where the expression is undefined or zero
The expression \(\frac{x + 2}{x + 1}\) is undefined when the denominator is zero. Therefore, we set the denominator equal to zero:

\[ x + 1 = 0 \implies x = -1 \]

This means \(x\) cannot be \(-1\).

Next, we find where the expression is equal to zero by setting the numerator equal to zero:

\[ x + 2 = 0 \implies x = -2 \]

### Step 2: Determine the intervals to test
The critical points we found are \(x = -2\) and \(x = -1\). These points divide the number line into the following intervals:

1. \((- \infty, -2)\)
2. \((-2, -1)\)
3. \((-1, \infty)\)

### Step 3: Test each interval
We will test a point from each interval to see if the inequality \(\frac{x + 2}{x + 1} < 0\) holds.

- **Interval 1: \(x < -2\)** (Choose \(x = -3\)):
\[ \frac{-3 + 2}{-3 + 1} = \frac{-1}{-2} = \frac{1}{2} > 0 \quad \text{(not part of the solution)} \]

- **Interval 2: \(-2 < x < -1\)** (Choose \(x = -1.5\)):
\[ \frac{-1.5 + 2}{-1.5 + 1} = \frac{0.5}{-0.5} = -1 < 0 \quad \text{(part of the solution)} \]

- **Interval 3: \(x > -1\)** (Choose \(x = 0\)):
\[ \frac{0 + 2}{0 + 1} = \frac{2}{1} = 2 > 0 \quad \text{(not part of the solution)} \]

### Step 4: Write the solution
From our tests, the inequality \(\frac{x + 2}{x + 1} < 0\) holds true in the interval \((-2, -1)\).

Since \(x = -1\) is not included in the solution (the expression is undefined there), we can write the final solution as:

\[ x \in (-2, -1) \]