Solve the following inequalities. `(|x+3|+x)/(x+2)gt 1`

Answer

To solve the inequality \((|x+3| + x)/(x+2) > 1\), we will break it down into two cases based on the expression inside the absolute value.

### Step 1: Identify Cases for the Absolute Value
The expression \(|x + 3|\) will behave differently depending on whether \(x + 3\) is non-negative or negative. Thus, we have two cases:
1. Case 1: \(x + 3 \geq 0\) (i.e., \(x \geq -3\))
2. Case 2: \(x + 3 < 0\) (i.e., \(x < -3\))

### Step 2: Solve Case 1 (\(x \geq -3\))
In this case, \(|x + 3| = x + 3\). The inequality becomes:
\[ \frac{(x + 3) + x}{x + 2} > 1 \]
This simplifies to:
\[ \frac{2x + 3}{x + 2} > 1 \]

### Step 3: Rearranging the Inequality
Subtract \(1\) from both sides:
\[ \frac{2x + 3}{x + 2} - 1 > 0 \]
Finding a common denominator:
\[ \frac{2x + 3 - (x + 2)}{x + 2} > 0 \]
This simplifies to:
\[ \frac{x + 1}{x + 2} > 0 \]

### Step 4: Finding Critical Points
The critical points occur when the numerator or denominator equals zero:
- Numerator: \(x + 1 = 0 \Rightarrow x = -1\)
- Denominator: \(x + 2 = 0 \Rightarrow x = -2\)

### Step 5: Test Intervals
We will test the intervals determined by the critical points \(-2\) and \(-1\):
1. Interval 1: \(x < -2\)
2. Interval 2: \(-2 < x < -1\)
3. Interval 3: \(x > -1\)

- For \(x < -2\) (e.g., \(x = -3\)):
\(\frac{-3 + 1}{-3 + 2} = \frac{-2}{-1} > 0\) (True)

- For \(-2 < x < -1\) (e.g., \(x = -1.5\)):
\(\frac{-1.5 + 1}{-1.5 + 2} = \frac{-0.5}{0.5} < 0\) (False)

- For \(x > -1\) (e.g., \(x = 0\)):
\(\frac{0 + 1}{0 + 2} = \frac{1}{2} > 0\) (True)

### Step 6: Conclusion for Case 1
Thus, for Case 1, the solution is:
\[ x < -2 \quad \text{or} \quad x > -1 \]

### Step 7: Solve Case 2 (\(x < -3\))
In this case, \(|x + 3| = -(x + 3)\). The inequality becomes:
\[ \frac{-(x + 3) + x}{x + 2} > 1 \]
This simplifies to:
\[ \frac{-3}{x + 2} > 1 \]

### Step 8: Rearranging the Inequality
Multiply both sides by \(x + 2\) (note that \(x + 2 < 0\) since \(x < -3\)):
\[ -3 > (x + 2) \]
This simplifies to:
\[ -3 > x + 2 \Rightarrow -5 > x \]

### Step 9: Conclusion for Case 2
Thus, for Case 2, the solution is:
\[ x < -5 \]

### Final Solution
Combining the results from both cases:
1. From Case 1: \(x < -2\) or \(x > -1\)
2. From Case 2: \(x < -5\)

The overall solution is:
\[ x < -5 \quad \text{or} \quad x > -1 \]