Solve `(|x-1|)/(x+2) gt 1`

Explanation

To solve the inequality \(\frac{|x-1|}{x+2} > 1\), we will break it down into cases based on the definition of absolute value.

### Step 1: Rewrite the inequality
We start with the inequality:
\[ \frac{|x-1|}{x+2} > 1 \]
This can be rewritten as:
\[ |x-1| > x + 2 \]

### Step 2: Consider the two cases for the absolute value
The expression \(|x-1|\) can be split into two cases:
1. Case 1: \(x - 1 \geq 0\) (i.e., \(x \geq 1\))
2. Case 2: \(x - 1 < 0\) (i.e., \(x < 1\))

### Step 3: Solve Case 1: \(x \geq 1\)
In this case, \(|x-1| = x - 1\). So the inequality becomes:
\[ x - 1 > x + 2 \]
Subtracting \(x\) from both sides:
\[ -1 > 2 \]
This is a contradiction, which means there are no solutions in this case.

### Step 4: Solve Case 2: \(x < 1\)
In this case, \(|x-1| = -(x - 1) = 1 - x\). So the inequality becomes:
\[ 1 - x > x + 2 \]
Rearranging gives:
\[ 1 - x - x > 2 \implies 1 - 2x > 2 \]
Subtracting 1 from both sides:
\[ -2x > 1 \]
Dividing by -2 (and reversing the inequality):
\[ x < -\frac{1}{2} \]

### Step 5: Combine the results
From Case 1, we found no solutions. From Case 2, we found that \(x < -\frac{1}{2}\) is a valid solution. However, we must also consider the domain of the original inequality, which is defined for \(x + 2 \neq 0\) or \(x \neq -2\).

### Step 6: Final solution
Thus, the solution set is:
\[ (-\infty, -2) \cup \left(-2, -\frac{1}{2}\right) \]

### Summary
The solution to the inequality \(\frac{|x-1|}{x+2} > 1\) is:
\[ x \in (-\infty, -2) \cup \left(-2, -\frac{1}{2}\right) \]