Solve the inequality `|x-1|+|2-x|gt3+x.`

Answer

To solve the inequality \( |x-1| + |2-x| > 3 + x \), we will consider different cases based on the values of \( x \).

### Step 1: Rewrite the inequality
The given inequality is:
\[ |x-1| + |2-x| > 3 + x \]
We can rewrite \( |2-x| \) as \( |-(x-2)| = |x-2| \). Thus, the inequality becomes:
\[ |x-1| + |x-2| > 3 + x \]

### Step 2: Identify critical points
The critical points where the expressions inside the absolute values change are \( x = 1 \) and \( x = 2 \). We will consider three cases based on these points:
1. \( x < 1 \)
2. \( 1 \leq x < 2 \)
3. \( x \geq 2 \)

### Step 3: Case 1: \( x < 1 \)
In this case, both \( x-1 \) and \( x-2 \) are negative:
\[ |x-1| = 1 - x \quad \text{and} \quad |x-2| = 2 - x \]
Substituting these into the inequality gives:
\[ (1 - x) + (2 - x) > 3 + x \]
Simplifying this:
\[ 3 - 2x > 3 + x \]
Subtracting 3 from both sides:
\[ -2x > x \]
Adding \( 2x \) to both sides:
\[ 0 > 3x \quad \Rightarrow \quad x < 0 \]
Thus, for this case, the solution is \( x < 0 \).

### Step 4: Case 2: \( 1 \leq x < 2 \)
In this case, \( x-1 \) is non-negative and \( x-2 \) is negative:
\[ |x-1| = x - 1 \quad \text{and} \quad |x-2| = 2 - x \]
Substituting these into the inequality gives:
\[ (x - 1) + (2 - x) > 3 + x \]
Simplifying this:
\[ 1 > 3 + x \]
Subtracting 3 from both sides:
\[ -2 > x \]
This condition does not hold for \( x \geq 1 \). Thus, there are no solutions in this case.

### Step 5: Case 3: \( x \geq 2 \)
In this case, both \( x-1 \) and \( x-2 \) are non-negative:
\[ |x-1| = x - 1 \quad \text{and} \quad |x-2| = x - 2 \]
Substituting these into the inequality gives:
\[ (x - 1) + (x - 2) > 3 + x \]
Simplifying this:
\[ 2x - 3 > 3 + x \]
Subtracting \( x \) from both sides:
\[ x - 3 > 3 \]
Adding 3 to both sides:
\[ x > 6 \]
Thus, for this case, the solution is \( x > 6 \).

### Final Solution
Combining the solutions from all cases, we have:
\[ x < 0 \quad \text{or} \quad x > 6 \]
In interval notation, the solution is:
\[ (-\infty, 0) \cup (6, \infty) \]