Solve : ` | x-3| + | x - 5| gt 9 `

Explanation

To solve the inequality \( |x - 3| + |x - 5| > 9 \), we will break it down into different cases based on the critical points where the expressions inside the absolute values change their signs. The critical points here are \( x = 3 \) and \( x = 5 \).

### Step 1: Identify the intervals
The critical points divide the number line into three intervals:
1. \( (-\infty, 3) \)
2. \( [3, 5) \)
3. \( [5, \infty) \)

### Step 2: Solve for each interval

#### Case 1: \( x < 3 \)
In this interval, both \( x - 3 \) and \( x - 5 \) are negative. Thus, we can rewrite the absolute values:
\[ |x - 3| = -(x - 3) = -x + 3 \]
\[ |x - 5| = -(x - 5) = -x + 5 \]
Substituting these into the inequality gives:
\[ (-x + 3) + (-x + 5) > 9 \]
Simplifying this:
\[ -2x + 8 > 9 \]
Subtract 8 from both sides:
\[ -2x > 1 \]
Dividing by -2 (remember to flip the inequality sign):
\[ x < -\frac{1}{2 \]
Thus, for this case, we have:
\[ x \in (-\infty, -\frac{1}{2}) \]

#### Case 2: \( 3 \leq x < 5 \)
In this interval, \( x - 3 \) is non-negative and \( x - 5 \) is negative. Thus:
\[ |x - 3| = x - 3 \]
\[ |x - 5| = -(x - 5) = -x + 5 \]
Substituting these into the inequality gives:
\[ (x - 3) + (-x + 5) > 9 \]
Simplifying this:
\[ 2 > 9 \]
This is a contradiction, so there are no solutions in this interval.

#### Case 3: \( x \geq 5 \)
In this interval, both \( x - 3 \) and \( x - 5 \) are non-negative. Thus:
\[ |x - 3| = x - 3 \]
\[ |x - 5| = x - 5 \]
Substituting these into the inequality gives:
\[ (x - 3) + (x - 5) > 9 \]
Simplifying this:
\[ 2x - 8 > 9 \]
Adding 8 to both sides:
\[ 2x > 17 \]
Dividing by 2:
\[ x > \frac{17}{2} \]
Thus, for this case, we have:
\[ x \in \left(\frac{17}{2}, \infty\right) \]

### Step 3: Combine the results
From the three cases, we have:
1. From Case 1: \( x \in (-\infty, -\frac{1}{2}) \)
2. From Case 2: No solutions
3. From Case 3: \( x \in \left(\frac{17}{2}, \infty\right) \)

Therefore, the final solution set is:
\[ x \in \left(-\infty, -\frac{1}{2}\right) \cup \left(\frac{17}{2}, \infty\right) \]