Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$.  What is $x$?

$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad\textbf{(E) }26$
1. We start with the given equation:
   \[
   x + y + xy = 80
   \]
   
2. To simplify this equation, we use Simon's Favorite Factoring Trick (SFFT). We add 1 to both sides of the equation:
   \[
   x + y + xy + 1 = 81
   \]
   
3. Notice that the left-hand side can be factored:
   \[
   (x + 1)(y + 1) = 81
   \]
   
4. Since \(x\) and \(y\) are integers and \(x > y > 0\), \(x + 1\) and \(y + 1\) must be factors of 81. The factors of 81 are:
   \[
   1, 3, 9, 27, 81
   \]
   
5. We need to find pairs \((x+1, y+1)\) such that \(x > y > 0\). We test the pairs:
   - \((1, 81)\): This gives \(x = 0\) and \(y = 80\), which does not satisfy \(x > y > 0\).
   - \((3, 27)\): This gives \(x = 2\) and \(y = 26\), which does not satisfy \(x > y > 0\).
   - \((9, 9)\): This gives \(x = 8\) and \(y = 8\), which does not satisfy \(x > y > 0\).
   - \((27, 3)\): This gives \(x = 26\) and \(y = 2\), which satisfies \(x > y > 0\).
   
6. Therefore, the valid pair is \((27, 3)\), which gives:
   \[
   x = 26 \quad \text{and} \quad y = 2
   \]

The final answer is \(\boxed{26}\)