If $ |x| + x + y = 10$ and $x + |y| - y = 12$, find $x + y$.

$ \textbf{(A)}\ -2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{18}{5}\qquad\textbf{(D)}\ \frac{22}{3}\qquad\textbf{(E)}\ 22 $
1. Consider the given equations:
   \[
   |x| + x + y = 10
   \]
   \[
   x + |y| - y = 12
   \]

2. Analyze the possible cases for \(x\) and \(y\) based on the absolute value functions.

3. **Case 1: \(x \geq 0\) and \(y \geq 0\)**
   - For \(x \geq 0\), \(|x| = x\).
   - For \(y \geq 0\), \(|y| = y\).
   - Substituting these into the equations:
     \[
     x + x + y = 10 \implies 2x + y = 10
     \]
     \[
     x + y - y = 12 \implies x = 12
     \]
   - Substituting \(x = 12\) into \(2x + y = 10\):
     \[
     2(12) + y = 10 \implies 24 + y = 10 \implies y = -14
     \]
   - This contradicts \(y \geq 0\). Therefore, this case is not possible.

4. **Case 2: \(x \geq 0\) and \(y \leq 0\)**
   - For \(x \geq 0\), \(|x| = x\).
   - For \(y \leq 0\), \(|y| = -y\).
   - Substituting these into the equations:
     \[
     x + x + y = 10 \implies 2x + y = 10
     \]
     \[
     x - y - y = 12 \implies x - 2y = 12
     \]
   - We now have the system of linear equations:
     \[
     2x + y = 10 \quad \text{(1)}
     \]
     \[
     x - 2y = 12 \quad \text{(2)}
     \]
   - Solve equation (2) for \(x\):
     \[
     x = 12 + 2y
     \]
   - Substitute \(x = 12 + 2y\) into equation (1):
     \[
     2(12 + 2y) + y = 10 \implies 24 + 4y + y = 10 \implies 24 + 5y = 10 \implies 5y = -14 \implies y = -\frac{14}{5}
     \]
   - Substitute \(y = -\frac{14}{5}\) back into \(x = 12 + 2y\):
     \[
     x = 12 + 2\left(-\frac{14}{5}\right) = 12 - \frac{28}{5} = \frac{60}{5} - \frac{28}{5} = \frac{32}{5}
     \]

5. Calculate \(x + y\):
   \[
   x + y = \frac{32}{5} + \left(-\frac{14}{5}\right) = \frac{32}{5} - \frac{14}{5} = \frac{18}{5}
   \]

The final answer is \(\boxed{\frac{18}{5}}\).