Let $a+\frac{1}{b}=8$ and $b+\frac{1}{a}=3$. Given that there are two possible real values for $a$, find their sum.

$\text{(A) }\frac{3}{8}\qquad\text{(B) }\frac{8}{3}\qquad\text{(C) }3\qquad\text{(D) }5\qquad\text{(E) }8$
1. Start with the given equations:
   \[
   a + \frac{1}{b} = 8
   \]
   \[
   b + \frac{1}{a} = 3
   \]

2. From the first equation, solve for \(\frac{1}{b}\):
   \[
   \frac{1}{b} = 8 - a
   \]
   Then, take the reciprocal to find \(b\):
   \[
   b = \frac{1}{8 - a}
   \]

3. Substitute \(b = \frac{1}{8 - a}\) into the second equation:
   \[
   \frac{1}{8 - a} + \frac{1}{a} = 3
   \]

4. Combine the fractions on the left-hand side:
   \[
   \frac{a + (8 - a)}{a(8 - a)} = 3
   \]
   Simplify the numerator:
   \[
   \frac{8}{a(8 - a)} = 3
   \]

5. Cross-multiply to clear the fraction:
   \[
   8 = 3a(8 - a)
   \]
   Expand and simplify:
   \[
   8 = 24a - 3a^2
   \]
   Rearrange into standard quadratic form:
   \[
   3a^2 - 24a + 8 = 0
   \]

6. Use the quadratic formula to solve for \(a\):
   \[
   a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \(a = 3\), \(b = -24\), and \(c = 8\):
   \[
   a = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3}
   \]
   Simplify inside the square root:
   \[
   a = \frac{24 \pm \sqrt{576 - 96}}{6}
   \]
   \[
   a = \frac{24 \pm \sqrt{480}}{6}
   \]
   \[
   a = \frac{24 \pm 4\sqrt{30}}{6}
   \]
   Simplify the fraction:
   \[
   a = 4 \pm \frac{2\sqrt{30}}{3}
   \]

7. The two possible values for \(a\) are:
   \[
   a_1 = 4 + \frac{2\sqrt{30}}{3}
   \]
   \[
   a_2 = 4 - \frac{2\sqrt{30}}{3}
   \]

8. Find the sum of these two values:
   \[
   a_1 + a_2 = \left(4 + \frac{2\sqrt{30}}{3}\right) + \left(4 - \frac{2\sqrt{30}}{3}\right)
   \]
   Simplify:
   \[
   a_1 + a_2 = 4 + 4 = 8
   \]

The final answer is \(\boxed{8}\).