Suppose that $a, b, c$, and $d$ are positive integers that satisfy the equations

$$
\begin{aligned}
& a b+c d=38 \\
& a c+b d=34 \\
& a d+b c=43
\end{aligned}
$$

What is the value of $a+b+c+d$ ?
(A) 15
(B) 16
(C) 17
(D) 18
(E) 19
Adding the second and third equations, we obtain

$$
\begin{aligned}
a c+b d+a d+b c & =77 \\
a c+a d+b c+b d & =77 \\
a(c+d)+b(c+d) & =77 \\
(a+b)(c+d) & =77
\end{aligned}
$$

Since each of $a, b, c$ and $d$ is a positive integer, then $a+b$ and $c+d$ are each positive integers and are each at least 2 .

Since the product of $a+b$ and $c+d$ is $77=7 \times 11$ (with 7 and 11 both prime), then one must equal 7 and the other must equal 11.

Therefore, $a+b+c+d=7+11=18$.

(We can check with some work that $(a, b, c, d)=(5,2,4,7)$ is a solution to the system.)

ANSWER: (D)