If $a c+a d+b c+b d=68$ and $c+d=4$, what is the value of $a+b+c+d$ ?
(A) 17
(B) 85
(C) 4
(D) 21
(E) 64
Solution 1

We factor out a common factor of $a$ from the first two terms and a common factor of $b$ from the last two terms, and so

$$
\begin{aligned}
a c+a d+b c+b d & =68 \\
a(c+d)+b(c+d) & =68 \\
a(4)+b(4) & =68 \\
4(a+b) & =68 \\
a+b & =17
\end{aligned}
$$

using the fact that $c+d=4$.

Then $a+b+c+d=(a+b)+(c+d)=17+4=21$.

## Solution 2

We let $c=d=2$, since $c+d=4$, and then substitute these values to find that

$$
a c+a d+b c+b d=68
$$

$$
\begin{array}{r}
2 a+2 a+2 b+2 b=68 \\
a+b=17
\end{array}
$$

Thus $(a+b)+(c+d)=17+4=21$.

ANSWER: (D)