Suppose that $a, b$ and $c$ are three numbers with

$$
\begin{aligned}
a+b & =3 \\
a c+b & =18 \\
b c+a & =6
\end{aligned}
$$

The value of $c$ is
(A) 2
(B) 11
(C) 3
(D) 6
(E) 7
Solution 1

If we add the second and third equations, we obtain

$$
\begin{aligned}
a c+b+b c+a & =18+6 \\
c(a+b)+(a+b) & =24 \\
(c+1)(a+b) & =24
\end{aligned}
$$

From the first given equation, $a+b=3$ and so we get $(c+1)(3)=24$ or $c+1=8$. Thus, $c=7$.

## Solution 2

From the first equation, $b=3-a$. (We could solve for $a$ in terms of $b$ instead.)

The second equation becomes $a c+(3-a)=18$ or $a c-a=15$.

The third equation becomes $(3-a) c+a=6$ or $-a c+3 c+a=6$.

Adding these two equations, we obtain $3 c=15+6=21$ and so $c=7$.

ANSWER: (E)