2. If $a, b, c$ and $d$ are real numbers such that
$$
\frac{b+c+d}{a}=\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}=r,
$$
find the value of $r$.
(A) 3
(B) 1
(C) -1
(D) 3 or 1
(E) 3 or -1
2. Answer: (E)
From the given equations, we obtain
$$
\begin{array}{l}
a+b+c+d=a(r+1), a+b+c+d=b(r+1), \\
a+b+c+d=c(r+1), a+b+c+d=d(r+1) .
\end{array}
$$

Adding these four equations gives
$$
4(a+b+c+d)=(a+b+c+d)(r+1) \text {, }
$$
that is,
$$
(3-r)(a+b+c+d)=0 \text {. }
$$

Thus $r=3$, or $a+b+c+d=0$. If $a+b+c+d=0$, then we see from the original given equations that $r=-1$. Hence the value of $r$ is either 3 or -1 .