Solve $a+b+c=d, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{d}$ in $\mathbb{R}$ From the first equation we have
$$\frac{1}{a+b+c}=\frac{1}{d}$$
So joining the second equation we get
$$\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
which leads to
$$(a+b)(a+c)(b+c)=0$$
Thus we know $a=-b$, $a=-c$ or $b=-c$, which further implies $c=d, b=d$ or $a=d$ respectively. How should I proceed from here to get all the solutions in $\mathbb{R}$?
 A: In short, you are right. You are looking for values $(a,b,c)\in \mathbb{R}^3$ so that those equations hold. One set of solutions comes when $a=-b$, and the same for the other two. All you have to do now is take the union of the sets that arise from each of these three conditions. Careful to make sure not to divide by 0!
So the final answer could look like $\{(a,b,c)\in\mathbb{R}^3\ : a,b,c \neq 0, $ and $a=-b$ or $a=-c$ or $b=-c\}$
A: There are two equations for four unknowns and your system is twice indeterminate. We have
$$a+b=d-c$$ and $$\frac{a+b}{ab}=\frac{c-d}{cd}$$
which imply
$$ab=(a+b)\frac{cd}{c-d}=-cd.$$
Now we know the sum and product of $a,b$ and by inspection we have
$$\begin{cases}a=d,\\b=-c\end{cases}\text{ or }\begin{cases}a=-c,\\b=d.\end{cases}$$
(A sum-product system has two solutions, obtained by swapping the unknowns.)
In this development $c,d$ are arbitrary but nonzero and we can write the solution set as
$$\{(v,-u,u,v):u,v\in\mathbb R_0\}\cup\{(-u,v,u,v):u,v\in\mathbb R_0\}.$$
