<s>[INST] Solve $\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$ for $x$ I have been scratching my head for solving this equation but I am unable to do this. Even I am unable to get how to use the hint. A way to solve this would be of great help

Solve for $x$ :-
$$\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$$
(Hint: $-3-4=-7$, if $\frac 1a+\frac 1b +\frac 1c \ne 0$)

 [/INST] \begin{align}
\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} &= -7\\
\frac{a+b-x}c+\frac{c}{c} + \frac{a+c-x}b+\frac{b}{b} + \frac{b+c-x}a+\frac{a}{a} -\frac{4abc}{a+b+c} &= -7+3\\
\frac{a+b+c-x}c + \frac{a+b+c-x}b + \frac{a+b+c-x}a -\frac{4abc}{a+b+c} &= -4\\
(a+b+c-x)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)&=\frac{4abc}{a+b+c}-4\\
a+b+c-x&=\frac{4abc(abc-a-b-c)}{(a+b+c)(ab+bc+ca)}
\end{align}
So, $\displaystyle x=a+b+c-\frac{4abc(abc-a-b-c)}{(a+b+c)(ab+bc+ca)}$.
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