What's the value of $x$ in the following equation? 
So this is how I approached this question, the above equations could be simplified to :
$$a = \frac{4(b+c)}{b+c+4}\tag{1!}$$
$$b = \frac{10(a+c)}{a+c+10}\tag{2}$$
$$c=\frac{56(a+b)}{a+b+56}\tag{3}$$
From above, we can deduce that $4 > a$ since $\frac{(b+c)}{b+c+4} < 1$ similarly $10 > b, 56 > c$  so  $a + b + c < 70$
Let, $$(a + b + c)k = 70\tag4$$
Now let, $$\alpha(b+c) = b+c+4\tag{1'}$$
$$\beta(a+c) = a+c+10\tag{2'}$$
$$\gamma( a+b ) = a+b+56\tag{3'}$$
Now adding the above 3 equations we get : 
$$2(a+b+c) + 70 = a(\gamma + \beta) + c(\beta + \alpha) + b(\alpha + \gamma) \rightarrow (2 + k)(a+b+c) = a(\gamma + \beta) + c(\beta + \alpha) + b(\alpha + \gamma)$$
Now from above we see that coefficient of $a,b,c$ must be equal on both sides so, $$(2 + k) = (\alpha + \beta) = (\beta + \gamma) = (\alpha + \gamma)$$
Which implies $\beta = \gamma = \alpha = 1+ \frac{k}{2} = \frac{2 + k}{2}$,
Now from $(1)$ and $(1')$ we get $a = \frac{4}{\alpha} = \frac{8}{2+k}$ similarly from $(2),(2')$ and $(3),(3')$ we find, $b = \frac{20}{2+k}, c = \frac{112}{2+k}$
Thus from above we get $a+b+c = \frac{140}{2+k}$ and from $(4)$ we get: $\frac{140}{2+k} = \frac{70}{k}$ from which we can derive $k = 2$
Thus we could derive $a = 2, b = 5, c = 28$ but, the problem now is $a, b, c$ values don't satisfy equation $(4)$ above for $k =2$
Well so, where do I err ? And did I take the right approach ? Do post the solution about how you solved for $x$.

You seem to be assuming through out that $a$,$b$,$c$ are positive integers (e.g. in your deduction that $a < 4$), which doesn't appear to be in the problem statement.
Given they're all integers, I would clear denominators so you don't have to deal with fractions. We then notice that the last equation becomes $x(a+b+c) = abc$, so you don't need to determine the actual values of $a,b,c$ but rather their sum and product.
I haven't checked what happens when you clear denominators in the top three equations but I would hope that by taking linear combinations of them you can obtain an expression for $a+b+c$ and $abc$.
