What's the value of $x$ in the following equation?

Question

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 $x$.

Answer 1

Your deductions are wrong and that is what is misleading you. Integers can be both positive and negative.

If you solve equations (1), (2) and (3) simultaneously you can find a, b and c. I did this to find $$a=3$$ $$b=5$$ $$c=7$$

You can then plug this into the forth equation given in the problem to solve for x. $$x = \frac{abc}{ a + b + c} $$

which solves to give $$x=\frac{105}{15}=7$$

Answer 2

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 $abc$.