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# Solve for x

$\frac {x-b^2}{a+c} + \frac {x-c^2}{a+b} + \frac {x-a^2}{b+c} = 4(a+b+c)$

How can I solve this equation for $$x$$? Please show me the process and also the basic idea of solving this type of equation.

Note by Omar Sayeed Saimum
1 year, 1 month ago

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x=(a+b+c)^2

- 1 year, 1 month ago

What have you tried? Where did you get stuck?

Given that this is a linear equation in $$x$$, it should be simple to shift terms around. (Of course, there may be other approaches.)

There is no guarantee that the answer you obtain will be "nice". Though, if you saw it in an olympiad, then it would likely be nice.

Staff - 1 year, 1 month ago

My answer is not "nice". A big one. But it's an Olympiad question according to my friend ( He gave me this problem to solve). So.....

- 1 year, 1 month ago

Can you state your answer? Ideally, make it as simplified as possible.

Staff - 1 year, 1 month ago

X= (4(a+b+c)^2(ab+bc+ca) + (a+b+c)(a^3+b^3+c^3))/((a+b+c)^2 + (ab+bc+ca))

- 1 year, 1 month ago

Hm, that doesn't seem quite right to me. Can you lay out all of your steps?

1. The terms should be degree 2, but your first term has degree 4.
2. In the 2nd term, you are multiplying by (a+b+c) and then dividing by (a+b+c)^2, so that could be simplified.

Staff - 1 year, 1 month ago

Please look at that again. the total term is divided by ((a+b+c)^2 + (ab+bc+ca)).

- 1 year, 1 month ago

Ah ic. Missed the brackets because none of the terms made sense to me.

I agree with the denominator being $$(a+b)(b+c) + (b+c)(c+a) + (c+a)(a+b)$$. It is more natural to write it as $$a^2 + b^2 + c^3 + 3ab + 3bc + 3ca$$.

I'm not sure how you arrived at the numerator though. To me, it should be $$b^2(b+c)(b+a) + c^2 (c+a)(c+b) + a^2(a+b)(a+c) + 4(a+b+c)(a+b)(b+c)(c+a)$$.

As always, if you lay out your steps, someone would be able to see what you did and provide feedback from there. But otherwise, we can only guess at what you're doing.

Staff - 1 year, 1 month ago

Yeah, the denominator is a^2 + b^2 + c^2 + 3(ab+bc+ca) I've just separated the term using (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) Anyway, if you prefer to lay out all my steps, then I'll do so. Oh, I think the answer is (a+b+c)^2. I was using some terms in the equation randomly. Then suddenly I noticed that if we use x = (a+b+c)^2, the equation becomes true.

- 1 year, 1 month ago

Yes, that's the answer. If you had the right numerator, you would be able to divide it out to arrive at that value.

Other than making a lucky observation about the answer, how else can we appraoch this problem?

Staff - 1 year, 1 month ago

That's what I'm asking to you. Still I'm trying to get this. But can't find anyway. Wil you kindly show me the process if you get that? Then I'll be relieved from this headache.😊

- 1 year, 1 month ago

tl;dr I would have expanded everything to get the answer because it's an obvious step that works. There are some observations that could be made that eventually arrive at the answer, but this might take longer than brute force expansion.

Here are some observations:

1. The LHS is linear in $$x$$.
2. The coefficient of $$x$$ is pretty ugly, like $$\frac{ a^2+b^2+c^2 + 3ab + 3bc + 3ac } { (a+b)(b+c)(c+a) }$$.
3. If we were to change the RHS to almost any other term, there is no nice solution that we can find. This suggests that the answer is "specially determined".
4. (I might be wrong here) There isn't a nice way to simplify the LHS. At the most, a substitution of $$s = a+b+c$$ might help, but I don't see how else to proceed. At this point in time, I would be tempted to brute force the calculations, which was what I did initially. We get the nice answer of $$(a+b+c)^2$$, which suggests something else might be going on.
5. Looking at the problem again, observe that if $$a+b = 0$$ then we must have $$x - c^2 \equiv 0 \pmod{a+b}$$. If we tried to solve the system of equivalence equations, namely $$x = c^2 + k_c (a+b) = a^2 + k_a (b+c) = b^2 + k_b( c+a)$$, we might eventually arrive at $$x \equiv (a+b+c)^2 \pmod{ (a+b)(b+c)(c+a) }$$. (Note: I say might because there isn't a standard approach to solving equivalence relations in variables.) We thus then plug in $$x = (a+b+c)^2 + K (a+b)(b+c)(c+a)$$ into the equation, and observe that we get the RHS when $$K = 0$$.

Staff - 1 year, 1 month ago

Thanks. But I didn't understand the 3rd line of the 5th step. x = (a+b+c)^2 (mod(a+b)(b+c)(c+a)) How?

- 1 year, 1 month ago

Note: I say might because there isn't a standard approach to solving equivalence relations in variables.

It is clear that $$x = (a+b+c)^2$$ satisfies $$x = c^2 + k_c (a+b) = a^2 + k_a (b+c) = b^2 + k_b( c+a)$$, but it's not clear how we can easily backtrack.

Staff - 1 year, 1 month ago

# Calvin Lin plz help me.

- 1 year, 1 month ago

Instead of asking people for help, tell us where you're stuck because we don't know what issues you're facing.

- 1 year, 1 month ago

I didn't solve this kind of equation before. I'm trying. But couldn't find any way. So please help me to solve this.

- 1 year, 1 month ago