# Explanation for Simultaneous Equations Solution

• The_Doctor
In summary, the conversation discussed the solution to a simultaneous equation problem involving three equations and three variables. The solution was found by considering a polynomial and substituting the given equations into it. It was noted that this method may not work for higher degree polynomials, and a broader strategy is needed to solve non-linear simultaneous equations. It was also mentioned that the solution to this problem is possible because of the symmetric nature of the equations and the limited number of whole number possibilities.
The_Doctor
Hi, there's this simultaneous equation problem, however I already know the solution, I just want to know more about the solution.
a+b+c=2,bc+ac+ab=-5, abc=-6
The solution is:
Consider the polynomial (x-a)(x-b)(x-c), where a,b,c are constants.
(x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(bc+ac+ab)x-abc
Substitute the given equations into the polynomial:
(x-a)(x-b)(x-c) = x^3-2x^2-5x+6
(x-a)(x-b)(x-c) = (x-1)(x+2)(x-3)
Therefore, a,b,c = 1,-2,3 in any order.

My question is, how can you know that the three equations relate to a cubic purely from inspection? And is this some broader strategy to solve non-linear simultaneous equations?

If the normal method of substituting is used to solve these simultaneous equations,
a=2-b-c
sub a in both equations:
b(2-b-c)+c(2-b-c)+bc=-5
b^2+bc+c^2-2b-2c=5
and
(2-b-c)bc=-6
bc(b+c-2)=6
b^2c+bc^2-2bc=6
There isn't much else you can do, unless get b+c-2 as the subject of both equations, then sub again, which doesn't yield anything, though it's not surprising that you can't solve this by normal methods, as if you could it would mean you could factorise any polynomial of degree 3, and by extension, of any degree easily, which you can't.

Also, just investigating, if you do a similar thing for a quadratic:
a+b=-3
ab=2
Again we can see see these are the coefficients of a polynomial, namely a quadratic:
(x-a)(x-b) = x^2-(a+b)x+ab
sub the given equations in:
=x^2+3x+2
=(x+1)(x+2)
so a, b=-1,-2 in any order
However, we can also solve this through normal methods:
a=2/b
2/b+b=-3
2+b^2=-3b
b^2+3b+2=0
(b+1)(b+2)=0
b=-1,-2, then solve for a
This could suggest you could do something similar for the cubic, however I can't see how.

The_Doctor said:
My question is, how can you know that the three equations relate to a cubic purely from inspection? And is this some broader strategy to solve non-linear simultaneous equations?

For the first question, they are just functions ('symmetric functions of roots') which become very familiar as soon as you study theory of algebraic equations. In a fairly obvious notation related to yours they are Ʃa, Ʃab, Ʃabc and for higher than cubic it goes on Ʃabcd etc.

For the second question the answer is no. You can do it in this case because there are nice whole numbers that guide you to a solution. e.g. abc = -6 = -1X3X2 = 1X-2X3 = ... = -1X1X6 =... - the whole number possibilities are limited. There is a nice answer to your natural question (it looks like you ought to be able to do it and then it always eludes you and seems like pinning mercury) on p 36-37 of Burnside and Panton "The Theory of Equations" vol 1 a bit too long for me to reproduce right now, and no doubt other algebra books. Crudely I might say that because they are symmetric functions there is nothing in them that distinguishes one root from another, there is nothing that distinguishes your a and your b etc., therefore you cannot get one and then from that get another, you would have to be able to get them all by the same procedure and that, except in special vases and although it is not self-evident, you cannot do.

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## What are simultaneous equations?

Simultaneous equations refer to a system of equations that contain multiple variables and must be solved together to find a solution. This type of problem often arises in mathematics and engineering.

## Why do we need to solve simultaneous equations?

Solving simultaneous equations allows us to find the values of the variables that satisfy both equations at the same time. This is important in order to find the unique solution to a problem or to check if the equations are consistent or inconsistent.

## What is the process for solving simultaneous equations?

The most common method for solving simultaneous equations is by substitution or elimination. In substitution, one equation is solved for a variable and then substituted into the other equation. In elimination, the equations are manipulated to eliminate one of the variables and then solve for the remaining variable.

## What does it mean for simultaneous equations to have no solution?

If a system of simultaneous equations has no solution, it means that there is no set of values for the variables that satisfies both equations at the same time. This can happen when the equations are parallel or when they do not intersect on a graph.

## Can simultaneous equations have more than one solution?

Yes, a system of simultaneous equations can have more than one solution. This is known as an infinite solution or a dependent system, where one of the equations can be derived from the other. In this case, there are an infinite number of solutions that satisfy both equations.

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