# Unsolvable maths problem

• Sucvicbil
Sucvicbil
TL;DR Summary: Isolating coefficient contribution to outcome change

This is a problem I've been trying to solve for almost 3 days now, without any solutions.
Say we have x + y + z = 5, and x is 1, y is 1, and z is 3. We have a second equation = 2x + 3y + 4z = 17. Using the outcome difference based on the coefficient change alone observed between the first and second equation, how can we predict the outcome of a new equation = 6x + 5y + 7z = ? without using the values of x, y or z at all. Seems deceptively simple at first, but it seems unsolvable.

Your last "equation" is not an equation--you have not specified the right hand side--so you have two equations in three unknowns. The system is under-determined, so there is no solution.

marcusl said:
Your last "equation" is not an equation--you have not specified the right hand side--so you have two equations in three unknowns. The system is under-determined, so there is no solution.
Hi! Thanks for responding. My thought process is - since the observable change between equation 1 and equation 2 is due to coefficients, then surely the coefficient changes contributes to the difference in outcome between 1 and 2, so if we know what this contribution is, for new equations - we can adjust this contribution based on the new coefficients to deduce new outcomes.

marcusl said:
so you have two equations in three unknowns. The system is under-determined, so there is no solution.
Correction: there is no unique solution.

Sucvicbil said:
My thought process is - since the observable change between equation 1 and equation 2 is due to coefficients, then surely the coefficient changes contributes to the difference in outcome between 1 and 2, so if we know what this contribution is, for new equations - we can adjust this contribution based on the new coefficients to deduce new outcomes.
This makes no sense to me. The most systematic way of solving systems of equations uses the linear algebra technique of writing the equations as an augmented matrix and using Gauss-Jordan elimination to find a solution.

The general solution of the two equations you provided is <x, y, z> = t<1, -2, 1> + <-2, 7, 0>, where t is any arbitrary real number. For example, if t = 0, x = -2, y = 7, and z = 0. Note that these values of x, y, and z satisfy both your equations.

Thank you for the clarification—no unique solution. You can see it this way: your first two equations eliminate one variable. For example, you can express z in terms of x and y. Substituting that into your third expression gives the right hand side, call it u, in terms of x and y. There’s no unique solution, as Mark44 points out.

Mark44 said:
This makes no sense to me. The most systematic way of solving systems of equations uses the linear algebra technique of writing the equations as an augmented matrix and using Gauss-Jordan elimination to find a solution.

The general solution of the two equations you provided is <x, y, z> = t<1, -2, 1> + <-2, 7, 0>, where t is any arbitrary real number. For example, if t = 0, x = -2, y = 7, and z = 0. Note that these values of x, y, and z satisfy both your equations.
Hi! Thanks for your response. My goal is not to calculate the values of x, y, z, my goal is to predict outcomes of new linearly independent equations by calculating an adjustable coefficient based on the coefficient differences between equation 1 and 2, and the contribution of this coefficient change to the outcome difference ( 17 - 5 ). This way, without seeing the values of x, y, z (again), and by using the coefficients in the new equation, I can simply adjust this deduced coefficient (using ratios or something along those lines) to predict the outcome of the new equation. I hope I managed to articulate that better.

What part of the responses you've gotten are you not following? We'd like to help you understand.

marcusl said:
What part of the responses you've gotten are you not following? We'd like to help you understand.
It seems we are defining two completely different problems. I understand the laws of needing as many linearly independent equations as the variables for a solution, but this is not what my problem is posing. My problem is around detangling the contribution of coefficients from the the contribution of positional weights in outcomes, solving this will enable me to be able design a system that predicts outcomes by multiplying an adjustable coefficient by a reference outcome. It's okay if you don't understand, I'll wait for someone who does! Have a lovely day.

Your description of the problem is insufficient. Either
1) this is a homework problem-in which case you need to write down the exact problem statement
or
2) you are trying to do something and you have decided on this approach- in which case you need to write down what you are trying to do

Sucvicbil said:
My goal is not to calculate the values of x, y, z, my goal is to predict outcomes of new linearly independent equations by calculating an adjustable coefficient based on the coefficient differences between equation 1 and 2, and the contribution of this coefficient change to the outcome difference ( 17 - 5 ). This way, without seeing the values of x, y, z (again), and by using the coefficients in the new equation, I can simply adjust this deduced coefficient (using ratios or something along those lines) to predict the outcome of the new equation.
Again, this makes very little sense to me. When you say that your goal is to "predict outcomes of new linearly independent equations" I don't know what this means. It would be helpful for you to explain why you want to do this.

If you have two linearly independent equations in three variables, this means that neither equation is a constant multiple of the other. Geometrically, in three dimensions, the two linearly independent equations represent planes that intersect in a line. This line of intersection is defined by the points that are on it, with specific values of x, y, and z. The planes can't be co-incident (same plane) and can't be parallel to each other. For three linearly independent equations, the geometry is that all three planes intersect in exactly one point. I'm not aware of any technique that allows you to fiddle with the coefficients of the first two equations to come up with the coefficients for a third equation.
Sucvicbil said:
I hope I managed to articulate that better.
Well, no.

I suggest you rewrite the problem using no extraneous information. If we are not allowed to use the fact that (1,1,3) is a solution, don't make that part of the problem. Tell us no more, and no less, than it takes to specify the problem.

PhDeezNutz
Sucvicbil said:
It's okay if you don't understand, I'll wait for someone who does!
Consider the possibility that your description isn't understandable. If none of us know what your asking, you might be waiting a long time.

Words often don't work well in describing math problems. Perhaps you could put together a more formal statement of the problem?

It seems to me you've used 2 equations to define a line in ##\mathbb R^3## and your asking what is the general result from a 1st order polynomial with arbitrary coefficients using a point on that line as the input. The answer is it could be anything.

To paraphrase Pauli, it's not even unsolvable.

Given <x, y, z> = t<1, -2, 1> + <-2, 7, 0>, then ax+by+cz = t<a, -2b, 1c> + <-2a, 7b, 0> = <(t-2)a, (7-2t)b, tc> ##\forall t \in \mathbb R##.
Is that what you wanted?

Maybe OP can consider the total derivative to evaluate how the coefficients affect the final output.

We're all playing "guess the question". Maybe we should wait for the OP to enter it verbatim.

PhDeezNutz, fresh_42 and Frabjous
We're all playing "guess the question". Maybe we should wait for the OP to enter it verbatim.
( To OP) This, is...NOT Jeopardy!!

Sucvicbil said:
My goal is not to calculate the values of x, y, z, my goal is to predict outcomes of new linearly independent equations by calculating an adjustable coefficient based on the coefficient differences between equation 1 and 2, and the contribution of this coefficient change to the outcome difference ( 17 - 5 ).
What you are overlooking is that if you have enough information to calculate the results of all such variations of coefficients then you have enough information to find x, y and z. Hence the two goals are the same.

What you can do is predict the results for certain combinations of coefficients, namely, those that can be produced by linear combinations of the results you have:
x + y + z = 5 and
2x + 3y + 4z = 17 so
(m+2n)x + (m+3n)y + (m+4n)z = 5m + 17n for any m and n.

MatinSAR

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