Finding the solution to system of 3 equations with 3 unknowns

AI Thread Summary
The discussion revolves around solving a system of three equations with three unknowns, where the user encounters an inconsistency leading to the result of 0 = 1. This indicates that there are no values for x1 or x3 that satisfy the equations. The user acknowledges that x2 is a leading variable in the third equation but is unsure how to proceed. Responses confirm that the solution space is empty, and no further steps can resolve the contradiction. The consensus is to accept the unexpected result as indicative of the system's inconsistency.
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1685829224949.png

I am trying to find solutions, however, I think I am getting a strange result that I am not too sure how to intercept.

I first multiply the first equation by 2 to get ##2x_1 - 8x_3 = 4## and then I add it to the second equation below to get ##0 = 1##. I think this means that there is no values for ##x_1## or ##x_3## that satisfy the equation. I am not too sure how to go from here but I know ##x_2## is a leading variable of the third equation.

Any help is greatly appreciated

Many thanks!
 
  • Like
Likes FactChecker
Physics news on Phys.org
ChiralSuperfields said:
I first multiply the first equation by 2 to get ##2x_1 - 8x_3 = 4## and then I add it to the second equation below to get ##0 = 1##. I think this means that there is no values for ##x_1## or ##x_3## that satisfy the equation.
Right. If you multiply the first line by ##-2## then you get ##-2x_1+8x_3= -4## and ##-2x_1+8x_3= -3.##
This cannot hold at the same time.
ChiralSuperfields said:
I am not too sure how to go from here but I know ##x_2## is a leading variable of the third equation.

Any help is greatly appreciated

Many thanks!
The solution space is empty. There is nothing left that can be done.
 
  • Like
Likes FactChecker and member 731016
ChiralSuperfields said:
get ##0 = 1##. I think this means that there is no values for ##x_1## or ##x_3## that satisfy the equation. I am not too sure how to go from here
Good work. Sometimes the hardest problems are the ones that give you an unexpected result. Just double-check your result and accept what it says. Your result is not just about ##x_1## and ##x_3##, it is about ##0 \ne 1##. So nothing more can be done.
 
  • Like
Likes member 731016 and fresh_42
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Replies
2
Views
2K
Replies
32
Views
2K
Replies
21
Views
2K
Replies
9
Views
2K
Replies
6
Views
2K
Replies
4
Views
1K
Replies
4
Views
1K
Back
Top