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.
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Homework Statement
Please see below
Relevant Equations
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For this,
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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!
 
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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.
 
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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.
 
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