Finding the solution to system of 3 equations with 3 unknowns

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SUMMARY

The discussion revolves around solving a system of three equations with three unknowns, specifically addressing the scenario where manipulating the equations leads to a contradiction, represented by the statement "0 = 1." The user initially multiplies the first equation by 2, resulting in the equation "2x_1 - 8x_3 = 4," which, when combined with the second equation, reveals an inconsistency. The conclusion drawn is that there are no values for x_1 or x_3 that satisfy the equations, indicating that the solution space is empty.

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Homework Statement
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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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