Finding the solution to system of 3 equations with 3 unknowns

In summary, the conversation discusses how the equations ##2x_1 - 8x_3 = 4## and ##-2x_1+8x_3= -4## cannot hold at the same time, leading to the conclusion that the solution space is empty and there is nothing more that can be done. The unexpected result of ##0 \ne 1## is accepted and no values for ##x_1## or ##x_3## will satisfy the equation.
  • #1
ChiralSuperfields
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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!
 
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  • #2
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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  • #3
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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