SOLUTION: Help!Is this a problem with no solution? x + y + 2z = 7 2x + 2y + 4z = -6 x+y+2z = 1 Thanks so much!

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Help!Is this a problem with no solution? x + y + 2z = 7 2x + 2y + 4z = -6 x+y+2z = 1 Thanks so much!      Log On


   



Question 607176: Help!Is this a problem with no solution?
x + y + 2z = 7
2x + 2y + 4z = -6
x+y+2z = 1

Thanks so much!

Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Yes. We can see this if we let w = x+y+2z

The first equation then becomes w = 7 and the third equation becomes w = 1.

But this is a contradiction since w is both 7 and 1 at the same time.

Since this is not possible, there are no solutions.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Multiply equation 2 by , then you have three equations with identical variable coefficients and different constants. Three mutually parallel lines. No point of intersection. No solution.

John

My calculator said it, I believe it, that settles it
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