Question 1198049: I am trying to figure this question out, but am having significant problems.
Given the linear system: x+y+z = α
6x-4y+5ßz = 31
5x+2y+2z = 13
Find the values of a and b such that the system:
a) has one single solution, and give the solution values of x,y,z.
b) has infinite solutions, and give the solution values of x,y,z.
c) has no solutions
The α and ß unknowns are really confusing me in this question. I understand the idea behind no solution, infinite solutions and one solution in the augmented matrix, but I can't seem to wrap my head around it when the α and ß are involved
Found 2 solutions by math_tutor2020, nsisson:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
There seems to be a "z" missing in the second equation.
Did you mean to say 6x-4y+5ßz = 31 for the second equation? Please let me know because the answer varies depending if the z is there or not.
Answer by nsisson(1) (Show Source):
You can put this solution on YOUR website! You are correct, the second equation is supposed to read 6x-4y+5ßz
I have edited the original question so it is presented properly
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