SOLUTION: For what values of k' the following system has 2x+y=0 x+ 2y + kz = -1 x k²z = 1 (a) Unique Solution (b) No solution.

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Question 1177247: For what values of k' the following system has
2x+y=0
x+ 2y + kz = -1
x k²z = 1
(a) Unique Solution (b) No solution.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

2x%2By=0 ..............1)
x%2B+2y+%2B+kz+=+-1..............2)
x+%2Bk%5E2%2Az+=+1..............3)
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2x%2By=0 ..............1), solve for y
y=-2x..............1a)
go to
x%2B+2y+%2B+kz+=+-1..............2), substitute+y from 1a)
x%2B+2%28-2x%29%2B+kz+=+-1.............., solve for x
x-4x%2B+kz+=+-1
-3x%2B+kz+=+-1
kz%2B1+=+3x
x=%28kz%2B1%29%2F3...........2a)

go to
x+%2Bk%5E2%2Az+=+1..............3), substitute x
%28kz%2B1%29%2F3+%2Bk%5E2%2Az+=+1........., solve for z
kz%2B1+%2B3k%5E2%2Az+=+3
kz%2B1+%2B3k%5E2%2Az+=3-1
z%28k%2B3k%5E2%29+=2
z+=2%2F%28k%2B3k%5E2%29
z+=2%2Fk%281%2B3k%29.........3a)

express x in terms of k

x=%28kz%2B1%29%2F3...........2a), substitute z
x=%28k%282%2Fk%281%2B3k%29%29%2B1%29%2F3
x=%28%282%2F%281%2B3k%29%29%2B1%29%2F3
x=%28%282%2F%281%2B3k%29%29%2B%281%2B3k%29%2F%281%2B3k%29%29%2F3
x=%28%282%2B1%2B3k%29%2F%281%2B3k%29%29%2F3
x=%28%283%2B3k%29%2F%281%2B3k%29%29%2F3
x=%283%281%2Bk%29%2F%281%2B3k%29%29%2F3
x=%281%2Bk%29%2F%281%2B3k%29.....................2b)

go to
y=-2x..............1a), substitute x
y=-2%281%2Bk%29%2F%281%2B3k%29
y=-%282%2B2k%29%2F%281%2B3k%29...............1b)



The solutions to the system of equations are:
x=%281%2Bk%29%2F%281%2B3k%29
y=-2%281%2Bk%29%2F%281%2B3k%29
z+=2%2Fk%281%2B3k%29+
-> for k%3C%3E0 and k%3C%3E-1%2F3