Alan G. answered • 03/18/16
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I posted a solution to this several days ago for another student. If you can search for the problem in WyzAnt, I suggest doing so. Otherwise, I can send you the old explanation when you post a reply.
Alan G.
Rola,
I could not locate the post I made, so I will send you another one.
If you solve the last two equations for y and z, you will get
-2y - 2z = -4 (multiplied the second equation by -2)
2y + cz = c.
Add these together: -2z + cz = -4 + c.
Solve for z: z(c - 2) = c - 4
z = (c - 4)/(c - 2).
Notice that c ≠ 2 if there is to be a solution.
Find y:
y + (c - 4)/(c - 2) = 2
y = 2 - (c - 4)/(c - 2) = c/(c - 2).
Find x from the first equation.
x + (c - 4)/(c - 2) = 1
x = 1 - (c - 4)/(c - 2) = 2/(c - 2).
The solution triple is:
(x,y,z) = ( 2/(c-2) , c/(c-2), (c-4)/(c-2) ), c≠2.
For c different from 2, this will produce a unique solution. When c = 2, there is NO solution. There is no value for c for which there is more than one (infinitely many) solutions.
I am so glad that you asked about this, because the first time I posted a solution, the question was incomplete. Thank you!
Please post a reply if you need more help with this.
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03/20/16
Rola M.
03/19/16