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Hi ,
Solve the system following in
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Can You Help me ??
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Do you need all solutions? There is an obvious one you can see.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Yes I need all solutions plz !
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Hi Shiiro;
From a computer search:
z = 0 there are only 2 solutions that I can find.
z = 1 there are an infinite number of solutions
z = 2 there are no solutions that I can find.
z = 3 there are only 2 solutions that I can find.
z = 4 there are no solutions that I can find.
z = 5 there are no solutions that I can find.
z = 6 there are only 2 solutions that I can find.
For 6 < z < 0 I can find no solutions.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Substitute x = 1-y-z into the 2nd equation, and you get
This gives the infinitely many solutions for z=1, and y can be chosen arbitrarily with x = 1-y-z. It also proves that if z is not 1, then 3 must divide z.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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When divided by 3, y can not leave a remainder of 2.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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y must be in the set {-3, -2, -1, 0, 1}
It should be easy from here. To come up with this restriction, use my previous post #6.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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