SOLUTION: X+Y+Z=4, X+Y-Z=4, 3X+3Y+Z=12 FIND VARIABLE XYZ

Question 1091457: X+Y+Z=4, X+Y-Z=4, 3X+3Y+Z=12
FIND VARIABLE XYZ
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The first equation says x+y+z is equal to 4; the second says x+y-z is also equal to 4. Either by logical reasoning or by formal algebra, that means z must be 0.

So now the first two equations both say x+y is 4; and the third says 3x+3y is 12. But that third equation is then equivalent to the other two.

We know z is 0; and all we know about x and y is that their sum is 4. So we can't find a single solution; there is an infinite family of solutions.
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So we can choose any value for x, and then the corresponding y value is 4 minus the x value we chose. Algebraically, there are an infinite set of solutions defined using parameter x as follows:

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