x+2yz=3

2xy+z=1

6x3yz=7

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1 Answer

Add first two equations:

-x+y=2 so y=x+2.

Substitute for y:

x+2(x+2)-z=3, 3x+4-z=3, 3x-z=-1; -2x-(x+2)+z=-1, -3x-2+z=-1, -3x+z=1, or 3x-z=-1; 6x-3(x+2)-z=-7, 3x-6-z=-7, 3x-z=-1, or z=3x+1.

So there is no unique solution, that is, there are many solutions.

We can find y and z given x. For example, x=0 gives us y=2 and z=1.

Other examples: x=1 gives y=3 and z=4; x=-1 gives y=1 and z=-2.

There are an infinite number of solutions. Any x will give solutions for y and z: y=x+2, z=3x+1, which can be written (x,x+2,3x+1).

 

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