help please I need to solve calling on Dr D & Duke, et al!?

Question

I am so surprised I have been trying for hours! have to give test
so, I figured since i cant stay up all morning due to promos in a few hours I felt it be best to ask the best ...thanks

z+y+x=6
2z+y=4
2x+y=8
am I correct please help! I am serious times running out

did I understand you correctly? that there are to many other solutions? Please there has to be a way to solve, I cant believe I stayed up thinking I could answer others with problems and at this stage in the game... am I to late ?

did I understand you correctly? that there are to many other solutions? Please there has to be a way to solve, I cant believe I stayed up thinking I could answer others with problems and at this stage in the game... am I to late ?

Z=1
Y=2
X=3
I think I found a solution will it pan out

Answer 1

You have indeed found a solution. z=1,y=2,x=3 is a solution to your system of equations.
So is x=2,y=4,x=0. So too is x=1,y=6,z= -1.
And x=-6,y=20,z=-8. And so is..............
What the answerer was telling you about not being able to solve this system was correct; you cannot find the 1 UNIQUE solution when you have more
unknowns than equations.
Here's what I'm getting at:
Pick any number you want( I chose 20) for a value
for y. Immediately, 2z+y=4 tells you z=-8, and
2x+y=8 says x=-6. Plug x=-6,y=20,and z=-8 into
x+y+z=6, and it balances perfectly.
Whatever number you choose, be it for x,y, or z, you
immediately define what the other 2 values will be.
Equation 1, x+y+z=6, is simply Equation 2 and 3
added together. 2z+y=4 added to 2x+y=8 is
2x+2y+2z+12. Divide both sides by 2, and you get
x+y+z=6. As Winston Churchill said, "a sheep in
sheep's clothing!".
Such systems of equations like yours are known
as "linearly dependent". Whenever that occurs, you
can always get A solution, but you can never get
THE single unique solution.
This was a sneaky question! I think the teacher or
text-book presented it to show that you need one equation for every unknown, and none of the
equations should be one of the other equations in
disguise.
Cheers!

Answer 2

The reason you are having trouble is that there are infinitely many solutions to this set of equations. To see this, notice that twice the first is equal to the second plus the third equations -- thus you essentially have two equations and three unknowns.

So the set of solutions to this is:
y = any real number
x = (8-y)/2
z = (4-y)/2