Question 1196171
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I'll get you started.


There are a multiple number of ways to approach elimination problems. 
It's mostly a trial and error type of thing, or something you develop a good eye for once you get enough practice.


Let's eliminate the variable z
The first two equations have +z in them
If we flip the signs of each term in equation (2), then we go from x+y+z = 0 to -x-y-z = 0


So we have
3x+2y+z = 1
-x-y-z = 0


Add the equations straight down<ul><li>The x terms add to 3x + (-x) = 2x</li><li>The y terms add to 2y + (-y) = y</li><li>The z terms cancel because z + (-z) = 0z = 0, which we intended (and why we did the sign flip in the second equation)</li><li>The right hand sides add to 1+0 = 1</li></ul>We end up with 2x+y = 1
I'll call this equation (4)


Return back to the original system
3x+2y+z=1
x+y+z=0
5x+3y-2z=-4



Temporarily delete the first equation to get
x+y+z=0
5x+3y-2z=-4


Now if we were to double everything in equation (2), then we go from x+y+z = 0 to 2x+2y+2z = 0


So this system
x+y+z=0
5x+3y-2z=-4


is the same as
2x+2y+2z=0
5x+3y-2z=-4


Add straight down:<ul><li>The x terms add to 2x + 5x = 7x</li><li>The y terms add to 2y + 3y = 5y</li><li>The z terms cancel because 2z + (-2z) = 0z = 0</li><li>The right hand sides add to 0 + (-4) = -4</li></ul>We end up with:
7x+5y = -4
which I'll refer to as equation (5)


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We now have a smaller system of equations
2x+y = 1
7x+5y = -4
which were equations (4) and (5) mentioned earlier.


I'll let you finish up the problem. Feel free to ask about any step, or if you are still stuck. 


Hint: multiply equation (4) by -5 so you can eliminate the y variable. 
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