3 problems 2x- y + z = 5 x + y -2z = -4 3x - y + 3z + 10 2x+3y - 2z = -11 3x - 2y + 3z = 7 x-4y + 4z = 14 x-y + z = 1 2x - 2y + 2z = 2 -3x + 3y - 3z = -3

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Solve each system by elimination of variables
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3 problems

2x- y + z = 5 ...........................(1)
x + y -2z = -4 ...........................(2)
3x - y + 3z =10 ...........................(3)

Solution. (1)+(2), we get
3x-z=1 ...............(4)
(2)+(3), we get
4x+z=6 .............(5)
Then (4)+(5), we get
7x=7
So, x=1
Then by (4), we get
z=2x-1=3*1-1=2
Then by (1), we have
y=2x+z-5=2*1+2-5=-1

So, x=1, y=-1, z=2

2x+3y - 2z = -11 .....................(1)
3x - 2y + 3z = 7 .....................(2)
x-4y + 4z = 14 .....................(3)

Solution. (1)+(3)*(-2), we get
11y-10z=-39 ....................(4)
(2)-3*(3), we get
10y-9z=-35 .................(5)
Then (4)*10-(5)*11, we get
-z=-39*10+35*11=-5
So, z=5

Then by (4), we get
11y=-39+10z=-39+10*5=11
So, y=1

Then by (3), we get
x=14+4y-4z=14+4*1-4*5=-2

So, x=-2, y=1, z=5

x-y + z = 1 ......................(1)
2x - 2y + 2z = 2 ..........(2)
-3x + 3y - 3z = -3 ...........(3)

Solution. We multiply by 0.5 on both sides of (2), we get
x-y+z=1
which is the same as (1).

We multiply by -1/3 on both sides of (2), we get
x-y+z=1
which is the same as (1).

So, we conclude that the system has infinitely many solutions.