help me solve this sysem of equations?

Question

answers must check out in the end

x-2y-z=3
x+y+2z=9
2x+3y=z=0





-3x+4y=-6
5x -3z =-22
3y+2z=-1

Answer 1

x-2y-z=3..........................(1)
x+y+2z=9........................(2)
2x+3y+z=0......................(3)
add 1 and 3
3x+y=3............................(4)
add (3)*-2 and 2
-3x-5y=9...........................(5)
add 4 and 5 -4y=12 so y=-3
3x-3=3 so x=2
2+6-z=3 so z=5
so x=2,y=-3 and z=5

-3x+4y=-6.............................(1)
5x -3z =-22............................(2)
3y+2z=-1...............................(3)
(1)*5+(2)*3
-15x+20y=-30
15x-9z=-66
adding 20y-9z=-96 *2
40y-18z=-192
(3)*9
27y+18z=-9
adding 67y=-201
y=-3
substituting in (1)
-3x-12=-6 x=-2 and
substituting in (3)
-9+2z=-1 and so z=4
x=-2,y=-3 and z=4

Answer 2

Pick one variable to solve first. For example Z. Call your 3 equations A, B and C, note that if you do C-B-A, x and y will disappear:

(2x - x - x) + (3y - y - 2y) + (z - 2z - z ) = (0 - 9 - 3)
0 0 -2z = -12
z = 6

Now replace the new found value of z in equations A and B, for example:

x - 2y - 6 = 3 -> x - 2y = 9
x + y + 12 = 9 -> x + y = -3

Say you want to get rid of Y now: A + 2B does the trick

(x + 2x) + (-2y + 2y) = 9 - 2*3
3x = 3
x = 1

Now calculate y with one of the equations

x + y = -3
1 + y = -3
y = -4

You do the second one :-)

Answer 3

check the 3rd eq frome the first one

Answer 4

you got a proble with your threed equation assuming is +z not =z then
x 2,333
y -4,000
z 7,333

x -2,000
y -3,000
z 4,000