SOLUTION: The directions are Solve each system by elimination. Check your answers. The first problem is x-y+z=-1 x+y+3z=-3 2x-y+2z=0 I've tried this like 15 times an

Click here to see ALL problems on Linear-equations

Question 238270: The directions are Solve each system by elimination. Check your answers. The first problem is
x-y+z=-1 x+y+3z=-3 2x-y+2z=0
I've tried this like 15 times and cannot figure it out. Most of the time I would check what I got and it would be 1 instead of -1 or 3 instead of -3... I know how to do them I think. But I just can't get it.
Found 2 solutions by alicealc, stanbon:
Answer by alicealc(293) (Show Source):
You can put this solution on YOUR website!
first, eliminate y from the 1st and 2nd equation:
x - y + z = -1
x + y + 3z = -3
--------------- (+)
2x + 4z = -4
divide both sides by 2:
x + 2z = -2

then, eliminate y from the 1st and 3rd equation:
x - y + z = -1
2x - y + 2z = 0
----------------- (-)
-x - z = -1
multiply both sides by (-1)
x + z = 1

then, eliminate x from both results:
x + 2z = -2
x + z = 1
------------ (-)
z = -3

subtitute z = -3 into x + z = 1
x + (-3) = 1
x = 1 + 3 = 4

substitute x = 4 and z = -3 into the 1st equation:
x - y + z = -1
4 - y + (-3) = -1
1 - y = -1
1 + 1 = y
2 = y

so,the result will be: x = 4, y = 2, z = -3

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Solve each system by elimination. Check your answers. The first problem is
x-y+z=-1
x+y+3z=-3
2x-y+2z=0
---------------------
Subtract 1st from the 2nd;
Subtract 2 times the 1st from the 3rd.
-----------
x-y+z=-1
0x+2y+2z=-2
0x+y+0z=2
-----------------
Solve the 3rd Eq. for y: y = 2
--------------------------
Substitute y=2 into the 2nd equation and solve for "z":
2*2 + 2z = -2
2Z = -6
z = -3
------------
Substitute into x-y+z=-1 and solve for "x"
x - 2 - 3 = -1
x = 4
===================================
Solution:
x = 4
y = 2
z = -3
00000000000000000000000000000000000
Cheers,
Stan H.