SOLUTION: How do I solve: x+y+z=4 x+y-z=6 2x-3y+z=-1 Thank you for your time.

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Question 469233: How do I solve:
x+y+z=4
x+y-z=6
2x-3y+z=-1
Thank you for your time.
Found 2 solutions by ewatrrr, lwsshak3:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Different ways to solve this system. EQs lend to elimination, substitution methods

x+y+z=4

x+y-z=6



 x +y+z=4  

-x -y+z =-6   |multiplying 2nd Eq thru by -1 to eliminate both x and y

    2z = -2  z = -1



2x-3y+(-1)=-1  |substituting z = -1

 x+y+(-1)=4

therefore:

2x-3y = 0

 x+y = 5



2x-3y = 0

-2x-2y = -10   Multiplying 2 EQ thru by -2 to eliminate x

   -5y = -10   y = 2  and x = 3   (x + y = 5)

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
1) x+y+z=4
2) x+y-z=6
3) x-3y+z=-1
...
1) x+y+z=4
2) x+y-z=6
add to eliminate z
4) 2x+2y=10
..
2) x+y-z=6
3) x-3y+z=-1
add to eliminate z
5) 2x-2y=5
..
4) 2x+2y=10
5) 2x-2y=5
add to eliminate y
4x=15
x=15/4
2y=2x-5=15/2-5=15/2-10/2=5/2
y=5/4
z=4-x-y=4-15/4-5/4=16/4-15/4-5/4=-4/4=-1
Check:
1) x+y+z=15/4+5/4-1=20/4-4/4=16/4=4
2) x+y-z=15/4+5/4+1=20/4+4/4=24/4=6
3) x-3y+z=15/4-3(5/4)-1=15/4-15/4-1=-1