Question 1016590: x+y+z+w=1
2x+3z+2w=5
3x-y+w=-12
2y+5z+3w=7
Answer by FrankM(1040)   (Show Source):
You can put this solution on YOUR website! E1 x+y+z+w=1
E2 x+0y+3z+2w=5
E3 3x-y+0z+w=-12
E4 0x+2y+5z+3w=7
4 equations, 4 unknowns. We need to combine 3 pairs of equation to eliminate a variable. Let's start with x. Note, I added the missing variables using 0 as a coefficient.
-3x+-3y+-3z+-3w=-3 E1*-3
3x-y+0z+w=-12 E3
0x-4y-3z-2w=-15 sum1
-3x+0y-9z-6w=-15 E2*-3
3x-y+0z+w=-12
0x-y-9z-5w=-27 sum2
2y+5z+3w=7 E4 x already gone
-4y-3z-2w=-15 sum1
-y-9z-5w=-27 sum2
4y+10z+6w=14 E4*2
-4y-3z-2w=-15 sum1
0y+7z+4w=-1 sum3
-4y-36z-20w=-108 sum2*4
4y+10z+6w=14 E4*2
0y-26z-14w=-94 sum4
49z+28w=-7 sum3*7
-52z-28w=-188 sum4*2
-3z=-195
z=65 go to sum 4
-26(65)-14w=-94
-1690-14w=-94
-14w=1596
w=-114
x+0y+3z+2w=5 E2
x+3(65)+2(-114)=5
x+195-228=5
x= 38
Last, E1
x+y+z+w=1
38+y+65-114=1
y= 12
(38,12,65,-114)
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