algebra
posted by Terri on .
solve the system
2xy+z=3
x+3y2x=11
3x2y+4z=1

double the first equation, then add that to the second to get
5x+y = 17
double the second, then add that to the third to get
5x + 4y = 23
now subtract those last two equations to get
3y = 6
y = 2
put that into 5x + y = 17 for
5x = 15
x = 3
now in the first
6  2 + z = 3
z = 1
I assumed your second equation had 2z at the end. typo ?? 
Thanks, yes, typo
I just finished solving it a long way, I solved the first two, then the second two,,etc etc. Your way seems much easier, I did get the same answers, yeah!!!! But, will your way always work?