math
posted by Paula on .
help with solving the system of equations
3x+4y+z=7
2y+z=3
5x+3y+8z=31

the middle equation is nice and gives us
z = 32y
sub that into the first and third
1) 3x + 4y + 3  2y = 7
3x + 2y = 4 (#1)
3) 5x + 3y + 8(32y) = 31
5x + 3y + 24  16y = 31
5x  13y = 55
5x + 13y = 55 (#3)
#1 by 5: 15x + 10y = 20
#3 by 3: 15x + 39y = 165
subtract them ...
29y = 145
y = 5
back in #1
3x + 2y = 4
3x + 10 = 4
x = 2
back into the middle
2y + z = 3
10 + z = 3
z = 7
x = 2, y = 5, z = 7