Ok so let us label our equations first 1- 3x+2y+z=7 2- 5x+5y+4z=3 3- 3x+2y+3z=1 subtracting equation 3 from equation 1 3x+2y+3z=1 (-). 3x+2y+z= 7 ---------------------- 2z=-6----->z=-3 since we already chose equation 1 and 3 we must involve equation 2 substituting the value if z in the second equation 5x+5y-12=3 5x+5y=15 choosing either the first or the third 3x+2y-3=7 3x+2y=10 solving the system 5x+5y=15 3x+2y=10 multipling the first equation by 2 and the second equation by 5 10x+10y=30 15x+10y=50 subtracting the two equations -5x=-20--->x=4 substituting for the value of x 40+10y=30--->10y=-10y--->y=-1 so our soultion is x=4,y=-1,z==-3 or (4,-1,-3)
