Algebra Question
posted by mike on .
I need to solve the system of equations letting z be the parameter
7x+y+7z=12
3x+4y+6z=1
This is suppose to be the right answer
{47/3122/31z43/3163/31z,z}
I keep getting
{22/314731z63/3143/31z,z}

7x+y+7z=12
3x+4y+6z=1
28x+4y+28z=48
3x+4y+6z=1
subtract bottom eq
31x22z=47
x= 47/31 +22/31 z
You can do the x. 
7x+4y=16
6x4y=36