Question

What is the solution of the system of the linear equations?

2x+3y–z=–16
–3x+3y+z=20
–x+2y+z=2
x=, y=, z=

Answer 1

`2x+3y-z=-16 \\ -3x+3y+z=20 \\ -x+2y+z=2 \\ \\ \hbox{the 1st and 2nd equation:} \\ 2x+3y-z=-16 \\ \underline{-3x+3y+z=20} \\ 2x-3x+3y+3y=20-16 \\ -x+6y=4 \\ \\ \hbox{the 1st and 3rd equation:} \\ 2x+3y-z=-16 \\ \underline{-x+2y+z=2 \ \ \ } \\ 2x-x+3y+2y=2-16 \\ x+5y=-14`


`\hbox{the two new equations:} \\ -x+6y=4 \\ \underline{x+5y=-14} \\ 6y+5y=4-14 \\ 11y=-10 \\ y=-(10)/(11) \\ \\ x+5y=-14 \\ x+5 * (-(10)/(11))=-14 \\ x-(50)/(11)=-(154)/(11) \\ x=-(104)/(11) \\ x=-9 (5)/(11)`


`-x+2y+z=2 \\ -(-(104)/(11))+2 * (-(10)/(11))+z=2 \\ (104)/(11)-(20)/(11)+z=(22)/(11) \\ (84)/(11)+z=(22)/(11) \\ z=-(62)/(11) \\ z=-5 (7)/(11) \\ \\ \boxed{x=-9 (5)/(11), y=-(10)/(11), z=-5 (7)/(11)}`