PLEASE SOMEONE HELP ME.
if you were to use substitution method to solve the following system, choose the new system of equations that would result if x was isolated in the second equation. 2x+5y-2z=-1 x+3y+z=-2 3x-2y+3x=-17
A. -y-4x=3
-11y=-11
B.11y=3
7y+6x=-11
C.4x+11y=-5
-11=-11
D.-y=-5
6x+7y=-11
so x from second equation will be
x=-2-3y-z
2x+5y-2z=-1
3x-2y+3z=-17 hope that this is hence right,correct with 3z not with 3x before =
so when x=-2-3y-z will get
2(-2-3y-z)+5y-2z=-1
3(-2-3y-z)-2y+3z=-17
-4-6y-2z+5y-2z=-1
-6-9y-3z-2y+3z=-17
-y-4z=3
-11y=-11 so from this result that y=1 so than
-1-4z=3 so -4z=4 so z=4/(-4)=-1 so z=-1 so than
x=-2-3(1)-(-1)=-2-3+1=-5+1=-4 so x=-4
x=-4
y=1
z=-1