#1 x + 3y + 2z = –16
#2 2x – y + 2z = –15
#3 2x – 2y – z = 1

#1 x + 3y + 2z = –16
x + 2z = -3y – 16
x=-3y – 2z – 16

#2 2(-3y – 2z – 16) – y + 2z = –15
-6y – 4z – 32 – y + 2z = -15
X= -7y – 2z = 17

#3 2(-3y – 2z – 16) – 2y – z = 1
-6y – 4z – 32 – 2y – z = 1
-8y – 5z = -31

-7y – 2z = 17
8y + 5z = 31
Y+3z = 48

This is what I have so far. Can anyone help me with?

Question

#1 x + 3y + 2z = –16
#2  2x – y + 2z = –15
#3  2x – 2y – z = 1
	
#1 x + 3y + 2z = –16
     x + 2z = -3y – 16
x=-3y – 2z – 16

       #2 2(-3y – 2z – 16) – y + 2z = –15
-6y – 4z – 32 – y + 2z = -15
X= -7y – 2z = 17

        #3 2(-3y – 2z – 16) – 2y – z = 1
-6y – 4z – 32 – 2y – z = 1
-8y – 5z = -31

-7y – 2z = 17
 8y + 5z = 31
Y+3z = 48

This is what I have so far.  Can anyone help me with?

anonymous · Answer

Its Solving a System of Equations with Three Variables

.sam. · Answer

You can solve this using vectors, did you learn that before?

anonymous · Answer

No i haven't

.sam. · Answer

From (1),
$$x + 3y + 2z = –16$$
$$x =-3y-2z-16$$
Sub (1) to (2),
$$2x – y + 2z = –15$$
$$2(-3y-2z-16) – y + 2z = –15$$
$$z=-\frac{7 y}{2}-\frac{17}{2}$$
Sub (z) and (x) to (3),
$$2(-3y-2(-\frac{7 y}{2}-\frac{17}{2})-16) – 2y – (-\frac{7 y}{2}-\frac{17}{2}) = 1$$
Solving it you'll get y=-1