The following system is solved by substitution. Which expression is substituted for z in the second and third equations?

x + 5y - z = 6
3x - y - 2z = 4
3x - 2y + 3z = 13

A x + 5y + 6
B x + 5y - 6
C -x - 5y + 6
D -x - 5y - 6
I did not mean to post this question but I realized Im unsure of how to do it so thanks anyone for help!!

Question

The following system is solved by substitution. Which expression is substituted for z in the second and third equations?

x + 5y - z = 6
3x - y - 2z = 4
3x - 2y + 3z = 13
 	
A	x + 5y + 6
B	x + 5y - 6
C	-x - 5y + 6
D	 -x - 5y - 6
 I did not mean to post this question but I realized Im unsure of how to do it so thanks anyone for help!!

anonymous · Answer

You are using the first equation and solving for z.  Just get z by itself on one side of the equals sign.  You will use 2 addition/subtraction operations to do that.

anonymous · Answer

Im not sure how to do that. What would I subtract from both sides? X?

anonymous · Answer

Think about it like you are "moving over" something.  For instance, you will move z over to the right (one way to solve).  You will add z to both sides.  Then, the -z on the left will drop out and a positive z will appear on the right.  Do similar for the 6 except now it's subtraction.

anonymous · Answer

So I would do this -->|dw:1348512298602:dw| Wouldnt there be too many variables? x y and z, i dont get how to get rid of them?