how do i do this Algebra 2 problem?

Question

x+y+2z=1
x-y+z=0
3x+3y+6z=4

all the variables cancel out. What are the answers?

Answer 1

When all the variables cancel out, then you have an infinite number of solutions, or no solutions. Here, you have none.

This problem shows you that you need to check for linear independence. That is, none of your equations should be linear combinations of any others. What is a linear combination? It's when you add equations which may or may not be multiplied by constants. If you get something like this, then you can't solve your system, as you'll have fewer equations than unknowns.

In your case, 3[equation1]-[equation3] yields 0=-1. This is false, so your system has no solution.

If there was a 3 instead of a 4, you would have an infinite number of solutions, where they satisfied the first and second equations.

Answer 2

x+y+2z =1
x-y+z=0
3x+3y+6z=4
rewrite the last equation as
3(x+y + 2z)=4
Then x+y+2z=4/3
But equation 1 says that x+y+2z=1
Therefore this system is inconsistent and cannot be solved.

Answer 3

x= -5/6
y= -1/6
z= 1

Answer 4

pretty sure it's dependent

{(x,y,z)|x+y+z=1}