Explanation:
what do we need to do with that ?
I assume we need to solve this (that means to get hopefully unique numbers for x, y, z or indicating that there are no or several or infinite solutions).
I also assume that the first equation is actually
2x + y - 3z = 0
you could transform it into
2x + y = 3z
but then this is NOT equal 0. if it were 0, then this would give us immediately
3z = 0
z = 0
that would be great, because then we could use the rest of the equations to simply solve for x and y.
so, let's do both cases, and you need to pick the approach for the equations that you actually have.
first case :
2x + y - 3z = 0
4x + 2y - 6z = 0
x - y + z = 0
when we look only at the first 2 equations, what do we see ?
we see that equation 2 is nothing else than equation 1 multiplied by 2.
that means that they are both the same equation giving us the same information.
we are left with only 2 different equations with 3 variables.
similarly to one equation with 2 variables this gives us infinitely many possible solutions.
because when we simply add the first and the third equation, we get
2x + y - 3z = 0
x - y + z = 0
-------------------------
3x 0 - 2z = 0
3x = 2z
x = 2z/3
this has infinitely many solutions, as for any value of z we can think of we get another fitting value of x.
second case :
2x + y = 3z = 0
4x + 2y - 6z = 0
x - y + z = 0
therefore, z = 0, and we get for the rest
2x + y = 0
4x + 2y = 0
x - y = 0
now the fact that equation 1 and 2 are the same did not bother us, as we have 2 different equations (either equation 1 or equation 2 and equation 3) with 2 variables.
let's add the first and the third equation :
2x + y = 0
x - y = 0
-----------------
3x 0 = 0
3x = 0
x = 0
x - y = 0
x = y
0 = y
so, the only solution here is
x = 0, y = 0, z = 0