Q:

System of linear equations: using an equation that is already gone?

If we have this system of equations:
$$x+4y+7z=0$$
$$2x+5y+8z=0$$
$$3x+6y+9z=0$$
my book says the first 2 steps for solving this system is to 1) subtract 2 times the first equation from the second equation to get $-3y-6z=0$ and 2) subtract 3 times the first equation from the last equation to get $-6y-12z=0$. 
I don't understand why you can use the first equation twice because by the time we get $-3y-6z=0$, the first equation $x+4y+7z=0$ is already gone. My thinking is that by subtracting 2 times the first equation from the second equation, this turns the equations $x+4y+7z=0$ and $2x+5y+8z=0$ into $-3y-6z=0$. So, we now have an equivalent system that consists of $-3y-6z=0$ and $3x+6y+9z=0$. These two equations are all we can work with now. So how are we able to use the first equation $x+4y+7z=0$ again (by multiplying it by 3 and subtracting it from $3x+6y+9z=0$) if we no longer have the first equation.

A:

The simplets  way of solving three linear simulteneous equations is to assume one of the variables as $p$, so let $z=p$. Then solve the fist two equations: $x+6y=-7p~~~~(1)$  and  $2x+5y=-8p$ get $y==2p$ and $x=p$. Now put these two values in third equation:
$$3x+6y+9y=0~~~(3) \Rightarrow 3p-12p+9p=0 \Rightarrow 0=0.$$ (I) If you get a true statement like $0=0$ 0r $5=5$. It means the equations have infinitely many solutions. Here, in this example this is the case you maget many sets of $(x,y,z)$ by assigning a values to $p.$
There may be two other cases(not in this example):
(II) If you get a false statement like $6=0$ or $7=8$, this means no solution.
(III) If you get a statement like $2p=10$, then $p$ gets fixed as 5. Then you can determine a unique triplet $(x,y,z)$ which will be the unique solution of your proble.
There are other ways too, but this one is the least expensive!

