Question 62306
To solve:
(1)  x-2y-3z=4
(2)  2x-4y+5z=-3
(3)  5x-6y+4z=-7

First, I will state the operations that we are allowed to perform on these equations without changing the solutions:
(a) We can interchange any two equations
(b) We can multiply an equation by a constant other than zero
(c) We can multiply an equation by a constant other than zero and add it to another equation, replacing that equation

We'll start by doing (c).  We'll multiply equation (1) by -2 and then add equation (1) to equation (2) and we will get:

(1) -2x+4y+6z=-8
(2)  0x+0y+11z=-11 from this equation, we see that z=-1
(3)  5x-6y+4z=-7

Next, we'll plug z=-1 into eq (1) and (3) and get:

(1)  -2x+4y=-2 
(3)   5x-6y=-3

Next, we'll multiply eq (1) by 5 and eq (3) by 2 and we get:

(1)  -10x+20y=-10
(3)   10x-12y=-6

Now add eq (1) to eq (3), replacing eq (3) and we have:

(1)  -10x+20y=-10
(3)    0x+ 8y=-16  and here we have y=-2

Next divide eq (1) by -10 
(1)  x-2y=1  now plug in y=-2 and we have:
(1)  x+4=1 ;x=-3
We now have 
x=-3
y=-2
z=-1

I'll leave the second problem up to you.  It looks to be fairly straightforward.  Hint: You know from eq (3) that y=z.  What happens if you plug that into equation (1), then add eq (1) and (2)?

Hope this helps----ptaylor