System of equations
(1) 3x+2y-z+w=10
(2) 4x-y+z-w=6
(3) x+y+z+3w=12
(4) 2x+7y-3z+8w=28
There are various methods that can be used to solve this system.
The method I use is this:
(5)=(1)+(2)=7x+y=16
(6)=3(1)-(3)=8x+5y-4z=18
(7)=8(1)-(4)=22x+9y-5z=52
This reduces 4 equations to 3. We can substitute y=16-7x in (6) and (7):
(6) 8x+5(16-7x)-4z=18⇒8x+80-35x-4z=18, (8) 27x+4z=62;
(7) 22x+9(16-7x)-5z=52⇒22x+144-63x-5z=52, (9) 41x+5z=92.
We have reduced 3 equations to 2. Now we get down to 1 equation with one variable:
(10)=5(8)-4(9)=29x=58⇒x=2⇒4z=62-54=8⇒z=2; x=2⇒y=16-14=2.
Let's substitute x=y=z=2 in (1): 6+4-2+w=10⇒w=2. So the solution is w=x=y=z=2.
CHECK
(1) 6+4-2+2=10 OK
(2) 8-2+2-2=6 OK
(3) 2+2+2+6=12 OK
(4) 4+14-6+16=28 OK