Question

How do you solve the system `w-2x+3y+z=3`, `2w-x-y+z=4`, `w+2x-3y-z=1`, `3w-x+y-2z=-4`?

Answer 1

`(w, x, y, z) = (2, 3, 1, 4)`

Explanation:

Given:

`{ (w-2x+3y+z=3),(2w-x-y+z=4),(w+2x-3y-z=1),(3w-x+y-2z=-4):}`

Adding the first and third equations together, we get:

`2w=4`

So:

`w=2`

Discarding the first equation, and substituting `w=2` into the others, we get:

`{ (-x-y+z=0),(2x-3y-z=-1),(-x+y-2z=-10):}`

From the first equation:

`z = x+y`

Substituting this into the second and third equations, we get:

`{ (x-4y=-1),(-3x-y=-10):}`

Multiplying the second equation by `-4` we get:

`12x+4y=40`

Adding this to the first equation, we get:

`13x = 39`

Hence:

`x = 3`

From the second equation:

`y = 10-3x = 10-9 = 1`

Then:

`z = x+y = 3+1 = 4`

So:

`(w, x, y, z) = (2, 3, 1, 4)`