Question

Precalc help?

Answer 1

`(x, y, z) = (1, -1, 2)`

Explanation:

Given:

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

Note that if we define `X=1/x`, `Y=1/y` and `Z=1/z`, then we get linear equations:

`{ (2X-Y+2Z=4), (X+2Y-2Z=-2), (3X+3Y+4Z=2) :}`

We can eliminate `Z` from the first equation by adding the second equation to it. We can eliminate `Z` from the third equation by adding twice the second equation to it. These two combinations give us:

`{ (3X+Y=2), (5X+7Y=-2) :}`

To eliminate `Y`, we can subtract `7` times the first equation from the second to get:

`-16X=-16`

Hence `X=1`.

Then substituting this value of `X` in `3X+Y=2` we find `Y=-1`.

Then substituting the values `X=1` and `Y=-1` in `2X-Y+2Z=4`, we find `Z=1/2`

Hence `x=1/X=1`, `y = 1/Y=-1` and `z=1/Z=2`