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x-y+z=4, x+y+z=2, 2x+y-3z=0...

`x-y+z=4`, `x+y+z=2`, `2x+y-3z=0`

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Given,
`x−y+z=4`.....(i)
`x+y+z=2`....(ii)
`2x+y−3z=0`....(iii)
Solving (iii), y=−2x+3z
Substituting this value in (i) and (ii),
we get, `3x−2z=4`...(iv)
`−x+4z=2`...(v) ...
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