# x+z=1 2x+y-z=-3 x+2y-z=-1

You need to solve for `x,y,z` the given system of equations, using substitution, hence, you need to use the top equation to write z in terms of `x` , such that:

`z = 1 - x`

You need to replace `1 - x` for `z` in the next two equations, such that:

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

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

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

You need to use the equation `3x + y = -2` to write y in terms of x, such that:

`y = -2 - 3x`

Replacing -`2 - 3x` for y in the second equation, yields:

`2x + 2(-2 - 3x) = 0`

`2x - 4 - 6x = 0 => -4x - 4 = 0 => -4(x + 1) = 0 => x + 1 = 0 => x = -1`

`y = -2 + 3 => y = 1`

`z = 1 + 1 => z = 2`

Hence, evaluating the solution to the given system of equation, using the requested substitution method, yields `x = -1, y = 1, z = 2.`

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