# solve simultaneous equations in 3 variables; i don't know how x+y+z=1 2xy-z^2=1

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### 1 Answer

You need to use the first equation to write z in terms of x and y such that:

`z = 1 - x - y`

You need to substitute 1 - x - y for z in the next equation such that:

`2xy - (1 - x - y)^2 =` 1

You need to expand the square such that:

`2xy - (1 + x^2 + y^2 - 2x - 2y + 2xy) - 1 = 0`

`2xy - 1 - x^2- y^2 + 2x + 2y - 2xy - 1 = 0`

Reducing the like terms yields:

-`1 - x^2 - y^2 + 2x + 2y - 1 = 0`

`x^2 + y^2 - 2x - 2y + 2 = 0`

You need to group the terms such that:

`(x^2 - 2x + 1) + (y^2 - 2y + 1) = 0`

Using the special products yields:

`(x - 1)^2 + (y - 1)^2 = 0`

Since the sum of squares is never zero, unless both terms are equal to zero, yields:

`(x - 1)^2 = 0 =gt x - 1 = 0 =gt x = 1`

`(y - 1)^2 = 0 =gt y - 1 = 0 =gt y = 1`

Evaluating z yields:

`z = 1 - 1 - 1 =gt z = -1`

**Hence, evaluating the solution to the system yields `x = 1 ;y = 1 ;z = -1.` **