Solution:

From equation (1), w + x = 2 – y

Substituting this into equation (3), we get (2 – y) y – 1 = z².

This rearranges to – y² + 2y – 1 = z², or – (y -1)²= z²

Therefore (y – 1)² + z² = 0.

As y – 1 and z must square to non-negative numbers, this is only possible if z = 0 and y = 1.

Therefore from equation (1), w + x = 1, and from equation (2), wx = 0.

If wx = 0, then either w = 0 which makes x = 1,

or x = 0 which makes w = 1.

Hence there are exactly two sets of values for w, x, y, and z which satisfy these equations.

w = 0, x = 1, y = 1, z = 0

and

w = 1, x = 0, y = 1, z = 0

Challenge 269: Awesome Algebra