Simultaneous equations

3x^2 = 2tx
-z = 2ty
-y = 2tz
x^2 + y^2 + z^2 = 2

Solve for x,y,z (if possible!?)

[1] 3x² = 2tx
[2] -z = 2ty
[3] -y = 2tz
[4] x² + y² + z² = 2

From [2], [3]

x² + y² + z² = x² + 4t² (y² + z²)
y² + z² = 4t² (y² + z²)

So either y=z=0 or t=±1/2

Suppose y=z=0

x=±√2 from [4], and t=±3/√2 from [1].

Solutions: (√2, 0, 0), (-√2, 0, 0)

Now if t = ±1/2, then the equations become

[1] 3x² = ±x
[2] -z = ±y
[3] -y = ±z
[4] x² + y² + z² = 2

(the sign of the ± chosen must be the same in all 3 cases). 

Either y=z or y=-z from [2] and [3], and either x = 0 or x = ±1/3 from [1].

If x = 0, we must have

2y² = y² + z² = 2

so y = ±1

Solutions: (0, 1, -1), (0, -1, 1)

If x = ±1/3, then

2 = x² + y² + z² = 1/9 + 2y²

and y = ±√(17/18)

Solutions:(1/3, √(17/18), -√(17/18)) (1/3, -√(17/18), √(17/18)) (-1/3, √(17/18), -√(17/18)) (-1/3, -√(17/18), √(17/18))


So altogether, eight solutions

(√2, 0, 0)
(-√2, 0, 0)
(0, 1, -1)
(0, -1, 1)
(1/3, √(17/18), -√(17/18))
(1/3, -√(17/18), √(17/18))
(-1/3, √(17/18), -√(17/18))
(-1/3, -√(17/18), √(17/18))