Simultaneous equations

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

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

z = -2ty
y = -2tz
-> z = -2t*-2tz
   z = 4ttz
   1 = 4tt
   t = .5
z = -2*.5*y
z = -y

3x^2 = 2tx
3x = 1
x = 1/3

x^2 + y^2 + z^2 = 2
1/9 + 2y^2 = 2
2y^2 = 17/9
y^2 = 17/18
y = sqrt(17/18)
z = -sqrt(17/18)

Thanks a lot, but how do you know t is 0.5 and not -0.5?

z = 4ttz

And how can you divide through by z on this line when z could be 0?

In the stated case, y,z are opposite signs.
For t = -1/2, x=-1/3, y,z are same signs.

>>5
So six combinations of x,y,z huh? thanks.

>>6
4.  Your choice of t uniquely determines x.  Your choice of y, then uniquely determines z.  That's 2x2 = 4.

[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))

>>8
I knew I shouldn't have trusted 4tran.

>>8

Oopsie, the last two should be


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

If the bottom sign is chosen, y=z

>>10

And actually

(0, 1, 1)
(0, -1, -1)

work as well (with t = -1).  So ten solutions in all.

Fuck I hate problems like this.

In case you wanted to know, everyone got the wrong answer. There are 7 solutions.