Three real numbers x, y, z are such that x2 + 6y = -􀀀17, y2 + 4z = 1 and z2 + 2x = 2. What is the value of x2 + y2 + z2? solve this if you can x2

I don't have a procedure for solving this, I just guessed, had to modify it once, and came up with:

x = -1     y = -3    z = -2

From these values, you can determine the value of x² + y² + z².

x^2 + 6y = -17   →    y = -[17 + x^2] / [6]   (1)

y^2 + 4z = 1  →  [1 - y^2] /[4] = z    (2)

z^2 + 2x = 2   →  x = [2 - z^2] /  [2]     (3)

Put (1) into (2)

z = [1 - (-[17 + x^2] / [6])^2 ] / 4  simplify

z = [1 - [(289 +34x^2 + x^4)]/36] / 4

z = [36 - 289 -34x^2 -x^4]/36]/4

z = [-253 - 34x^2 -x^4]/ 144

z =   -[(x^4 + 34x^2 +253)/144]     sub this into (3)

[-[(x^4 + 34x^2 +253)/144]]^2 + 2x = 2 .....  this would be diificult to solve by hand....using a Computer Algebra System, we have that x = -1    ...so x^2 = 1

And y = -[17 +(-1)^2'/[-6]   =  -(18)/6  = -3      ...so y^2  = 9

And z = [1 - (-3)^2]/ 4 = (-8)/ 4  = -2    so z^2  = 4

So

x^2 + y^2 + z^2 =   1 + 9 + 4 = 14