设6x-y-2z=0,3x+y-28z=0(z不等于0),求(9x^2-3xy)/(y^2+6z^2)的值

Question

设6x-y-2z=0,3x+y-28z=0(z不等于0),求(9x^2-3xy)/(y^2+6z^2)的值

Answer 1

6x - y - 2z = 0 ___i
3x + y - 28z = 0___ii
用 i 加 ii 形成,
9x - 30z =0
3x = 10z
z = 3x/10___iii
用 i 减 2(ii) 形成,
-3y + 54z = 0
y = 18z
z = y/18___iv
iii = iv,
3x/10 = y/18
54x = 10y
x = 10y/54___v
(9x² - 3xy)/(y² + 6z²)___vi
把 iv 和 v 放入 vi 里,
[9(10y/54)² - 3(10y/54)(y)] / [y² + 6(y/18)²]
= [(900y²/2916) - (30y²/54)] / [y² + 6y²/324]
= [(-20/81)y²] / [(55/54)y²]
= -1080/4455
= -8/33