precal
posted by katie .
Solve for y:
(y(10/y))^2+6(y(10/y))27=0

(y(10/y))^2+6(y(10/y))27=0
his is just like the other one: Let u=y1/y and you have
u^2 + 6u 27 = 0
(u+9)(u3) = 0
(y  10/y + 9)(y  10/y  3) = 0
(y^2 + 9y  10)(y^2  3y  10) = 0
(y+10)(y1)(y5)(y+3) = 0
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