Not sure where last two equasions comes from. Question is rewritten from a book preparing to A-levels. Thanks for help in advance.

Sketch the curve given by y=(3x+3)/(x(3-x)) ***(1)***

"Range of values of y":

from (1): 3xy - (x^2)y = 3x +3 ***(2)***

from (2): (x^2)y + 3x(1-y)+3=0 ***(3)***

so for y=0 => 3x+3=0 => x=-1

Now for y!=-1 this is a quadratic in x.

Thus, for real x,

[3(1-y)]^2 - 4y(3)>= 0

i.e. (3y-1)(y-3) >= 0

...question continues with sketching the curve and is easy apart from last two lines.

Kind regards,

aski