How do you solve #x/(x+2)>=2#?

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Answer 1 · 2017-01-01T16:15:24

The answer is #x in [-4, 2[ #

You cannot do crossing over

#x/(x+2)>=2#

#x/(x+2)-2>=0#

#(x-2(x+2))/(x+2)>=0#

#(x-2x-4)/(x+2)>=0#

#(-x-4)/(x+2)>=0#

Let #f(x)=(-x-4)/(x+2)#

We can do a sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaaa)##-4##color(white)(aaaaaaa)##-2##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##-x-4##color(white)(aaaaa)##+##color(white)(aaaa)##0##color(white)(aaa)##-##color(white)(aa)##∥##color(white)(aa)##-#

#color(white)(aaaa)##x+2##color(white)(aaaaaaa)##-##color(white)(aaaa)####color(white)(aaa)##-##color(white)(aaa)##∥##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##-##color(white)(aaaa)####color(white)(aaa)##+##color(white)(aaa)##∥##color(white)(aa)##-#

Therefore,

#f(x)>=0#, when #x in [-4, 2[ #