How do you solve #x/(x+3)>=2#?

1 Answer
Dec 27, 2016

Answer:

The answer is #x in [-6, -3[#

Explanation:

We cannot do crossing over.

We rewrite the inequality as #x/(x+3)-2>=0#

#(x-2(x+3))/(x+3)>=0#

#(x-2x-6)/(x+)=(-x-6)/(x+3)>=0#

Let #f(x)=-(x+6)/(x+3)#

We can make a time chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-6##color(white)(aaaaa)##-3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##-x-6##color(white)(aaaa)##+##color(white)(aaaaaa)##-##color(white)(aa)##∣∣##color(white)(aa)##-#

#color(white)(aaaa)##x+3##color(white)(aaaaaa)##-##color(white)(aaaaaa)##-##color(white)(aa)##∣∣##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaaaa)##+##color(white)(aa)##∣∣##color(white)(aa)##-#

Therefore,

#f(x)>=0#, when #x in [-6, -3[ #