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

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Answer 1 · 2017-08-09T17:13:08

The solution is #x in (-oo,-6] uu(-2,+oo)#

We cannot do crossing over, let's rewrite the inequality

#(x-2)/(x+2)<=2#

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

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

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

#(-x-6)/(x+2)<=0#

#-(x+6)/(x+2)<=0#

Let #f(x)=-(x+6)/(x+2)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaaa)##-oo##color(white)(aaaa)##-6##color(white)(aaaaaaaa)##-2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##-(x+6)##color(white)(aaaa)##+##color(white)(aaaa)##0##color(white)(aaaa)##-##color(white)(aaaaa)##-#

#color(white)(aaaa)##(x+2)##color(white)(aaaaaa)##-##color(white)(aaaaaaaaa)##-##color(white)(aa)##||##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaa)##-##color(white)(aaa)##0##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(aa)##-#

Therefore,

#f(x)<=0# when #x in (-oo,-6] uu(-2,+oo)#

graph{(x-2)/(x+2)-2 [-22.8, 22.81, -11.4, 11.42]}