How do you solve #(3x)/(x + 2)<5#?

1 Answer
Jun 18, 2017

Answer:

The solution is #x in (-oo,-5) uu (-2, +oo)#

Explanation:

We cannot do crossing over, we rearrange the inequality

#(3x)/(x+2)<5#

#(3x)/(x+2)-5<0#

#(3x-5(x+2))/(x+2)<0#

#(3x-5x-10)/(x+2)<0#

#(-2x-10)/(x+2)<0#

#(-2(x+5))/(x+2)<0#

Let #f(x)=(-2(x+5))/(x+2)#

Let's build a sign chart

#color(white)(aaaa)##x##color(white)(aaaaaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaaaaa)##-2##color(white)(aaaaaaa)##+oo#

#color(white)(aaaa)##-(x+5)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##-#

#color(white)(aaaa)##(x+2)##color(white)(aaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##-#

Therefore,

#f(x)<0#, when #x in (-oo,-5) uu (-2, +oo)#
graph{(3x)/(x+2)-5 [-36.53, 36.52, -18.28, 18.27]}