How do you solve #(x+10)/(x+3)<2#?

1 Answer
Jul 31, 2017

Answer:

The solution is #x in (-3,4)#

Explanation:

We cannot do crossing over

#(x+10)/(x+3)<2#

Let's rewrite the inequality

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

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

#((x+10-2x-6))/(x+3)<0#

#((4-x))/(x+3)<0#

Let #f(x)=(4-x)/(x+3)#

We build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-3##color(white)(aaaaaaaa)##4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

Therefore,

#f(x)<0# when #x in (-3,4)#

graph{(x+10)/(x+3)-2 [-16.02, 16.02, -8.01, 8.02]}