How do you solve #4/(x-3)<2# using a sign chart?

1 Answer
Feb 15, 2017

Answer:

The answer is #x in ]-oo, 3[uu]5, +oo[#

Explanation:

Let's rearrange the equation

#4/(x-3)-2<0#

#(4-2(x-3))/(x-3)<0#

#(4-2x+6)/(x-3)<0#

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

Let #f(x)=(10-2x)/(x-3)#

The domain of #f(x)# is #D_f(x)=RR-{3}#

Now, we build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaaa)##3##color(white)(aaaaaaa)##5##color(white)(aaaaaa)##+oo#

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

#color(white)(aaaa)##10-2x##color(white)(aaaa)##+##color(white)(aa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#

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

Therefore,

#f(x)<0# when #x in ]-oo, 3[uu]5, +oo[#