How do you solve #3/(x-2)<5/(x+2)# using a sign chart?

1 Answer
Mar 6, 2017

Answer:

The solution is #x in ]-2,2 [ uu ]8, +oo [#

Explanation:

We cannot do crossing over

Let's simplify the inequality

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

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

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

#(3x+6-5x+10)/((x-2)(x+2))<0#

#(16-2x)/((x-2)(x+2))<0#

#(2(8-x))/((x-2)(x+2))<0#

Let #f(x)=(2(8-x))/((x-2)(x+2))#

We, now, build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaaaaaa)##2##color(white)(aaaaaa)##8##color(white)(aaaaaaa)##+oo#

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

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

#color(white)(aaaa)##8-x##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(aaa)##+##color(white)(aaaa)##-#

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

Therefore,

#f(x)<0# when #x in ]-2,2 [ uu ]8, +oo [#