How do you solve #(x+7)/(x-3)>0#?

1 Answer
Feb 6, 2017

Answer:

The answer is #x in ]-oo, -7 [uu] 3, +oo[#

Explanation:

Let #f(x)=(x+7)/(x-3)#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-7##color(white)(aaaa)##3##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+7##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(x)>0# when #x in ]-oo, -7 [uu] 3, +oo[#