How do you solve #m/(m-5)+3/(m-1)>0# using a sign chart?

1 Answer
Apr 11, 2017

Answer:

The solution is #m in (-oo,-5)uu(1,3)uu(5,+oo)#

Explanation:

Let's do some simplification

#m/(m-5)+3/(m-1)>0#

#(m^2-m+3m-15)/((m-5)(m-1))>0#

#(m^2+2m-15)/((m-5)(m-1))>0#

#((m+5)(m-3))/((m-5)(m-1))>0#

Let #f(m)=((m+5)(m-3))/((m-5)(m-1))#

We can build the sign chart

#color(white)(aaaa)##m##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaaaaa)##1##color(white)(aaaaa)##3##color(white)(aaaaaaa)##5##color(white)(aaaa)##+oo#

#color(white)(aaaa)##m+5##color(white)(aaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aa)##+##color(white)(aaaa)##||##color(white)(aa)##+#

#color(white)(aaaa)##m-1##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aa)##+##color(white)(aaaa)##||##color(white)(aa)##+#

#color(white)(aaaa)##m-3##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aa)##+##color(white)(aaaa)##||##color(white)(aa)##+#

#color(white)(aaaa)##m-5##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aa)##-##color(white)(aaaa)##||##color(white)(aa)##+#

#color(white)(aaaa)##f(m)##color(white)(aaaaa)##+##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aa)##-##color(white)(aaaa)##||##color(white)(aa)##+#

So,

#f(m)>0# when #m in (-oo,-5)uu(1,3)uu(5,+oo)#