How do solve #(x+2)/(x+5)>=1# algebraically?

1 Answer
Jan 23, 2017

The answer is #x<-5#

Explanation:

Multiply LHS and RHS by #(x+5)^2#

Therefore,

# (x+2) / (x+5) * (x+5)^2>=1*(x+5)^2#

#(x+2)(x+5)>=(x+5)^2#

#(x+2)(x+5)-(x+5)^2>=0#

#(x^2+7x+10)-(x^2+10x+25)>=0#

#7x+10-10x-25>=0#

#-3x-15>=0#

#3x<=-15#

#x<=-5#

We have to remove the equal sign, as we divide by #0#

So,

#x<-5#