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

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Answer 1 · 2017-01-23T12:09:53

The answer is #x<-5#

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#