How do you solve #x/(x+2)>=2#?

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Answer 1 · 2017-07-29T09:15:42

We need to multiply both sides by #(x + 2)^2# in order to preserve the inequality sign

i.e. #x(x + 2) >= 2(x + 2)^2#

so, #2(x + 2)^2 - x(x + 2) <= 0#

=> #(x + 2)[2(x + 2) - x] <= 0#

i.e. #(x + 2)(x + 4) <= 0#

Hence, #-4 <= x < -2#

Note x cannot equal -2 as this is excluded from the domain

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