How do you solve # | 5 / (2x-1) | >= | 1 / (x - 2) |#?

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Answer 1 · 2017-10-18T12:34:52

See below.

For #x ne 2# and #x ne 1/2#

# | 5 / (2x-1) | >= | 1 / (x - 2) | rArr 5abs(x-2) ge abs(2x-1)# or equivalently

#5 sqrt((x-2)^2)-sqrt((2x-1)^2) = epsilon^2 > 0# and squaring both sides

#5^2(x-2)^2-10 sqrt((x-2)^2(2x-1)^2)+(2x-1)^2=epsilon^4# and again

#(5^2(x-2)^2+(2x-1)^2-epsilon^4)^2=10^2(x-2)^2(2x-1)^2#

or

#(11 + epsilon^2 - 7 x) (9 + epsilon^2 - 3 x) (epsilon^2 + 3 x-9) ( epsilon^2 + 7 x-11)=0#

and then the solution sets

#{(11-7x le 0),(9-3x le 0),(3x-9 le 0),(7x-11 le 0):}#

giving the solution set #x = {11/7,3}#