How do you solve #|x-1| - |x-3| >= 5#?

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Answer 1 · 2017-10-13T17:20:54

See below.

The solution for

#sqrt((x-1)^2)-sqrt((x-3)^2) = 5+e^2# is contained in the solutions for

#(x-1)^2+(x-3)^2-2 sqrt((x-1)^2(x-3)^2) = (5+e^2)^2#

and also in the solutions for

#((x-1)^2+(x-3)^2-(5+e^2)^2)^2=(x-1)^2(x-3)^2# or

#(3 + e^2) (7 + e^2) (9 + e^2 - 2 x) (1 + e^2 + 2 x)=0# and then we obtain the conditions

#{(9 + e^2 - 2 x=0),(1 + e^2 + 2 x=0):}#

which is equivalent to

#2x ge 9# and #2x le -1#

Thus we conclude that the inequality has no solution.