Solve for x: a) `||x-1|+2|le4`, b) `(x-4)/(x+2) le |(x+2)/(x-1)|`

Explanation

a) `||x-1|+2|le4| rArr -4 le|x-1|+2le4`
`rArr -6 le|x-1|le2`
`rArr |x-1|le2 rArr -2 le-1 le2`
`rArr -1 lexle3 rArr x in [-1.3]`
b) Case I: Given inequation will be satisfied for all x such that
`(x-4)/(x+2) le0 rArr x in (-2,4)-{1}`............(i)
(Note: {1} is not in domain of RHS)
Case 2:
`(x-4)/(x+2) lt 0 rArr x in(-infty, -2) cup (4, infty)`.................(ii)
Given inequation becomes
`(x-2)/(x-1) ge (x-4)/(x+2)` or `(x-2)/(x-1) le (x-4)/(x+2)`
On solving we get on solving we get
`x in (-2,4//5) cup (1, infty)` `x in (-2,0) cup (1,5//2)`
taking intersection with (ii) we get taking intersection with (ii), we get
`x in (4, infty)`...........(iii) `x in phi`
Hence, solution of the original inequation : `x in (-2,infty)-{1}` (taking union of (i) and (ii)