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Mathematics 22 Online
OpenStudy (anonymous):

Determine solution set of the inequality \[|x-1|\geq 3|x+1|\]

terenzreignz (terenzreignz):

Interesting :) We can always bring the 3 inside the absolute value... \[\Large |x-1 |\ge |3x + 3|\]

terenzreignz (terenzreignz):

Or... wait... rethinking...

terenzreignz (terenzreignz):

Might be better to divide both sides by |x+1| instead \[\Large \left|\frac{x-1}{x+1}\right|\ge 3\]

OpenStudy (anonymous):

hmm

terenzreignz (terenzreignz):

So that this means either \[\Large \frac{x-1}{x+1}\ge 3\]or \[\Large \frac{x-1}{x+1}\le -3\]

OpenStudy (anonymous):

the solution is -2<x<-1/2 but i don't know how we got it ..

terenzreignz (terenzreignz):

Patience. Solve this one: \[\Large \frac{x-1}{x+1}\ge 3\]

OpenStudy (anonymous):

ok..

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