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

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Answer 1 · 2018-06-19T09:45:09

The solution is #x in (2,4)#

You cannot do crossing over.

The inequality is

#x/(x-2)>2#

#=>#, #x/(x-2)-2>0#

Placing on the same denominator

#=>#, #(x-2(x-2))/(x-2)>0#

#=>#, #(x-2x+4)/(x-2)>0#

#=>#, #(4-x)/(x-2)>0#

Let #f(x)=(4-x)/(x-2)#

Let's build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##2##color(white)(aaaaaa)##4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-2##color(white)(aaaaaa)##-##color(white)(aa)##||##color(white)(aa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##4-x##color(white)(aaaaaa)##+##color(white)(aaaa)####color(white)(a)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaa)####color(white)(a)##+##color(white)(aaaa)##-#

Therefore,

#f(x)>0# when #x in (2,4)#

graph{(4-x)/(x-2) [-20.27, 20.27, -10.14, 10.14]}