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

1 Answer
Oct 12, 2015

Answer:

Reformulate without multiplying either side by a potentially negative expression to find:

#x < 1#

Explanation:

This could be a little fiddly, since if you multiply both sides by #x - 1#, you don't know the sign of #x - 1#, so you have to split into cases where you do or do not reverse the inequality.

So let's use a different approach:

#1 > x / (x - 1) = (x - 1 + 1) / (x - 1) = 1 + 1/(x - 1)#

Subtract #1# from both ends to get:

#0 > 1/(x-1)#

That is:

#1/(x-1) < 0#

The left hand side will be negative when #x - 1 < 0#, that is when #x < 1#.

In general you can perform any of the following operations on an inequality and preserve its truth:

(1) Add or subtract the same value on both sides.
(2) Multiply or divide both sides by the same positive value.
(3) Multiply or divide both sides by the same negative value and reverse the inequality (i.e. #<# becomes #>#, #<=# becomes #>=#, etc.).
(4) Apply the same strictly increasing monotonic function to both sides (e.g. #f(x) = e^x#).
(5) Apply the same strictly monotonically decreasing function to both sides and reverse the inequality.