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

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.

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

Subtract #1# from both ends to get:

#0 > 1/(x-1)#

#1/(x-1) < 0#

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

(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.

To solve the inequality \( \frac{x}{x - 1} < 1 \), first, find the values of \(x\) that make the expression \(x/(x - 1)\) less than 1. Begin by setting up the equation \( \frac{x}{x - 1} = 1 \) and solving for \(x\). This gives us \(x = x - 1\). Next, subtract \(x\) from both sides to isolate \(1\) on one side, resulting in \(0 = -1\). Since this equation is false, we conclude that the solution does not include \(x = 1\). Now, examine the behavior of \( \frac{x}{x - 1} \) for \(x\) values greater and less than \(1\). If \(x < 1\), then \(x - 1\) will be negative, causing the fraction to be negative. If \(x > 1\), then \(x - 1\) will be positive, making the fraction positive. Therefore, the solution to \( \frac{x}{x - 1} < 1 \) is \( x < 1 \) or \( x > 1 \), excluding \( x = 1 \). To summarize, the solution is \( x < 1 \) or \( x > 1 \), with \(x\) not equal to \(1\).