Inequality with (1-x) as denominator How do I solve $\frac{1}{x-1}>0$ for $x$?
If I multiply both sides with $x-1$ then becomes $1\gt 0$. I know it's wrong. How do I solve it?
 A: Hint: ${x\over{y}}>0$ where $x>0$ if and only if $y>0$ (if it confuses you, try to see what happens if $y\le0$!)
In other words, ${1\over{x-1}}>0$ if and only if $x-1>0$
A: A basic fact is that a number $a\neq 0$ and its inverse have the same sign, hence
$$\frac1{x-1}>0\stackrel{x\neq 1}\iff x-1>0.$$
A: The sign of the fraction is given by the multiplication of the sign's of the numerator and the denominator. In your case you want the fraction be positive, and since the numerator is $1 > 0$, then the denominator must also have the same sign. Thus, you want $ x-1 > 0$, which turns out to give you $ \boxed{x >1}$.
A: Remember that all fractions $\dfrac{a}{b}$ is greater than zero, if $a>0$ and $b>0$ or $a<0$ and $b<0$. In the case of your problem, we can take the same approach. In $\frac{1}{x-1}$, $a=1$ and $b=x-1$. We know that $1>0$ therefore the answer is $x-1>0$
A: The quotient of real numbers is positive $\iff$ numerator and denominator are of the same sign. Numerator is $1$, that's positive, so denominator must be positive, too: $$x-1>0$$ so the solution is $$x>1$$ Done.
A: Divide this into cases. The expression is not defined if $x = 1$.  If $x > 1$, you can multiply both sides by $x - 1$ to get $1 >0$  So, if $x > 1$ the inequality is satisfied.  
If $x < 1$, multiplying both sides by $x - 1$ reverses the inequality and you hae
$1 < 0$.  This is never true, so if $x < 1$, the inequality does not hold.
Hence the solution is $x > 1$.
A: Instead of multiplying both sides by $x-1$, multiply instead by $(x-1)^2$
