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?
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7 Answers
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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 $x > 1$.
answered Apr 22, 2015 at 12:44
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5$\begingroup$ put in $x = 0$ and you will see that that answer is in error. $\endgroup$ Apr 22, 2015 at 12:46
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1$\begingroup$ lets see 1/(x-1)<0 we dont know if x-1 is more or less than zero but we do know that (x-1)^2 is more than zero and wont affect the inequality hence, multiply both sides by (x-1)^2 giving x-1 < 0 rearranging gives x<1. My friend sent this to me . $\endgroup$– MathxxApr 22, 2015 at 13:08
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1$\begingroup$ @Mathxx How come
>in your friend solution became<? Btw your friend solution is correct (with the<changed) by chance. Actually with that procedure you study when the denominator is> 0, which, along with studying when the numerator is> 0(in this case always), is the correct way of studying when the whole inequality is> 0$\endgroup$– NarmerApr 22, 2015 at 13:43 -
$\begingroup$ @Mathxx Your question reads $\frac{1}{x-1} > 0$ but your comment here starts with $\frac{1}{x-1} < 0$. Which one is it? $\endgroup$– KyleApr 22, 2015 at 16:31
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Instead of multiplying both sides by $x-1$, multiply instead by $(x-1)^2$
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1$\begingroup$ This is a lot simpler than splitting into cases $\endgroup$ Apr 22, 2015 at 14:47
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1$\begingroup$ That discards information when $(x-1)^2 = 0$, and the only way to deal with that is to split into cases anyway . . . $\endgroup$– ruakhApr 22, 2015 at 15:11
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3$\begingroup$ @ruakh: It's pretty trivial to see that the LHS of the original inequality is undefined iff $x-1=0 \iff (x-1)^2 = 0$. $\endgroup$ Apr 22, 2015 at 16:31
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1$\begingroup$ while this approach doesn't eliminate the need for splitting into cases, it reduces its number (in this way we need only consider the cases where $x-1$ is zero or nonzero). $\endgroup$– John JoyApr 22, 2015 at 23:03
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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 $x-1>0$
answered Apr 22, 2015 at 12:42
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A basic fact is that a number $a\neq 0$ and its inverse have the same sign, hence $$\frac1{x-1}>0\stackrel{x\neq 1}\iff x-1>0.$$
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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}$.
answered Apr 22, 2015 at 13:07
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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 $x-1>0$
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2$\begingroup$ Beware! $\frac{a}{b}$ is greater than zero also when $a<0$ and $b<0$! $\endgroup$– NarmerApr 22, 2015 at 13:51
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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.
x < 1is the right answer if the question is about1/(1-x) > 0.x > 1is the right answer if the question is about1/(x-1)>0. $\endgroup$