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I know that when solving inequalities, you reverse the sign when dividing or multiplying by a negative number, but right now I need to solve $\frac{x+9}{x-9}≤2$. Do I need to reverse the sign when I multiply $x-9$ across?

I'm sorry if this is outrageously obvious...

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    $\begingroup$ You have to split into two cases $x>9$ and $x<9$. $\endgroup$ – user99914 Mar 12 '16 at 23:16
  • $\begingroup$ @JohnMa But the point of this is to solve for $x$ so why would we set cases for $x$? In this case, what would the answer, $x$, be? $\endgroup$ – Spica Mar 12 '16 at 23:27
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    $\begingroup$ We know that there are only three cases that $x$ could be, $x-9 <0, =0, $or $>0$. The second case is not possible as now $x-9$ is at the bottom. So we split into two cases, in order to solve the inequality. For example, when $x- 9>0$, then you multiply both side by $x-9$ (this won't change the inequality sign as $x-9 >0$) to get $x+9 \le 2(x-9)$, which is the same as $x\ge27$. So $\{ x-9 >0\}\cap\{ x\ge 27\}$ is part of the solution. Now do the same for $x-9<0$. $\endgroup$ – user99914 Mar 12 '16 at 23:34
  • $\begingroup$ @JohnMa When I do the same for $x-9<0$, I get to $x+9≥2(x-9)$ which gives me $x≤27$. But I still don't understand. For example, in order for $x-9<0$ to be true, $x<9$; and for $x-9>0$ to be true, $x>9$; which would be saying when $x>9$, $x≥27$, and when $x<9$, $x≤27$... How can you define the values of $x$ based on a defined value of $x$ itself? $\endgroup$ – Spica Mar 12 '16 at 23:43
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    $\begingroup$ First we conclude that if a fixed value $x>9$ solves our inequality then it has to be $>=27$ (As far as I see here you would agree with me). Until now we do not know yet if $x>=27$ really does solve our inequality - but if you read now your transformations backwards, you will see that $x>=27$ does indeed solve the inequality you started with (here, during the second step, we do not assume $x>9$, what confused you a bit). $\endgroup$ – CHwC Mar 13 '16 at 0:36
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This is an inequality. You should not multiply across by $(x-9)$ at all. Instead you should bring the $2$ over to the left and combine the fraction through a common denominator. Verify you arrive at: $\frac{-x+27}{x-9}≤0$ Now one should make separate numberlines for numerator and denominator and verify the signs on the numberlines. NUM:$$___________plus_______27__minus______$$ DENOM: $$___minus____9____plus_____________$$ When you "divide" the numberlines you look for the negative interval, which is $x<9$ or $x\geq27$

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  • $\begingroup$ You should use \geq instead of => $\endgroup$ – Nikunj Mar 13 '16 at 4:02
  • $\begingroup$ @Nikunj Ha! For the less or equal sign I copy-pasted the Op's inequality symbol. Now I needed the greater or equal symbol and I didn't know it...(but now I do) $\endgroup$ – imranfat Mar 13 '16 at 4:07
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I suppose it depends on what $x$ is an element of. If we assume that $x\in\mathbb{R}$, we have to break this inequality into separate cases.

There is the case where $x>9$, in which case the sign does not change, and there is the case of $x<9$, in which the sign does change.

Of course, this is undefined for $x=9$.

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  • $\begingroup$ But the point of this is to solve for $x$ so why would we set cases for $x$? In this case, what would the answer, $x$, be? $\endgroup$ – Spica Mar 12 '16 at 23:27
  • $\begingroup$ We need to split this problem into cases, because there is no solution for this problem without assuming $x<9$ or $x>9$. The result is that we get one solution for one assumption and a different solution for the other. $\endgroup$ – Alekos Robotis Mar 12 '16 at 23:40
  • $\begingroup$ In order for $x-9<0$ to be true, $x<9$; and for $x-9>0$ to be true, $x>9$; which would be saying when $x>9$, $x≥27$, and when $x<9$, $x≤27$... How can you define the values of $x$ based on a defined value of $x$ itself? $\endgroup$ – Spica Mar 12 '16 at 23:46
  • $\begingroup$ I don't understand what you mean. The issue is that $x$ is arbitrary. So, in separating into two cases, we make sure to not lose generality of our solution. $\endgroup$ – Alekos Robotis Mar 13 '16 at 0:00
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    $\begingroup$ It is rather like that: First we conclude that if a fixed value $x>9$ solves our inequality then it has to be $>=27$ (As far as I see here you would agree with me). Until now we do not know yet if $x>=27$ really does solve our inequality - but if you read now your transformations backwards, you will see that $x>=27$ does indeed solve the inequality you started with (here, during the second step, we do not assume $x>9$, what confused you a bit). $\endgroup$ – CHwC Mar 13 '16 at 0:34

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