Now let me reclassify my problem:

I was solving some inequality until I stopped at this step $-1\le\frac{2}{x}$ Why did I stop?

Because if I do this next step $-x\le 2$ and then multiply both sides by $-1$, I will come out with this $x\ge-2$

I try some inputs on the main inequality and figure out that I am wrong even though My algebra had no problems?

So I remember something and go back in time and do this $-1\le\frac{2}{x}$ swap denominator and numerator for both sides $-1\ge\frac{2}{x}$ then $-x\ge2$ then $x\le-2$

And by the power of the nature this one is correct even though I made a paradoxical step as I assumed that $-\frac{2}{x}>0$ so that I become able to reverse the inequality. At last $x$ is less than $-2$ which means that my assumption was right .

I now have a problem with the fact that Algebra fooled me up there giving me a wrong answer,or did it?

Put in mind that I put in mind that $x$ is never equal to zero, but that's not what I am here for.

Also to note I do this stuff on the number line and that's what matters and then I can use the most suitable notation for my answer.

  • $\begingroup$ Hard to understand what you're asking here. $\endgroup$ – Hongyu Wang Sep 27 '18 at 13:38
  • $\begingroup$ I want to add that you could also multiply both sides of the equation with $x^2$ which is always positive (except when $x=0$). Now it remains to solve the quadratic equation: $$-x^2 - 2x \leq 0$$ where you only should be careful with $x=0$. This is fairly easy since it factorizes quickly. The other options are also valid of course but this is a more direct method :) $\endgroup$ – Stan Tendijck Sep 27 '18 at 14:18

A principled way of solving this is as follows. Start with $-1 \leq 2/x$. There are now two cases: either $x$ is positive or $x$ is negative (as you say, it cannot be zero). Consider these cases individually.

  • If $x$ is positive, then after multiplication of both sides by $x$ we get $-x \leq 2$ or $x \geq -2$. But since we assume $x$ is positive, this really means that the correct condition is $x > 0$.
  • If $x$ is negative then when we multiply by $x$ we have to remember to flip the inequality, i.e., we then get $-x \geq 2$ or $x \leq -2$.

So the complete set of solutions is $x>0$ or $x \leq -2$.

  • $\begingroup$ So now you are saying that X is negative , how did you know that X is negative before solving it (sorry if it is a dumb question ) , In other words why did you take x<=-2 as your answer $\endgroup$ – user597368 Sep 27 '18 at 13:43
  • 1
    $\begingroup$ @user597368 $x$ can be either positive or negative, so you have to consider both possibilities individually as their properties alter the answer $\endgroup$ – MRobinson Sep 27 '18 at 13:45
  • $\begingroup$ Ok so in this case for what reason is the solution x<=-2 and not x>=-2 ,again sorry if it is a dumb question $\endgroup$ – user597368 Sep 27 '18 at 13:48
  • $\begingroup$ @user597368: that's it ! $\endgroup$ – Yves Daoust Sep 27 '18 at 15:11

If you multiply both sides of an inequality by a positive factor, the direction of the inequality remains. By a negative factor, it gets reversed. This rule is enough for you to solve the problem.

Starting from $$-1\le\frac2x$$

we multiply by $-x$. Then

  • $x<0\implies x\le-2,$

  • $x>0\implies x\ge-2$.

This is summarized by

$$x\le-2\lor x>0.$$

  • $\begingroup$ Can you please do the same but multiply by x instead $\endgroup$ – user597368 Sep 27 '18 at 14:12
  • $\begingroup$ @user597368: do it if you like, the result will be identical (and you will probably multiply by $-1$ after multiplying by $x$). $\endgroup$ – Yves Daoust Sep 27 '18 at 14:18
  • $\begingroup$ When I do it I get X>=-2 When X>0 $\endgroup$ – user597368 Sep 27 '18 at 14:21
  • $\begingroup$ @user597368 So do I, so what's the problem ? $\endgroup$ – Yves Daoust Sep 27 '18 at 14:32
  • $\begingroup$ Look at your summarization it is not the same , I don't get that summarization then $\endgroup$ – user597368 Sep 27 '18 at 14:40

Write $$-\frac12\leq\frac1x,$$ that is, the reciprocal of $x$ must be at least $-1/2$. Surely all positive number are solutions and the negative ones must be at most $-2$.


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