# Solving $-1\le \frac{2}{x}$, I came up with 2 opposite solutions

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.

• Hard to understand what you're asking here. – Hongyu Wang Sep 27 '18 at 13:38
• 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 :) – 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 $$x \leq -2$$.

• 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 – user597368 Sep 27 '18 at 13:43
• @user597368 $x$ can be either positive or negative, so you have to consider both possibilities individually as their properties alter the answer – MRobinson Sep 27 '18 at 13:45
• 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 – user597368 Sep 27 '18 at 13:48
• @user597368: that's it ! – 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

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

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