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

Question

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.

Answer 1

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 $x \leq -2$.

Answer 2

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

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

Answer 3

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 $-2$.