Easy way to show $\frac{2x}{x-1}>x$

Question

In my book I found this following inequality : $$\frac{2x}{x-1}>x$$

In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both sides by $x$ and then solve? Is there any other not so difficult way to solve this?

Answer 1

You can not divide by $x$ because $x$ can be positive or $x$ can be negative.

It's $$\frac{2x}{x-1}-x>0$$ or $$\frac{x(3-x)}{x-1}>0,$$ which gives the answer: $$(-\infty,0)\cup(1,3).$$ I used the intervals method.

We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.

Now, on $(3,+\infty)$ the expression $\frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.

Thus, the sign on $(1,3)$ is $+$.

If we pass a point $1$ then the sign changes again.

Thus, the sign on $(0,1)$ is $-$.

And if we pass a point $0$ then the sign still changes.

Thus, the sign on $(-\infty,0)$ is $+$.

Now, we can write the answer.

Answer 2

Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.

Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.

Once you take care about these points, you should find easily the full solution.

Answer 3

You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.

In this example: $$ \frac{2x}{x-1}-x>0 $$ $$ \frac{3x-x^2}{x-1}>0 $$ $$ \frac{x(3-x)}{x-1}>0 $$ A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?

Answer 4

We require $\frac{2x}{x-1}-x>0$ or $\frac{x(x-3)}{x-1}<0$, which is true for $x\in(-\infty,0)\cup(1,3)$. It can be solved using the wavy-curve method.