Solving an inequality $\frac{3}{x-1}\lt -\frac 2x$

Question

I've picked up my old book on Calculus, and going through the introductory examples of the preliminaries, I fail to see my mistake for the following exercise:

Solve the inequality $\frac 3{x-1} < -\frac2x$

In the walkthrough solution in the book, there's a note that we would like to multiply by $x(x-1)$ to clear the fractions in the inequality, but that would require us to consider three cases separately, so instead they transpose and combine the two fractions into a single one:

$$ \frac3{x-1}<-\frac2x\\ \frac3{x-1}+\frac2x<0\\ \frac{5x-2}{x(x-1)}<0 $$

Then goes on to solve the inequality based on the fact that they look for when the numerator and denominator have opposite signs. That is:

CASE I

$5x-2>0$ and $x(x-1)<0$ which gives the interval $\left(\frac25,1\right)$.

CASE II

$5x-2<0$ and $x(x-1) >0$ which gives the interval $(-\infty, 0)$

The solution set is then the union of the two intervals: $$(-\infty,0)\cup\left(\frac25,1\right)$$

Question

I first made an attempt solving this without looking at the walkthrough, and multiplied by $x$ (considering two cases, one for $x<0$ and the other for $x>0$) followed by adding $2$:

CASE I $x<0$

$$ \frac3{x-1}<-\frac2x\\ \frac{3x}{x-1}+2>0\\ 5x-2>0\\ x>\frac25 $$

Not a solution as there is no interval for $x>\frac25$ when $x<0$

CASE I $x>0$

$$ \frac3{x-1}<-\frac2x\\ \frac{3x}{x-1}+2<0\\ 5x-2<0\\ x<\frac25 $$

Gives the interval $(0,\frac25)$.

Well, it is clear I've done something wrong here, but I fail to understand what it is.

I (thought I) was using the rules $$ a < b \,\text{and}\, c > 0 \;\rightarrow\; ac < bc\\ a < b \,\text{and}\, c < 0 \;\rightarrow\; ac > bc $$

Answer 1

If $x\lt 0$, then we have $x-1\lt -1\lt 0$.

So, multiplying the both sides of $$\frac{3x}{x-1}+2\gt 0$$ by $x-1\lt 0$ gives $$3x+2(x-1)\color{red}{\lt }0.$$

If $x\gt 0$, then we have $x-1\gt -1$. So, note that the sign of $x-1$ is not determined.

Answer 2

$$\frac{5x-2}{x(x-1)}<0 \Leftrightarrow (5x-2)x(x-1)<0$$ Let $f(x)=(5x-2)x(x-1)$

$$f(x)<0 \Leftrightarrow x \in (-\infty;0) \cup (\frac25;1)$$

Answer 3

In your solution, from the step $ \frac{3x}{x-1} + 2 < 0 $ to $ 5x-2<0 $, you missed the denominator.

Hence, the real inequality for that case is $$\frac{5x-2}{x-1} < 0$$ which is the same as in their solution.

Answer 4

By direct case analysis,

$$\frac3{x-1}<-\frac2x$$

• for $x<0$, the inequality is obviously true ($-<+$).

• for $1<x$, the inequality is obviously false ($+\not \lt-$).

• for $0<x<1$, you can multiply by $-(x-1)x$, yielding $-3x<2(x-1)$ or $5x>2$.

Hence

$$(-\infty,0)\cup\left(\frac25,1\right).$$

I guess that this is a more economical solution.