I must solve following inequation:


Now the text says that I have to solve the inequation "direct" without solving the according equations.

What does that mean?

I would say that I have to multiply by $(1-2x)$ then I get

$x-3<0$ and

$L_1 = [x \le 2]$

$L_2 = [x >0] $

but I have the feeling that I'm doing something wrong. Especially I don't understand what the paragraph about solving the inequation "direct" means. Could someone maybe explain that to me?


If you mulitply an inequality by a negative number, it will exchange the signs $>$ and $<$ (because for example $1<2$ but $-1>-2$...) So it is not the best to multiply by the denominator unless you're sure it is positive!

I think, by 'direct', they meant as starting point to think about, when a fraction $\displaystyle\frac AB$ will be negative in general.. (If $A>0$ and $B<0$ or $A<0$ and $B>0$.)


Besides to above solutions see the table below:

enter image description here

It shows that your desire interval would be $(-\infty, 1/2)\cup (3,+\infty)$


Since $$\frac{x-3}{1-2x}=\frac{(x-3)(1-2x)}{(1-2x)(1-2x)} \ \ \ x\neq \frac{1}{2}$$

and $(1-2x)(1-2x)=(1-2x)^2>0$


We allowed multiply that inequality by $(1-2x)^2$ and get,


Can you solve that?

  • $\begingroup$ Yes, I can solve that. But why do we have to "allow" the multiplication? Can't I just look at the fraction and see when it is 0 and these are the borders for my solution? $\endgroup$ – Lukas Plommer Oct 17 '12 at 9:38
  • $\begingroup$ Is it possible that the solution is L = {x<(1/2) v x>3} ? Is that a valid way of writing the solution for an inequation? $\endgroup$ – Lukas Plommer Oct 17 '12 at 9:44
  • $\begingroup$ Solution: $x>3 \ \ \text{or} \ \ x<\frac{1}{2}$ $\endgroup$ – Salech Rubenstein Oct 17 '12 at 9:49
  • $\begingroup$ We multiply the numerator and denominator by $(1-2x)$, therefor, the sign is not changing. $\endgroup$ – Salech Rubenstein Oct 17 '12 at 9:55
  • $\begingroup$ @LukasPlommer: yes, $L=\{x \mid x<1/2 \, \lor \, x>3\}$. $\endgroup$ – Berci Oct 17 '12 at 10:55

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