1
$\begingroup$

Sorry for the basic question. But take for example:

$$3x-8\leq 0$$

$$-2+3x-x^2\leq 0$$

If we sum these two inequalities we obtain: $$0\leq x^2-6x+10$$

The solution of this inequality is of course any $x \in \mathbb{R}$. However, we can also attempt to solve them separetly and obtain: $$x \leq 8/3$$ $$(x-2)(x-1)\geq 0$$

which of course implies that $x \in (-\infty,1]\cup[2,8/3]$.

$\endgroup$
  • $\begingroup$ Solving separately is the right way. The first method doesn't give us new information. $\endgroup$ – Ofek Gillon Mar 11 '17 at 14:08
2
$\begingroup$

We have, for all real numbers $a,b$, $$ a\le0,\quad b\le0 \implies a+b \le 0 $$ and we don't have $$ a\le0,\quad b\le0 \iff a+b \le 0 $$ since for example $$ -11+1=-10\le0 \quad\text{but}\quad 1>0. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.