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 $x \in (-\infty,1]\cup[2,8/3]$.

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

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 $$ -11+1=-10\le0 \quad\text{but}\quad 1>0. $$


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