# Why does summing up two inequalities alter the solution?

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

• Solving separately is the right way. The first method doesn't give us new information. – 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 $$-11+1=-10\le0 \quad\text{but}\quad 1>0.$$