What the "inequality is right" statement means? A problem for example:
$a > b$. Is $a - b > -3$ right? If yes, then does it mean that it's solution is $(-3;\infty)$, which means $a - b$ can be equal to $-2$, for example?
So "inequality is right" in this problem means that plot can be found on $(-3; \infty)$ or it is $(-3; \infty)$?
Thanks.
 A: If $a>b$, then $a-b > 0$, not $-3$. Of course, $a-b>0$ also implies that, since $0>-3$, we have that $a-b>-3$. However, $a-b>0$ is a stronger assumption (for example, $a-b>-3$ may mean that $a-b=-2$, but that cannot be because $-2<0$.
As for the second part of the question: If you know that $x>-3$, then you know that $x$ is some element of the set $(-3,\infty)$. It does not mean that $x$ itself is the set $(-3,\infty).$
For example, it is correct to say:

For all values $x$ in the set $(-3,\infty)$, the inequality $x>-3$ is correct.

It is not correct to say:

The inequality $(-3,\infty)>-3$ is correct.

Things may be true for each element of a set but are not true for the set itself. For example, all humans have two legs, but the set of all humans does not have legs at all.
A: It depends a bit what you mean by right.
First we note that $a\gt b$ is equivalent to $a-b\gt 0$. Since we also have $0\gt -3$ and the order is transitive by definition of order we have that $a\gt b$ implies $a-b\gt -3$, and if we know $a\gt b$ then $a-b\gt-3$ is true.
The reverse implication is not true, so the two statements are not equivalent - which would be another interpretation of "right".
A third interpretation would say that $a-b\gt -3$ is not the sharpest inequality of its kind which can be derived from the original. To illustrate that this last idea is different from the others note that $a\gt 2$ implies that $a^2\gt 1$, but we can get a sharper inequality $a^2\gt 4$. However $a^2\gt 4$ does not imply $a\gt 2$ because we could have $a=-3$.
People who work with inequalities work with all three ideas. When you ask a question you need to be clear which idea is in view.
