What's the difference of the answers between "or" and "and" compound inequalities?
Matthew Harrington 
I'm seeing two kinds of compound inequalities like : with "or" and "and". For that I'm confused to solve them. What's the difference of them and how can I understand in order to solve easily?
$\endgroup$4 Answers
$\begingroup$and means they both have to be true, or means at least one of them has to be true. $x \gt 0$ and $x \lt 1$ is satisfied only on the interval $(0,1)$. $x \gt 0$ or $x \lt 1$ is satisfied for all real $x$ because at least one of them is true.
$\endgroup$ $\begingroup$For example, $x\neq1$ and $x\neq2$ it's $x\in\mathbb R\setminus\{1,2\}$,
but $x\neq1$ or $x\neq2$ it's $x\in\mathbb R$.
$\endgroup$ $\begingroup$The key difference is with "or", $x$ only needs to satisfy one of the inequalities. With "and", $x$ needs to satisfy both.
Let's do an example with "or". Say we have "$x>6$ or $x <3$". It turns out $x=7$ satisfies the compound inequality. This is because $x$ satisfies the first inequality $7>6.$ Notice that $7\not< 3$ so it doesn't satisfy the second inequality, but that doesn't matter since it only needs to satisfy one of them. Similarly $x=2$ qualifies because $2<3$, even though $2$ does not satisfy $x>6.$ $x=5$ does not qualify because it fails to satisfy either: we have $5\not>6$ and $5\not< 3.$
For an example with "and", try "$x<7$ and $x >1$." For $x$ to qualify it must satisfy both inequalities. So $x=12$ does not qualify because while it satisfies $12>1$ it does not satisfy $12<7.$ $x=6$, however, does qualify since both $6<7$ and $6>1.$
$\endgroup$ $\begingroup$Adding to the answer of Ross, if you have $x<-3$ and $x>5$ then this statement cannot be true for any x$\in \Bbb R$. If it were $x<-3$ or $x>5$ then this statement is satisfied $\forall x \in (- \infty,-3) \cup (5,+\infty)$
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