How do you read $\pm$? Why does $|x|=3$ imply $x=\pm 3$? Q1) How is this symbol read as $±$? 
"Plus AND Minus" or Plus OR Minus"?
All this time I've been reading it as Plus-Minus.
I asked my math teacher and he said it's the former one. Then my question is if it's "Plus AND Minus" then when we take square roots on both sides of an equation, like,
$$x^2 = 9 \implies x= ± 3 $$
How can we say $x$ equals $2$ values? It should be $+3$ or $-3$ right?
Q2) My teacher said, $$|x| ≠ ± x$$ but $$|x| = 5 \implies x = ±5$$
Wait... what? What is going on here? I think I lack the basic math knowledge here. 
EDIT: I forgot to mention another thing.
He says, $$\sqrt{9} = 3$$
And then he says, $$\sqrt{x^2}= |x|$$ but wouldn't that mean of $$\sqrt{9} = |3|$$ & $$ |3| = ± 3$$ Didn't he just contradict himself?
Thanks :)

The symbol $\pm$ should be read plus or minus.
Background
By convention, if $x \geq 0$, then $\sqrt{x}$ is the principal square root of $x$, meaning the nonnegative square root of $x$.  Hence, $\sqrt{9} = 3$.  If we want to denote the negative square root, we write $-\sqrt{x}$.  Thus, the negative square root of $9$ is expressed in the form $-\sqrt{9} = -3$.  Notice that the principal square root of a number is uniquely defined.
The absolute value of a real number $x$, denoted $|x|$, is its distance from $0$ on the real number line.  Since distances are nonnegative, the absolute value of a number is never negative.  If $x \geq 0$, then $|x| = x$.  For instance, $|5| = 5$ since $5$ is five units from $0$ and $|0| = 0$ since $0$ is zero units from $0$.  If $x < 0$, then $|x| = -x$.  For instance, $|-4| = 4$ since $-4$ is four units from $0$.  Consequently, we can write a piecewise definition of the absolute value as follows:
$$|x| = \begin{cases}
x & \text{if $x \geq 0$}\\
-x & \text{if $x < 0$}
\end{cases}$$
Observe that
\begin{align*}
\sqrt{5^2} & = \sqrt{25} = 5 = |5|\\
\sqrt{0^2} & = \sqrt{0} = 0 = |0|\\
\sqrt{(-4)^2} & = \sqrt{16} = 4 = -(-4) = |-4|
\end{align*}
Observe that if $x \geq 0$, then $\sqrt{x^2} = x = |x|$.  If $x < 0$, then $\sqrt{x^2} = -x = |x|$.  Therefore, we obtain the alternative definition of absolute value:
$$\sqrt{x^2} = |x|$$

Solve the equation $x^2 = 9$.

\begin{align*}
x^2 & = 9\\
\sqrt{x^2} & = \sqrt{9} && \text{take the principal square root of each side of the equation}\\
|x| & = 3 && \text{evaluate the principal square roots}
\end{align*}
The equation $|x| = 3$ means that $x$ is a real number whose distance from $0$ on the real number line is $3$.  There are two such numbers.  They are $3$ and $-3$.  However, $x$ cannot simultaneously be both $3$ and $-3$.  Hence, we say that $x = 3$ or $-3$.  Our solution set is $S = \{3, -3\}$.  Thus, this equation has two distinct solutions that happen to be opposite in sign.  When this occurs, the solution is sometimes expressed in the form
$$x = \pm 3$$
to indicate that if $x$ is a solution of the equation, then $x = 3$ or $x = -3$.  To put is another way, either $3$ or $-3$ could be substituted for $x$ to make the equation $x^2 = 9$ true.
Clearly, it does not make sense to say $x = 3$ and $x = -3$ since $3 \neq -3$.  Consequently, it also does not make sense to read the symbol $\pm$ as plus and minus.
The symbol $\pm$ indicates that the symbol may be substituted by either the positive or negative values of the variable (see this definition).  Therefore, it does not make sense to write
$$|x| = \pm x$$
since there is only a single value which may be substituted for $|x|$.  If $x \geq 0$, that value is $x$; if $x < 0$, that value is $-x$.
