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 $-x$.
