Is $\pm$ distributive over addition?

Question

I realize that the title is pretty vague without some context.

I am trying to show that $$-\frac{i\hbar}{\sqrt2}\left(\pm i\color{blue}{\left(g_y+\frac{yz^2f^{\prime}}{r}\right)}-\pm i\left(\frac{yz^2f^{\prime}}{r}\right)-g_z\right)=\frac{1}{\sqrt2}\left(i\hbar g_z \pm \hbar g_y\right)\tag{*}$$

In order to try to get close to the answer on the RHS, the bracket in blue must distribute over addition. However, if we had simply $$\pm (7+5)\stackrel{?}{=}\pm 7 \pm 5\tag{?}$$

When considering just the positive and negative signs independently $(?)$ seems to hold as $$+7+5=12$$ & $$-7-5=-12$$ and the LHS ($\pm 12$) of $(?)$ is satisfied.

Could someone please explain whether or not $(?)$ holds?

Answer 1

Yes, you are correct. $$\pm (7+5) = \pm 7 \pm 5 = \pm 12$$ since both the numbers $5$ and $7$ hold the same positive sign.

Just for the information: Care is needed only in cases like: $$\pm (7 - 5) = \pm 7 \mp 5 = \pm 2$$ and $$\pm(8-9) = \pm 8 \mp 9 = \mp 1$$

Answer 2

When one expression contains multiple plus-minus/minus-plus signs, it is considered equivalent to a pair of expressions, one with all upper versions of those signs, and the other one with lower versions.

For examples: $$\cos(A \pm B) = \cos(A) \cos(B) \mp \sin(A) \sin(B)$$ is equivalent to $$\begin{cases} \cos(A + B) = \cos(A) \cos(B) - \sin(A) \sin(B) \\ \cos(A - B) = \cos(A) \cos(B) + \sin(A) \sin(B) \end{cases}$$ and $$x^3 \pm 1 = (x \pm 1)\left(x^2 \mp x + 1\right)$$ is equivalent to $$\begin{cases} x^3 + 1 = (x + 1)\left(x^2 - x + 1\right) \\ x^3 - 1 = (x - 1)\left(x^2 + x + 1\right) \end{cases}$$

As both plus and minus, resulting from $\pm$ and $\mp$, are commutative and distributive, both $\pm$ and $\mp$ are commutative and distibutive.