Halmos Finite-Dimensional Vector Spaces Sec. 36 Ex. 5 on invertible transformations The exercise

If $A, B, C, D$ are linear transformations (all on the same vector space), and if both $A+B$ and $A-B$ are invertible, then there exist linear transformations $X$ and $Y$ such that
$$AX + BY = C$$
$$BX + AY = D$$

What I tried
I found these values for $X$ and $Y$, which I believe make the equations work. $Y$ is identical to $X$ except for a minus sign in the middle instead of a plus.
$$X = \frac{1}{2} (A+B)^{-1} (C+D) + \frac{1}{2} (A-B)^{-1} (C-D)$$
$$Y = \frac{1}{2} (A+B)^{-1} (C+D) - \frac{1}{2} (A-B)^{-1} (C-D)$$
However, I'm concerned, first, that these expressions involve $\frac{1}{2}$ (i.e. $(1+1)^{-1}$) so they do not work if the field has characteristic 2. Second, they're just a bit unwieldy. Is there a better solution?
 A: We can apply Gaussian elimination reasonably well. The equations:
$$AX + BY = C \tag{1}$$
$$BX + AY = D \tag{2}$$
are equivalent to:
$$AX + BY = C \tag{1}$$
$$(A + B)X + (A + B)Y = C + D. \tag{3}$$
As you can see, $(3) = (1) + (2)$. We can recover $(2)$ by computing $(3) - (1)$. As $A + B$ is invertible, $(3)$ is equivalent to
$$X + Y = (A + B)^{-1}(C + D), \tag{4}$$
which implies (not necessarily equivalent to)
$$AX + AY = A(A + B)^{-1}(C + D). \tag{5}$$
Together, $(5)$ and $(1)$ imply $(5) - (1)$:
$$(A - B)Y = A(A + B)^{-1}(C + D) - C, \tag{6}$$
which is equivalent to
$$Y = (A - B)^{-1}(A(A + B)^{-1}(C + D) - C). \tag{8}$$
No (scalar) division is necessary, so there's no assumption about the characteristic of the field! If we instead take $B(4) - (1)$, we get the only possible solution for $X$:
$$X = (B - A)^{-1}(B(A + B)^{-1}(C + D) - C). \tag{9}$$
We now need to check these formulas. Substituting $(8)$ and $(9)$ into the left hand side of $(1)$,
\begin{align*}
AX + BY &= A(B - A)^{-1}(B(A + B)^{-1}(C + D) - C) + B(A - B)^{-1}(A(A + B)^{-1}(C + D) - C) \\
&= (B(A - B)^{-1}A - A(A - B)^{-1}B)(A + B)^{-1}(C + D) + (A - B)(A - B)^{-1}C \\
&= (B(A - B)^{-1}A - (A - B)(A - B)^{-1}B - B(A - B)^{-1}B)(A + B)^{-1}(C + D) + C \\
&= (B(A - B)^{-1}(A - B) - B)(A + B)^{-1}(C + D) + C \\
&= 0(A + B)^{-1}(C + D) + C = C.
\end{align*}
Substituting $(8)$ and $(9)$ into the left hand side of $(2)$,
\begin{align*}
AY + BX &= A(A - B)^{-1}(A(A + B)^{-1}(C + D) - C) + B(B - A)^{-1}(B(A + B)^{-1}(C + D) - C) \\
&= (A(A - B)^{-1}A - B(A - B)^{-1}B)(A + B)^{-1}(C + D) - C \\
&= (A(A - B)^{-1}A + (A - B)(A - B)^{-1}B - A(A - B)^{-1}B)(A + B)^{-1}(C + D) - C \\
&= (A(A - B)^{-1}(A - B) + B)(A + B)^{-1}(C + D) - C \\
&= (A + B)(A + B)^{-1}(C + D) - C \\
&= C + D - C = D.
\end{align*}
If you follow/believe my argument so far, these are the only possible solutions to the equation. Thus, my solution and your solution must agree when $\operatorname{char} F \neq 2$, though it seems to be painful to show directly. But, my solution is well-defined without having to divide scalars, so it works over a general field.
