given symmetric invertible matrices $A, B \in \mathbb{R}^{n \times n}$ and vectors $a, b \in \mathbb{R}^n$. Assume we know the solution of $Ax_1=a$ and $Bx_2=b$ and $A+B$ is an invertible matrix. Are there any statements about the solution of $(A+B)x_3=a+b$?
At first I observed that simply setting $x_3=x_1+x_2$ does only yield $(A+B)(x_1+x_2)=a+b+AB^{-1}b+BA^{-1}a$ and therefore $x_3=x_1+x_2$ is only a solution if $AB^{-1}b+BA^{-1}a=0$ which is a very restrictive constraint.
Afterwards I tried to use Woodburys matrix identity stating that $(A+B)^{-1} = A^{-1}-(A+AB^{-1}A)^{-1}$ and therefore we compute $x_3=(A+B)^{-1}(a+b)=x_1+A^{-1}b-(A+AB^{-1}A)^{-1}(a+b)$ . We can apply the formula the other way around and obtain $x_3=(B+A)^{-1}(a+b)=x_2+B^{-1}a-(B+BA^{-1}B)^{-1}(a+b)$, but then I get stuck.
Do you know how I can proceed or any other results that might help state something about the solution $x_3$?
Thank you very much in advance!