Exact solution:
Assume we agree that for
$y_1:=IRR(CF1)$, $y_2:=IRR(CF2)$, $y:=IRR(CF1+CF2)$, the following equations hold by definition:
$$-1000+\frac{100}{1+y_1}+\frac{100}{(1+y_1)^2}+\frac{1100}{(1+y_1)^3}=0$$
$$-200+\frac{20}{1+y_2}+\frac{30}{(1+y_2)^2}+\frac{1}{(1+y_2)^3}=0$$
$$-1200+\frac{120}{1+y}+\frac{130}{(1+y)^2}+\frac{1001}{(1+y)^3}=0$$
These equations represent the definition of IRR and must hold. It is known that a system of equations can be solved by adding the equations, so it follows:
$$-1000-200+\frac{100}{1+y_1}+\frac{20}{1+y_2}+\frac{100}{(1+y_1)^2}+\frac{30}{(1+y_2)^2}+\frac{1100}{(1+y_1)^3}+\frac{1}{(1+y_2)^3}\stackrel{!}{=}-1200+\frac{120}{1+y}+\frac{130}{(1+y)^2}+\frac{1001}{(1+y)^3}$$
So we need to solve
$$\frac{100}{1+y_1}+\frac{20}{1+y_2}+\frac{100}{(1+y_1)^2}+\frac{30}{(1+y_2)^2}+\frac{1100}{(1+y_1)^3}+\frac{1}{(1+y_2)^3}=\frac{120}{1+y}+\frac{130}{(1+y)^2}+\frac{1001}{(1+y)^3}$$
Call the left hand side "$X$" and multiply the equation by $(1+y)^3$:
$$X(1+y)^3=120(1+y)^2+130(1+y)+1001$$
This is a cubic equation:
$$X(1+y)^3-120(1+y)^2-130(1+y)-1001=0$$
The real-valued solution to such equation given by here is:
$$1+y= - \frac{1}{3a}\left(b\ +\ C\ +\ \frac{\Delta_0}{C}\right)$$
where $C = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}$, $\Delta_0 = b^2-3 a c $, $\Delta_1 = 2 b^3-9 a b c+27 a^2 d$.
It follows:
$$y= - \frac{1}{3a}\left(b\ +\ C\ +\ \frac{\Delta_0}{C}\right)-1$$
with $a=X$, $b=-120$, $c=-130$, $d=1001$.
This solution incorporates the two IRRs $y_1,y_2$. It happens to have the same expression as when calculating the "normal" IRR based on the cashflows since $X=1200$ by definition. Note that this must be the case, because we are looking for the same $y$ that solves both equations.
As requested by the OP and moderator, I will write out this solution:
$$y= - \frac{1}{3X}\left(-120\ +\ \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}\ +\ \frac{\Delta_0}{\sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}}\right)-1$$
$$= - \frac{1}{3\left(\frac{100}{1+y_1}+\frac{20}{1+y_2}+\frac{100}{(1+y_1)^2}+\frac{30}{(1+y_2)^2}+\frac{1100}{(1+y_1)^3}+\frac{1}{(1+y_2)^3}\right)}\left(-120\ +\ \sqrt[3]{\frac{2 b^3-9 a b c+27 a^2 d + \sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2 - 4 \Delta_0^3}}{2}}\ +\ \frac{\Delta_0}{\sqrt[3]{\frac{2 b^3-9 a b c+27 a^2 d + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}}\right)-1$$
I stop writing out the solution here as it would not add more information. This equation infers the PIRR from the IRRs, all other values are constants. It is the exact solution.
However, it shows that the solution must incorporate the cashflows, it is not possible to separate out the solution to depend only on $y_1,y_2$. The third root term in the equation immediately implies that the cashflows cannot be canceled out.
Approximate Solution:
The IRR $y$ of the sum of the cashflows must lie withing the range $[y_1,y_2]$ where $y_1\leq y_2$. If $y$ is higher than $y_1$ and $y_2$, then the above equation has no solution since all cash flows on the right were discounted at a higher rate than the left such that the equation could not hold. By the same argument it follows that $y$ must be larger than the minimum of $y_1,y_2$.
A natural guess would be to take the average of the two:
$$y\approx (y_1+y_2)/2=-0.225$$
Since we have a sum of CF1 and CF2, we may further weight the average by the absolute sum of cash flows to get:
$$y\approx \frac{2200}{2200+251}0.1+\frac{251}{2200+251}\cdot(-0.55)=0.03345$$
This estimate is very close to the exact value of $0.046$.