Common ion effect

Let $ce{HA}$ ($x$ molar) be a strong electrolyte and $ce{HB}$ ($c$ molar) — weak electrolyte. As $ce{HA}$ is strong electrolyte, it dissociates completely:

$$
begin{array}{lcc}
ce{&HA &-> &H+ &+ &A-} \
text{initially} & x && 0 && 0 \
text{after} & 0 && x && x
end{array}
$$

As HB is weak electrolyte

$$
begin{array}{lcc}
ce{&HB &-> &H+ &+ &B-} \
text{initially} & c && 0 && 0 \
text{equilibrium} & c – cα && cα && cα
end{array}
$$

Then

$$K_mathrm{a} =frac{cαcdot (cα + x)}{c – cα}$$

Now my doubt is how can we add the concentration of $ce{H+}$ together as their volume can be different. Suppose $x = pu{1 M}$, and $cα = pu{0.01 M},$ then how can be the resultant concentration be $pu{1.01 M}?$ Can someone explain in detail?

Edit:
$alpha$ is the degree of dissociation
Now let us suppose we had a ‘x’ molar strong electrolyte and a ‘c’ having a common ion let us suppose H. Now while writing Ka for the dissocition of weak electrolyte as given in equation
$$
begin{array}{lcc}
ce{&HB &-> &H+ &+ &B-} \
text{initially} & c && 0 && 0 \
text{equilibrium} & c – cα && cα && cα
end{array}
$$

Then
$$K_mathrm{a} =frac{[B] cdot [H] }{[HB] }$$
Now due to common ion effect we have to consider the total concentration of H we get which gives us

$$K_mathrm{a} =frac{cαcdot (cα + x)}{c – cα}$$
Now how did the concentration of H become $$(cα + x)$$
Let us now suppose instead of x we had a 2 molar strong electrolyte (HA) so the concentration of H that comes from HA is 2 molar. Now for the weak electrolyte let us suppose we get 0.01M as the concentration of H$(cα) $ that comes from HB now according to the equation
$$K_mathrm{a} =frac{cαcdot (cα + x)}{c – cα}$$
We get the total concentration of H is (cα + x). Using our supposed value we get total concentration of H as 2+0.01=2.01M but how can we just add the two concentration. Just like in if we had a 1M solution1 and a 2M solution2 we would not get the total concentration of the two solutions after mixing as 3M