What is the pH of a solution of a strong acid and weak acid?

All the acids have a pH of less than 7. The pH of the solution would be somwhere between 0-7 depending upon the amount of the strong acid and weak acid mixed

Strong acids have a ph of 1-3 and weak acids ph 6-7 after seven base series starts

Keep in mind that drink water is a 7 on the pH scale. 7 Is neutral
pH scale goes from 1 - 14 with 1 being the strongest possible acid and 14 being the strongest possible base. If you've got a weak acid whose dissociation contributes to the pH it would probably be a 4 or a 5 on the pH. Any smaller number is too strong and a bigger number would not contribute that much to the pH.

Without numbers it is difficult to give anything but general advice.

First consider the strong acid - It is taken that this dissociates completely , so it is easy to get the [H+] from the molarity of the acid in the final solution.

In general , the [H+] from the weak acid is discounted because it makes no contribution to the [H+] in the final solution . But you present a case where this contribution has to be taken into account.This presents a complication.
I will have to resort to using numbers in order to explain this easily.

Let us assume that you want to calculate the [H+] of a solution consisting of equal volumes of 0.5M HCl and 0.5M weak acid HW which has Ka = 8.5*10^-2. Note the comparatively high Ka - this will indicate that it is a medium weak acid and will dissociate to contribute a reasonably high concentration of H+ ions.
If you mix equal volumes of these two solutions you will have : [HCl ] = 0.25M  and [HW] = 0.25M

We know that HCl will completely dissociate , and from the HCl [H+] = 0.25M

Now the weak acid HW. This will partially dissociate:
 HW ? H+ + W-

The [H+] in a solution containing ONLY  HW can be calculated from the Ka equation:
 Ka = [H+]² / [HW]
Note that [HW] is the concentration of the undissociated acid HW in solution. In this case you may not use the short cut and take [HW] = 0.25.
 I assume that you know how to calculate [H+] in this case. This is only part of the general explanation - there is no need to do this calculation . As long as you understand that this is the equation that will give you [H+] of the acid solution.

But what is the situation if you add HCl to this solution , as you require in your question . The additional H+ from the HCl dissociation causes the equilibrium
HW ? H+ + W-
 to shift to the left. The concentration  of H+ from HW is reduced.
How to calculate this:

Let the [H+] from HW = A
The total [H+] in the solution = 0.25 + A ( that is from the HCl plus the HW.)
The [HW] in the solution =( 0.25 - A ) This accounts for the reduced concentration of HW from dissociation .

But the Ka of HW remains at 8.5*10^-2
Substitute:
 8.5*10^-2 = ( 0.25+A)² / (0.25-A)
simplify and solve for A
8.5*10^-2 =  0.0625 + 0.5A + A² / (0.25 - A)
0.02125 - 0.085A  = 0.0625 +0.5A + A²
A² + 0.5A + 0.0625 - 0.02125 + 0.085A = 0
A² + 0.0585A - 0.04125 = 0

Solve quadratic for A : + Root = 0.176

You have [H+] from dissociation of HW = 0.176M added to [H+] from HCl = 0.25M :
Total [H+] = 0.426
pH = -log 0.426
pH = 0.37