For example when an electrum wallet is crypted we got the following:
2 + 3 = X, we know that x is 5 cause we know the first 2 numbers, same with electrum when doing encryption:
unencrtyped_wallet + password_encryption = X,
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but in this case we got something like this:
2 + X = 5, we can also find X because X = 5 - 2
same should be applicable to my situation:
unecrypted_wallet + X = encrypted_wallet
It would be more accurate to describe the decryption key (the password) as being on a 3-dimential plane. In your example, you can determine the solution to "X" however you do not know where on the "Z" plane the solution is, and there is an unlimited number of places on the Z axis the solution could be. Even having multiple plaintext sets will not help you.
I am going to speculate you are in a position that can be described as below:
You have an electrum wallet ("Wallet 1") that you do not have the password to
You have a second electrum wallet ("Wallet 2") that you know has the same password as Wallet 1
You have a decrypted copy of Wallet 2, possible from an old backup.
Wallet 1 has a lot of coin in it
Depending on how complex your password is, and how much you know about your password, you may be able to brute force the password via automated means. There are a number of software programs and service providers that can try to guess your password, and know if it figures it out.
For example, if you know your password is 10 digits, and is a combination of uppercase, lowercase letters and numbers that are something close to "random" there are a 8.39 * 10
17 possibilities of what your password could be. However if you know your password starts with the uppercase letter "P" followed by 9 numbers that are something close to random, the number of possibilities drops to 1 * 10
8.