math help i need urgently

A resturant has 161L of a cleaning solution which consists of 22% water and the remaining amount is soap. How much water is needed to be added to obtain a 55% water mixture.

If there are 6 litres of a drink and 80% of it is the special mixture, and 20% of it is just water, how many litres of water must be added so that the drink has 50% of it's special mixture PS - Thanks math helper for the other question, but i dont get one...

thanks, but i still dont know why you did 0.78(161)

just saw your reply - thanks!

ok so this is what i devised out - 0.80(6) + 0x = 0.50(6 + x) Hence, i get x = 3.6 I dont know if this is right, for the answer sheet gives x = 2 thanks

please reply :sobbing face:

wtf

what the f