Puzzle-Syrup water mixture-Rejoinder

This refers to my previous blog wherein I had posted a question-From a container containing a syrup-water mixture in the ratio 5/3,how much should be removed and replaced by an equal quantity of water so that the syrup and water are contained equally
If  the syrup content is to be changed from 5/8 to 1/2 we have to remove 1/8 content of syrup which will need 1/5 of the total mixture--{5/8}-{(1/5)*(5/8)}=5/8-1/8=1/2.
The water content then will become 3/8-(1/5ths of 3/8)+1/5=3/8-3/40+1/5=    that is (15-3+8)/40=20/40=1/2 as required.
Other variations  ratio 5/7..1/7ths of the mixture to be replaced
                                    5/9..2/9 ths of the mixture to be replaced
                                    5/11..3/11 ths of the mixture to be replaced as follows
syrup 7/12-1/7 ths of 7/12=1/2  ..water   5/12-(5/12*1/7)+1/7=(35-5+12)/84=42/84=1/2
         9/14-2/9 ths of 9/14=1/2                 5/14-(5/14*2/9)+2/9=(45-10+28)/126=63/126=1/2
         11/16-3/11ths of 11/16=1/2            5/16-(5/16*3/11)+3/11=(55-15+48)/176=88/176=1/2