A question related to ratio and proportions

Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days. Find the time taken by A alone to complete the work?
answer is 25/2 days
give the solution please

A = rate of A
B = rate of B

given they both finish in 5 days
5A + 5B = 1

new rate of A is 2A
new rate of B is B/3

given they finish in 3 days
(2A)(3) + (B/3)(3) = 1
6A + B = 1

6A + 1/5 - A = 1
A = 4/25

since A works twice as fast, his rate is 2 * 4/25 = 25/2, and thus it takes A 1/(2/25) = 25/2 days to finish the task

edit:
1 means 100% of the job, and 100% = 100/100 = 1

oky set two equalities:
A+B=5 work done by A and B took 5 days
2A+1/3B=3 work done when A worked 2 times as hard and B worked 1/3 times as hard took 5 days
now change equation number one to B=5-A
then plug it into equation 2 for the B value
2A+1/3(5-A)=3
since i don't like fractions ill multiplay everything by 3 makeing the equation like this
6A+5-A=9 Solve for A
5A=4 so A=4/5 now we found how many days A worked when he was 2 times as Hard so we divide by 2 to get his normal rate A=2/5
now in order to find days needed for A to do it alone we divide the required days which is 5 by the amount A normally works which is 2/5
so it becomes 25/2