WHAT IS THE BEST APPROACH TO SOLVING THIS PROBLEM:

ALICE'S TAKE-HOME PAY LAST YEAR WAS THE SAME EACH MONTH, AND SHE SAVED THE SAME FRACTION OF HER TAKE-HOME PAY EACH MONTH. THE TOTAL AMOUNT OF MONEY THAT SHE HAD SAVED AT THE END OF THE YEAR WAS 3 TIMES THE AMOUNT OF THAT PORTION OF HER MONTHLY TAKE-HOME PAY THAT SHE DID NOT SAVE. IF ALL THE MONEY SHE SAVED LAST YEAR WAS HER TAKE-HOME PAY, WHAT FRACTION OF HER TAKE-HOME PAY DID SHE SAVE EACH MONTH?

THE ANSWER IS 1/5.

Alice's total take home per month = x
Alice's saving per month = y
Alice's saving for the entire yar = 12y
What Alice did not save = x-y
As per the stem , Alice's annual saving = 3 time what she did not save pre month
i.e. 12y = 3(x-y)
4y = x- y
y = x/5.

you can also do this problem by picking numbers, because you don't have a figure for the monthly pay (and, apparently, it doesn't matter, because the problem would be impossible to solve if it did matter).

so just let the monthly pay be '1'.
let the monthly savings be 'S', which is both a fraction and an actual amount of money. (it's both because the monthly pay is '1', so the fraction is the same as the dollar amount)
then she has saved 12S at the end of the year.
each month, she didn't save (1 - S)
so
12S = 3(1 - S)
12S = 3 - 3S
15S = 3
S = 3/15 = 1/5

faster, and probably less confusing, with only one variable