santisd1 wrote:

Hi, after reading all these I see the logic in keeping the "sum of the triplets" in mind..... but isn't that already accounted for in the original J = x+3L formula? As long as you're adding 20 to each side, aren't you already counting for each of the 3 "L"s? (ie --- J+20 = x+3(L+20) )

I'm running into the same problem as the OP and these explanations only half a little sense - I know 20x3 = 60... that's literally the only sensical party I understand from the jump of if L = (J -x)/3 , then in 20 years, that will be J-x + 60/3.... Especially since that 3 there SHOULD be counting (at least in my mind) for all the triplets....

Can someone break this down to an elementary level of how 20 years later becomes +60 in the equation?

My final formula was L+20 = ((J+20)-x)/3

leading to... L = (J-40-x)/3

Thank you in advance!

Hi

First of all you need to be clear what is the meaning of 'L' in your solution. Are you taking L to be age of each triplet? OR are you taking L to be the 'sum of ages of 3 triplets'?

So lets say we take L to mean age of each of Lee's triplets. Then sum of their ages right now = L + L + L = 3L. So now we have J = x + 3L or 3L = J - x or L = (J - x)/3

Or we can say that age of each triplet right now = (J - x)/3.

This means age of each triplet after 20 years = (J - x)/3 + 20

Now this (J-x)/3 + 20 can also be written as: (J-x)/3 + 60/3 (since 20 = 60/3)

So this becomes = (J - x + 60) / 3