"Barb Knox" <see@[EMAIL PROTECTED]
> wrote in message
news:see-884DEA.15444013112007@[EMAIL PROTECTED]
> In article <4738fefa$0$19614$4c368faf@[EMAIL PROTECTED]
>,
> "Marvel" <Martel@[EMAIL PROTECTED]
> wrote:
>
>> I cant figure it out any help appreciated!
>>
>> I am not smarter than a fifth grader!
>> ------
>>
>> Ben had 50 more coins than Joan.
>
> Let B and J be Ben's and Joan's initial coins, respectively.
> So [1]: B = 50 + J
>
>> After Joan gave 29 of her coins to Ben, she had 1/3 as many coins as
Ben.
>
> So [2]: (J - 29) = (B + 29) ...
>
> (You fill in the "...")
>
> That gives 2 equations in 2 unknowns.
> Solve by substituting 50+J (from [1]) for every B in [2], and then
> re-arranging terms to get all the 'J's on one side.
>
>> How many coins did Ben have at first?
>
> Having solved for J, use [1] to calculate B.
>
>
I have not figured out how to convert it
> --
> ---------------------------
> | BBB b \ Barbara at LivingHistory stop co stop
uk
> | B B aa rrr b |
> | BBB a a r bbb | Quidquid latine dictum sit,
> | B B a a r b b | altum viditur.
> | BBB aa a r bbb |
> -----------------------------
|
