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.
--
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| 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 |
-----------------------------
|
