In article <see-884DEA.15444013112007@[EMAIL PROTECTED]
>, Barb Knox
<see@[EMAIL PROTECTED]
> wrote:
> 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.
--
Paul Sperry
Columbia, SC (USA)
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