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From: Ankush Learner on 5 Aug 2010 06:16
z=15+.50+x+y

where
x = z (a/100)
y = z {b/(b+100)}
a = 5
b = 10.2

Thanks in advance

Regards
Ankush
From: alainverghote on 5 Aug 2010 06:33
On 5 août, 12:16, Ankush Learner <ankushlear...(a)gmail.com> wrote:
> z=15+.50+x+y
>
> where
> x = z (a/100)
> y = z {b/(b+100)}
> a = 5
> b = 10.2
>
> Thanks in advance
>
> Regards
> Ankush

Quite easy!

five equations and five unknown (letters)
Just work with z=15+.50x+y
substitute until z= const. + times z

....

Alain
From: Frederick Williams on 6 Aug 2010 09:14
Ankush Learner wrote:
>
> z=15+.50+x+y
>
> where
> x = z (a/100)
> y = z {b/(b+100)}
> a = 5
> b = 10.2

Answered in alt.algebra.help. It really would be better if you didn't
ask the same question all over Usenet.

--
I can't go on, I'll go on.
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