janet has 83 nickels and dimes that total 6.90 how many of each type does she have

I think this should be interpreted as:

n + d = 83                         (1)

5n + 10d = 690                 (2)

From (1) we have: n = 83 - d           (3)

Put (3) into (2)

5(83 - d) + 10d = 690

415 -5d + 10d = 690

5d = 275

d = 55

Put this into (3)

n = 83 - 55

n = 28

So there are 28 nickels and 55 dimes

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Sorry, that's a little too parochial for me. You're going to have to remind me how many cents there are in a nickel, and how many there are in a dime.

janet has 83 nickels and dimes that total 6.90 how many of each type does she have

YES i am with you anon.

I think that there are 5c in a nickel and 10c in a dime.  

Yep, I checked, that is right. WE are not all yanks you know!   

Anyway, let there be N dimes.

nickels = 83*5 = 415 c

dimes = N*10 = 10N cents

415+10N = 690     remember we are working in cents

10N=690-415

10N=275

N=27.5 dimes

but you cannot have half a dime so your question has a bad smell to it.

What I mean is that it is an impossible situation.    

Yes you could be right Alan.