SOLUTION: joe has a nickles and dime that is worth $2.15. if the number of dimes was doubled and the number of nickles was increased by 28 , the value of the coins would be $6.00. How many n

Algebra ->  Absolute-value -> SOLUTION: joe has a nickles and dime that is worth $2.15. if the number of dimes was doubled and the number of nickles was increased by 28 , the value of the coins would be $6.00. How many n      Log On


   

Question 387717: joe has a nickles and dime that is worth $2.15. if the number of dimes was doubled and the number of nickles was increased by 28 , the value of the coins would be $6.00. How many nickles and dimes does he has?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let n = number of nickels
Let d = number of dimes
given:
(1) 5n+%2B+10d+=+215 (in cents)
(2) 10%2A%282d%29+%2B+5%2A%28n+%2B+28%29+=+600
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There are 2 equations and 2 unknowns,
so it's solvable
(2) 20d+%2B+5n+%2B+140+=+600
(2) 5n+%2B+20d+=+460
Now subtract (1) from (2)
(2) 5n+%2B+20d+=+460
(1) -5n+-+10d+=+-215
10d+=+245
d+=+24.5 impossible
In (1), n has to be odd to get 215
In (2), n+%2B+28 ends up being odd again
and that can't give me 600
These numbers are wrong somehow, I think.