SOLUTION: Sean has 103 coins consisting of nickels,dimes and quarters. the number of dimes is one less than twice the number of nickels and the number of quartes is two more than three times

Question 84951This question is from textbook elementary and intermediate algebra
: Sean has 103 coins consisting of nickels,dimes and quarters. the number of dimes is one less than twice the number of nickels and the number of quartes is two more than three times the number of nickels. how many ocins of each kind does he have? This question is from textbook elementary and intermediate algebra

Answer by tutorcecilia(2152) (Show Source): You can put this solution on YOUR website!
Nickels + Dimes + Quarters = 103
Nickels = = N
Dimes = (2) x Nickels - 1 = 2N-1
Quarters = (3) x Nickles + 2 = 3N + 2
Total coins = 103
.
(N) + (2N-1) + (3N+2) = 103 [combine like-terms]
N+2N-1+3N-1=103
6N+1=103 [solve for the N-term]
6N=103-1
6N=102
N=17
.
checking
Nickels = 17
Dimes = 2(N)-1=(2)(17)-1=33
Quarters = (3)(17)+ 2=53
.
So, 17+33+53=103
103=103 [checks out]

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