SOLUTION: Suppose that 25 nickels, dimes, and quarters are worth $2.75 and there are twice as many dimes as nickels. How many of each denomination are there? What's the formula that's

Question 128767:
Suppose that 25 nickels, dimes, and quarters are worth $2.75 and there are twice as many dimes as nickels. How many of each denomination are there?
What's the formula that's the thing I don't understand with story problems?
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
N+D+Q=25
.05N+.10D+.25Q=2.75
D=2N TWICE AS MANY DIMES AS NICKELS.
NOW REPLACE D WITH 2N IN EACH OF THE EQUATIONS & SOLVE.
N+2N+Q=25
3N+Q=25 OR Q=25-3N
.05N+.10*2N+.25Q=2.75
.05N+.2N+.25Q=2.75
.25N+.25Q=2.75 NOW REPLACE Q WITH (25-3N) & SOLVE FOR N
.25N+.25(25-3N)=2.75
.25N+6.25-.75N=2.75
-.5N=2.75-6.25
-.5N=-3.50
N=-3.50/-.5
N=7 NUMBER OF NICKELS.
D=2*7=14 NUMBER OF DIMES.
7+14+Q=25
21+Q=25
Q=25-21
Q=4 NUMBER OF QUARTERS.
PROOF
.05*7+.10*14+.25*4=2.75
.35+1.40+1.00=2.75
2.75=2.75
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