Michael's bank contains only nickels, dimes and quarters.
N = number of nickels
D = number of dimes
Q = the number of quarters
There are 63 coins in all
N + D + Q = 63
valued at $5.50.The N nickels are worth $0.05N
The D dimes are worth $0.10D
The Q quarters are worth $0.25Q
0.05N + 0.10D + .25Q = 5.50
Multiply every term by 100 moves the decimals two places over:
5N + 10D + 25Q = 550
Divide every term by 5
N + 5D + 5Q = 110The number of nickels is 5 short of being 3 times the sum of
the number of dimes and quarters.the sum of number of dimes and quarters = D+Q
3 times the sum of number of dimes and quarters = 3(D+Q)
5 short of being 3 times the sum of the number of dimes and quarters = 3(D+Q)-5
So
N = 3(D+Q) - 5
N = 3D + 3Q - 5
N - 3D - 3Q = -5
So we have the system of 3 equations in 3 unknowns:
N + D + Q = 63
N + 2D + 5Q = 110
N - 3D - 3Q = -5
Solve that system by elimination.
How many dimes are in the bank?
Solve the system and see. If you can't post again
asking how to solve it.
Edwin
