Answer:
There are 38 dimes, 22 nickels and 16 quarters.
Step-by-step explanation:
Let n, d and q represent the # of nickels, dimes and quarters respectively.
Then n + d + q = 76
The value of a nickel is $0.05; that of a dime is $0.10, and that of a quarter is $0.25.
Thus, the value of n nickels is $0.05n (and so on).
The total value of the coins is $0.05n + $0.10d + $0.25q = $8.90.
d = n + q allows us to eliminate d.
First, n + d + q = 76 becomes n + (n + q) + q = 76, and second:
$0.05n + $0.10(n + q) + $0.25q = $8.90. Here we have succeeded in eliminating d from two different equations, and now we have these two different equations in two unknowns (n and q), which is solvable.
Simplifying both equations, we get:
2n + 2q = 76 and
5n + 10n + 10q + 25q = 890, or 15n + 35q = 890
Let's use the substitution method of solving linear equations:
Rewrite 2n + 2q = 76 as n + q = 38, or n = 38 - q. Substituting this result into the second equation, we get:
15(38 - q) + 35 q = 890, or
570 - 15q + 35q = 890, or
570 + 20 q = 890. Then 20q = 890 - 570 = 320, and q = 320/20 = 16.
There are 16 quarters. Thus, the number of nickels is n = 38 - 16 = 22.
Finally, since n + d + q = 76, 22 + d + 16 = 76, or:
22 + d = 60, or d = 60 - 22 = 38.
There are 38 dimes, 22 nickels and 16 quarters.
