Respuesta :
Let n = number of nickels.
Let d = number of dimes.
Let q = number of quarters.
Number of coins:
n + d + q = 52
Value of coins:
0.05n + 0.1d + 0.25q = 7.8
Info about number of coins:
n + d = q + 6
n + d - q = 6
We have a system of three equations:
n + d + q = 52
0.05 + 0.1d + 0.25q = 7.8
n + d - q = 6
Subtract the third equation from the first equation:
n + d + q = 52
- n + d - q = 6
-------------------------
2q = 46
q = 23
Now replace q in the first ans second equations with 23.
n + d + 23 = 52
n + d = 29 Eq.1
0.05n + 0.1d + 0.25(23) = 7.8
0.05n + 0.1d + 5.75 = 7.8
0.05n +0.1d = 2.05 Eq. 2
Now solve equations 1 and 2 as a system of two equations.
n + d = 29
0.05n + 0.1d = 2.05
d = 29 - n
0.05n + 0.1(29 - n) = 2.05
0.05n + 2.9 - 0.1n = 2.05
-0.05n = -0.85
n = 17
Answer: There are 17 nickels.
Let d = number of dimes.
Let q = number of quarters.
Number of coins:
n + d + q = 52
Value of coins:
0.05n + 0.1d + 0.25q = 7.8
Info about number of coins:
n + d = q + 6
n + d - q = 6
We have a system of three equations:
n + d + q = 52
0.05 + 0.1d + 0.25q = 7.8
n + d - q = 6
Subtract the third equation from the first equation:
n + d + q = 52
- n + d - q = 6
-------------------------
2q = 46
q = 23
Now replace q in the first ans second equations with 23.
n + d + 23 = 52
n + d = 29 Eq.1
0.05n + 0.1d + 0.25(23) = 7.8
0.05n + 0.1d + 5.75 = 7.8
0.05n +0.1d = 2.05 Eq. 2
Now solve equations 1 and 2 as a system of two equations.
n + d = 29
0.05n + 0.1d = 2.05
d = 29 - n
0.05n + 0.1(29 - n) = 2.05
0.05n + 2.9 - 0.1n = 2.05
-0.05n = -0.85
n = 17
Answer: There are 17 nickels.