25q + 10d + 5n = 250
q + d + n = 18
Let's eliminate d by multiplying the
second equation through by -10 and
adding the two equations:
25q + 10d + 5n = 250
-10q - 10d - 10n = 180
----------------------
15q - 5n = 70
Divide through by 5
3q - n = 14
3q - 14 = n
So the smallest possible number of quarters is 5
Substitute q = 5
n = 3q - 14 = 3(5) - 14 = 15 - 14 = 1
Substitute q = 5 and n = 1 in
q + d + n = 18
5 + d + 1 = 18
d + 6 = 18
d = 12
So 1 solution is 5 quarters, 1 nickel and 12 dimes
----------------------------------------------------
Substitute q = 6
n = 3q - 14 = 3(6) - 14 = 18 - 14 = 4
Substitute q = 6 and n = 4 in
q + d + n = 18
6 + d + 4 = 18
d + 10 = 18
d = 8
So 1 solution is 6 quarters, 4 nickels and 8 dimes
----------------------------------------------------
Substitute q = 7
n = 3q - 14 = 3(7) - 14 = 21 - 14 = 7
Substitute q = 7 and n = 7 in
q + d + n = 18
7 + d + 7 = 18
d + 14 = 18
d = 4
So 1 solution is 7 quarters, 7 nickels and 4 dimes
----------------------------------------------------
Substitute q = 8
n = 3q - 14 = 3(8) - 14 = 24 - 14 = 10
Substitute q = 8 and n = 10 in
q + d + n = 18
8 + d + 10 = 18
d + 18 = 18
d = 0
So 1 solution is 8 quarters, 10 nickels and 0 dimes
----------------------------------------------------
We can't use 9 quarters for that's $2.25, and we'd
have to make the remaining 25 cents with 9 nickels
and dimes, which is impossible because even 9
nickels alone is 45 cents!!
----------------------------------------------------
So there are 4 solutions:
5 quarters, 1 nickel, and 12 dimes.
6 quarters, 4 nickels, and 8 dimes.
7 quarters, 7 nickels, and 4 dimes.
8 quarters, 10 nickels, and 0 dimes.
Edwin
