Answer :
the value of the coin times the number of coins equals the total value
so
0.05n + 0.10d + 0.25q = 4.65
the we are given that there are "three more quarters than dimes"
this translates to
q = d + 3
or
d = q - 3
then we are also given that there are "twice as many nickels as quarters"
this translates to
n = 2q
we plug d = q - 3 and n=2q
into the first equation and solve for q
0.05(2q) + 0.10(q-3) + 0.25q = 4.65
we distribute/multiply
0.10q + 0.10q - 0.30 + 0.25q = 4.65
then add like terms
0.45q - 0.30 = 4.65
then add 0.30 to both sides
0.45q = 4.95
then divide both sides by 0.45
and we get
q = 4.95/0.45 = 11
therefore there are 11 quarters
so
0.05n + 0.10d + 0.25q = 4.65
the we are given that there are "three more quarters than dimes"
this translates to
q = d + 3
or
d = q - 3
then we are also given that there are "twice as many nickels as quarters"
this translates to
n = 2q
we plug d = q - 3 and n=2q
into the first equation and solve for q
0.05(2q) + 0.10(q-3) + 0.25q = 4.65
we distribute/multiply
0.10q + 0.10q - 0.30 + 0.25q = 4.65
then add like terms
0.45q - 0.30 = 4.65
then add 0.30 to both sides
0.45q = 4.95
then divide both sides by 0.45
and we get
q = 4.95/0.45 = 11
therefore there are 11 quarters