from all this information we can come up with a few equations to work with...
first we have that the worth of each coin times the number of those coins all added together equals 4.65
\(0.05n + 0.10d + 0.25q = 4.65\)
here I have n = nickels, d = dimes, and q = quaters
then we are told that there are "three more quarters than dimes"
which translates to the equation
\(q = d + 3\)
or
\(d = q - 3\)
we are also told that there are "twice as many nickels as quarters"
this translates to
\(n = 2q\)
to solve for quarters, we plug in d (in terms of q) and n (in terms of q) into the first equation we created.
\(0.05(2q) + 0.10(q - 3) + 0.25q = 4.65\)
then we distribute and start trying to solve for \(q\)
\(0.10q + 0.10q - 0.30 + 0.25 = 4.65\)
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
\(q = \frac{4.95}{0.45} = 11\)
therefore our answer is B) 11 quarters
