Question 53602
We know that a nickel can be represented as .05, and similarly for a dime and a quarter:  .10 and .25; so the equation becomes {{{.05x+.10y+.25z=5.50}}}, where x is the number of nickels, y is the number of dimes, and z is the number of quarters, and furthermore, {{{x+y+z=37}}}.  But that has 3 unknowns, so let's go further:  Jo has 4 more quarters than nickels.  Because the number of nickels is x, that means she has (x+4) quarters; so our equation is now:  {{{.05x+.10y+.25(x+4)=5.50}}}.  Expanding that equation we get {{{.05x+.10y+.25x+1=5.50}}}.  Combining like terms and simplifying we get {{{.30x+.10y=4.50}}}.  Going back to the number of coins equation:  {{{x+y+z=37}}}, we substitute in (x+4) for z (number of quarters) to get {{{x+y+x+4=37}}} and combining like terms and simplifying to get {{{2x+y=33}}}, or {{{y=33-2x}}}.  Now let's substitute that value of y into our previous equation to get {{{.3x+.10(33-2x)=4.5}}}.  Expanding gives us {{{.3x+3.3-.2x=4.5}}}.  Combining like terms and simplifying we get {{{.1x=1.2}}}, or {{{x=12}}}.  That means we have 12 nickels, or $.60; because we have 4 more than 12 for the number of quarters, we have 16 quarters, or $4.00.  So far we have 28 coins (12 nickels and 16 quarters); that means we must have 9 dimes (37-28=9), or $.90; adding up all the money, we get $.60+$4.00+$.90=$5.50.  So that means are equations and answer of 9 dimes is correct.