SOLUTION: Here is another problem! Jo has 37 coins(all nickles, dimes and quarters) worth $5.50. She has 4 more quarters than nickles. How many dimes does Jo have? I can't figure out wh

Algebra ->  Expressions-with-variables -> SOLUTION: Here is another problem! Jo has 37 coins(all nickles, dimes and quarters) worth $5.50. She has 4 more quarters than nickles. How many dimes does Jo have? I can't figure out wh      Log On


   

Question 104577This question is from textbook algebra Structure and Method
: Here is another problem!
Jo has 37 coins(all nickles, dimes and quarters) worth $5.50. She has 4 more quarters than nickles. How many dimes does Jo have? I can't figure out what the variable for dimes is.
Thanks! This question is from textbook algebra Structure and Method

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Jo has 37 coins(all nickles, dimes and quarters) worth $5.50. She has 4 more quarters than nickles. How many dimes does Jo have? I can't figure out what the variable for dimes is.
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Let number of nickels be "x" ; value of these is 5x cents
# of quarters is "x+4" ; value of these is 25(x+4) = 25x+100 cents
# of dimes = 37-(x+x+4)="33-2x"; value is 10(33-2x)=330-20x cents
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EQUATION:
value + value + value = 550 cents
5x + 25x+100 + 330-20x = 550
10x = 120
x = 12 (# of nickels)
x+4 = 16 (# of quarters)
33-2x= 9 (# of dimes)
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Cheers,
Stan H.