# SOLUTION: Jo has37 coins(all nickels, dimes, andquarters) worth \$5.50. She has 4 more quarters than nickels. How many dimes does she have?

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 Question 171665This question is from textbook : Jo has37 coins(all nickels, dimes, andquarters) worth \$5.50. She has 4 more quarters than nickels. How many dimes does she have?This question is from textbook Answer by ptaylor(2052)   (Show Source): You can put this solution on YOUR website!Let x=number of nickels Then x+4=number of quarters So the number of dimes would be: 37-x-(x+4)= 37-2x-4= 33-2x----number of dimes Now we are told that: (lets deal in pennies) 5x+25(x+4)+10(33-2x)=550 get rid of parens (distributive) 5x+25x+100+330-20x=550 collect like terms 10x+430=550 subtract 430 from each side 10x+430-430=550-430= 10x=120 divide each side by 10 x=12-----------------number of nickels x+4=12+4=16-------------number of quarters 33-2x=33-24=9---------------number of dimes CK 5*12+25*16+9*10=550 60+400+90=550 550=550 We could also solve this problem using three unknowns: x=number of nickels y=number of dimes z=number of quarters x+y+z=37-----------------------eq1 5x+10y+25z=550------------------eq2 z=x+4----------------------------eq3 Substitute z=x+4 into eq1 and eq2: 2x+y=33-------------eq1a 30x+10y=450-----------eq2a Multiply eq1a by 15 and subtract eq2a from it: 5y=45 y=9---------------number of dimes Hope this helps---ptaylor