# SOLUTION: A jar with pennies, nickels, and dimes contains a total of 55 coins worth \$3.23. How many of each are there if the number of pennies is 10 less than the sum of nickels and dimes to

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 Question 181055: A jar with pennies, nickels, and dimes contains a total of 55 coins worth \$3.23. How many of each are there if the number of pennies is 10 less than the sum of nickels and dimes together?Answer by ptaylor(2048)   (Show Source): You can put this solution on YOUR website!YOU NEED TO RECHECK YOUR PROBLEM. UNLESS I MADE A MISTAKE, WE ARE COMING OUT WITH FRACTIONS OF COINS AND WE CAN'T HAVE FRACTIONS OF COINS. Let x=number of pennies y=number of nickels z=number of dimes Then: x+y+z=55-----------------eq1 Lets deal in pennies x+5y+10z=323------------------eq2 and x+10=y+z or x-y-z=-10 or x-(y+z)=-10----------------------eq3 but from eq1, y+z=55-x. Substitute this into eq3 x-(55-x)=-10 or x-55+x=-10 add 55 to each side 2x=45-----------------------------------------NO GOOD!!!! LET'S TRY ANOTHER APPROACH: subtract eq1 from eq2 and we get: 4y+9z=268-----------------------------------eq1a subtract eq3 from eq2 and we get: 6y+11z=333------------------------------------eq2a Multiply eq1a by 3: 12y+27z=804----------------------eq1b Multiply eq2a by 2: 12y+22z=666--------------------------eq2b subtract eq2b from eq1b: 5z=138-------------------------------------------NO GOOD!!!!! Hope this helps---ptaylor