SOLUTION: I have 30 total coins that total $3.55. There are 3 more dimes than nickels. How many quarters do I have. 5n+10(n+3)+25(30-2n+3)=355

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Question 949308: I have 30 total coins that total $3.55. There are 3 more dimes than nickels. How many quarters do I have.
5n+10(n+3)+25(30-2n+3)=355
Answer by Alan3354(69443) About Me  (Show Source):
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I have 30 total coins that total $3.55. There are 3 more dimes than nickels. How many quarters do I have.
5n+10(n+3)+25(30-2n+3)=355
5n+10(n+3)+25(30-(2n+3))=355
5n+10(n+3)+25(30-2n-3)=355 **************** The -1 applies to the 3
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n = # of nickels
Solve for n
5n+10(n+3)+25(30-2n-3)=355
5n + 10n+30 + 750-50n-75 = 355
-35n + 705 = 355
-35n = -350
n = 10 nickels
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Solve for q