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Betty has 10 more dimes than quarters. If she has $3.45, how many coins does she have?
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D = Q + 10 25D + 10D = 345, D and Q in cents.
Solve by substitution.
its asking how many dimes and how many quarters
25Q + 10D = 345.
Solving the system D = Q + 10 10D + 25Q = 345 gives you the number of dimes and the number of quarters.
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x=dimes x-10+quarters .10(x)+.25(x-10)=3.45 10x+25x-250=345 35x=595 x=17 17 dimes and (17-10) quarters -> 17+7=24 total coins
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