# SOLUTION: Joe has a collection of nickels and dimes that is worth \$6.10. If the number of dimes were tripled and the number of nickels were decreased by 49, the total value of the coins woul

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 Question 408040: Joe has a collection of nickels and dimes that is worth \$6.10. If the number of dimes were tripled and the number of nickels were decreased by 49, the total value of the coins would be \$14.05. How many nickels and dimes does he have?Answer by ankor@dixie-net.com(15660)   (Show Source): You can put this solution on YOUR website!Let n = no. of nickels Let d = no. of dimes : Write an equation for each statement: : "Joe has a collection of nickels and dimes that is worth \$6.10." .05n + .10d = 6.10 .05n = 6.10 - .10d divide by .05 n = (122-2d); use this form for substitution : " If the number of dimes were tripled and the number of nickels were decreased by 49, the total value of the coins would be \$14.05." .05(n-49) + .10(3d) = 14.05 .05n - 2.45 + .3d = 14.05 .05n + .3d = 14.05 + 2.45 .05n + .3d = 16.50 Replace n with (122-2d) .05(122-2d) + .3d = 16.50 6.10 - .10d + .3d = 16.50 -.10d + .3d = 16.50 - 6.10 .20d = 10.40 d = d = 52 dimes originally and n = 122 - 2d n = 122 - 2(52) n = 122 - 104 n = 18 nickels originally : : This satisfies the equations but looking at the statement: "If the number of dimes were tripled and the number of nickels were decreased by 49, the total value of the coins would be \$14.05." : How can you decrease the number of nickels by 49 if you only have 18!!! : Plunging ahead though: .10(3*52) + .05(18-49) = .10(156) + .05(-31) = 15.60 - 1.55 = 14.05; the math is happy with it, but how about real life?