Question 1049416
n = no. of nickels
d = no. of dimes
q = no. of quarters
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Write an equation for each statement, 
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A coin bank contains 27 coins, all nickels, dimes, or quarters.
n + d + q = 27
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 The combined value of the nickels and dimes is $0.90.
.05n + .10d  = .90
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 The combined value of the dimes and quarters is $4.00
.10d + .25q = 4.0
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How many coins of each type are in the bank?
 Use elimination on the last two equations
 0n + .10d + .25q = 4.00
.05n + .10d + 0q = 0.90
--------------------------Subtraction eliminates d
-.05n +.25q = 3.10
.25q = .05n + 3.10
multiply equation by 4
1q = .20n + 12.40
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Substitute for q in the first equation
n + d + .2n + 12.40 = 27
1.2n + d = 14.60
d = -1.2n + 14.60
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Substitute for d in the 2nd equation
.05n + .10(-1.2n+14.60) = .90
.05n - .12n - 1.46 = .90
-.07n = .90 - 1.46
-.07n = -.56
n = -.56/-.07
n = + 8 nickels
then
q = .20(8) + 12.40
q = 1.6 + 12.4
q = 14 quarters
and using the 1st equation
8 + d + 14 = 27
d = 27 - 22
d = 5 dimes
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Check this in the 3rd equaton
.10(5) + .25(14) = 4
.50 + 3.50 = 4
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A long journey but we finally got it. Hope this made sense to you. CK