You can put this solution on YOUR website! Warren has 40 coins; nickels dimes and quarters worth $4.05. He has seven more nickels than dimes. How many quarters does he have?
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Write 3 equations, one for each statement:
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"Warren has 40 coins;"
n + d + q = 40; eq1
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" nickels dimes and quarters worth $4.05."
.05n + .10d + .25q = 4.05; eq2
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"He has seven more nickels than dimes."
n = d + 7
n - d = 7; eq3
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Add equations 1 and 3:
n + d + q = 40
n - d + 0 = 7
------------------adding eliminates d
2n + q = 47
q = (47-2n)
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Multiply eq2 by 10, subtract eq1
.5n + 1d + 2.5q = 40.5
1n + 1d + 1q = 40
------------------------subtraction eliminates d again
-.5n + 0 + 1.5q = .5
-.5n + 1.5q = .5
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Substitute (47-2n) for q in the above equation, find n
-.5n + 1.5(47-2n) = .5
-.5n + 70.5 - 3n = .5
-.5n - .3n = .5 - 70.5
-3.5n = -70
n =
n = +20 nickels
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Find d;
n - d = 7
20 - d = 7
d = 13 dimes
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Obviously, that leaves us with 7 quarters.
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You should check these solutions using eq2:
