SOLUTION: joe has $4.70 in half-dollars, quarters, and dimes. The number of quarters is 10 less than twice the number of dimes. The number of half-dollars decreased by the number of quarters

Question 65380: joe has $4.70 in half-dollars, quarters, and dimes. The number of quarters is 10 less than twice the number of dimes. The number of half-dollars decreased by the number of quarters is 2. Find the umber of coins of each type.
Answer by Zizzle(2) (Show Source): You can put this solution on YOUR website!
Let the number of dimes = d
Let the number of quarters = 2d-10
Let the number of half dollars = 2d-8
.10(d) + .25(2d-10) + .50(2d-8) = 4.70
.10d + .5d-2.5 + 1d-4 = 4.70
1.6d - 6.5 = 4.70
1.6d = 4.70+6.5
1.6d = 11.20
d=7
The number of dimes = 7
The number of quarters = 2(7)-10, which is equal to 4
The number of half dollars = 2(7)-8, which is equal to 6
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