Question 108375
An equation for each if the 3 statements:
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1.Moira has $2.95 half dollars, dimes, and nickels.
.05n + .10d + .5h = 2.95
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2.The number of half-dollars is 1 less than twice the number of dimes.
  h = 2d - 1
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3.The sum of the number of half-dollars and the number of nickels is 8.
   n + h = 8
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Find the number of coins of each type.
Let's try to put n and d in terms of h
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Rearrange equation 2:
2d - 1 = h
2d = h + 1
d = {{{(h+1)/2}}}
or
d = .5(h+1)
d = (.5h+.5)
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Rearrange equation 3:
n + h = 8
n = (8-h)
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Back to eq 1: .05n + .10d + .5h = 2.95
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Substitute (.5h+.5) for d & (8-h) for n; solve for h: 
.05(8-h) + .10(.5h+.5) + .5h = 2.95
:
.4 - .05h + .05h + .05 + .5h = 2.95; group and combine like terms:
:
-.05h + .05h + .5h + .4 + .05 = 2.95
:
.5h + .45 = 2.95
:
.5h = 2.95 - .45
:
.6h = 2.50
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h = 2.5/.5
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h = 5 half dollars
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Use n = 8 - h to find n
n = 8-5
n = 3 nickels
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Use d = .5h + .5 to find d
d = .5(5) + .5
d = 2.5 + .5
d = 3 dimes
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Check solution in the 1st equation:
.05(3) + .10(3) + .5(5) =
.15 + .30 + 2.5 = 2.95; confirms our solutions
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Did this help you? A lot of steps, but each one is logical, not hard to understand, is it?