SOLUTION: Find the number of nickels, dimes, and quarters in a collection of 80 such coins if the nickels and the quarters are worth $4.50 and the value of the quarters and the dimes is $5.5

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Question 933517: Find the number of nickels, dimes, and quarters in a collection of 80 such coins if the nickels and the quarters are worth $4.50 and the value of the quarters and the dimes is $5.50.
My work:
.5n + .25q = 4.50
.10d + .25q = 5.50
.10d - .5n = 1.00
n + .5q = 9
d + 2.5q = 55
d - 5n = 10
d + n + q = 80 coins
Ive tried everything, even made a chart - please help!!
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the number of nickels, dimes, and quarters in a collection of 80 such coins if the nickels and the quarters are worth $4.50 and the value of the quarters and the dimes is $5.50.
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Quantity:: n + d + q = 80 coins
Value:: 5n + 25q = 450 cents
Value:: 10d + 25q = 550 cents
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Using the bottom 2 equations,
Subtract and solve for 10d-5n::
10d-5n = 100
Modify::
2d - n = 20
n = 2d-20
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Substitute for "n" in equation #1 to get::
2d-20 + d + q = 80
3d + q = 100
q = -3d+100
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Substitute for "n" and for "q" in 5n+25q=450 to solve for "d"
5(2d-20) + 25(-3d+100) = 450
10d - 100 -75d + 2500 = 450
-65d = 450-2400
-65d = -1950
d = 30 (# of dimes)
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Solve for "q" using q = -3d+100
q = -3*30+100 = 10 (# of quarters)
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Solve for "n" using n = 2d-20
n = 2*30-20 = 40 (# of quarters)
----
Checking with the 1st equation:
30 + 10 + 40 = 80 coins
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Cheers,
Stan H.
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Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

nickels=n,
dimes=d, and
quarters+=q
given:
n%2Bd%2Bq=80 .....eq.1
if the nickels and the quarters are worth $4.50
0.05n%2B0.25q=4.50 .....eq.2
the value of the quarters and the dimes is $5.50
0.25q%2B0.10d=5.50 .....eq.3

solve the system:
n%2Bd%2Bq=80 .....eq.1
0.05n%2B0.25q=4.50 .....eq.2
0.25q%2B0.10d=5.50 .....eq.3
_________________________________
0.05n%2B0.25q=4.50 .....eq.2...multiply by 100
0.25q%2B0.10d=5.50 .....eq.3...multiply by 100
____________________________
5n%2B25q=450 .....eq.2
25q%2B10d=550 .....eq.3
__________________________________

n%2Bd%2Bq=80 .....eq.1..solve for n
n=80+-d-q.......substitute in eq.2

5%2880+-d-q%29%2B25q=450 .....eq.2 ...
400+-5d-5q%2B25q=450
+20q-5d%2B400+=+450
+20q-5d=50...solve for d
+20q-50=5d
+20q%2F5-50%2F5=5d%2F5
+d=4q-10...................

go to 25q%2B10d=550 .....eq.3....substitute d
25q%2B10%284q-10%29=550
25q%2B40q-100=550
65q=550%2B100
65q=650
q=650%2F65
highlight%28q=10%29


go to +d=4q-10. ........substitute q and find d
+d=4%2A10-10
+d=40-10
highlight%28+d=30%29
go to
n=80+-d-q.......substitute q and d, find n
n=80+-30-10
highlight%28+n=40%29