SOLUTION: I can not figure this out!!! Suppose a meter accepts nickels, dimes and quarters. You inserted a total of 32 coins with a value of $6.80. How many of each coin did you insert?

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Question 373815: I can not figure this out!!!
Suppose a meter accepts nickels, dimes and quarters. You inserted a total of 32 coins with a value of $6.80. How many of each coin did you insert?
This is in the Systems of Equations section of my textbook but I've also been told that it maybe a diophantine equation. I know for a system of equations with three varibles you need 3 equations. I have figured out two of the equations.
n+d+q=32 and .05n+.10d+.25q=6.80 but from there it goes blank! I appreciate any help anyone can offer!! Thanks in advance!
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
Suppose a meter accepts nickels, dimes and quarters. You inserted a total of 32 coins with a value of $6.80. How many of each coin did you insert?
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n+d+q=32
5n+10d+25q=680
multiply first equation by -25
-25n-25d-25q=-800
add it to the second equation
-20n-15d=-120
20n+15d=120
/5
4n+3d=24
4n = 24-3d
4n= 6-0.75d
1 nickel = 5 cents
..
n= 6 -0.75d
n has to be an integer.
so 0.75d has to be an integer
the smallest value of d should be an integer and less than 6
For d = 4
(0.75 d = 3 is an integer)
n= 6-3
n=3
..
d = 4, n =3 so q= 25
4*10+3*5+25*25=680
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m.ananth@hotmail.ca