SOLUTION: Kelly has a collection of nickels, dimes and quarters with a total value of $9.60. There are 5 more dimes than nickels and 2 times more quarters than nickels, how many of each coin

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Question 948773: Kelly has a collection of nickels, dimes and quarters with a total value of $9.60. There are 5 more dimes than nickels and 2 times more quarters than nickels, how many of each coin does she have?
Please help solve this equation!
So far I have done these steps:
5(x)+10(x+5)+25(2x)=960
5x+10x+50+50x=960
65x+50-50=960-50=910
65x/910=14
I got that far into the equation, but now I don't know what coin does 14 stand for and I have NO IDEA how to get the numbers for 'dimes' and 'quarters.'
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
NOTE: Fixed Mistake!

First examine the description and formulate a system of equations.
Assigning n, d, q, for the coin counts of the expected coins,
this is the first way you may form a system of equations:

system%280.05n%2B0.1d%2B0.25q=9.60%2Cd-n=5%2Cq%2Fn=2%29

Simplify that system, including dividing the money count equation members by 0.05.

You could obtain this equivalent system:
system%28n%2B2d%2B5q=192%2C-n%2Bd=5%2C2n=q%29
a much simpler system.

Think carefully about this simplified system to plan a strategy. The latter two equations each
have two variables and both contain n. Solve the second equation for d in terms of n,
and solve the third equation for q in terms of n. Now substitute these for d and q in the simplified
money count equation. You will be able to solve for the value of n.

The rest should be clear what to do.
(Fixed mistake)