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This problem is to be solved using one single equation in one unknown.
I will show you on how to do it.
The major step is to choose the leading unknown reasonably.
Let x be the number of quarters.
Then the number of dimes is (x-6), and the number of nickels is 2x.
By knowing it, you can write the "money" equation momentarily
5*(2x) + 10*(x-6) + 25x = 435 cents.
Simplify
10x + 10x - 60 + 25x = 435
45x = 435 + 60 = 495
x = = 11.
ANSWER. 11 quarters; 11-6 = 5 dimes, and 2*11 = 22 nickels.
CHECK. 5*22 + 5*10 + 25*11 = 435 cents, which is precisely correct (!)
Solved.
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The lesson to learn from my post is THIS :
This problem can be solved using one single equation in one unknown.
It is designed and intended to be solved in this way.
It is intended for 6-7 grade students who are not familiar yet with systems of equations.
To see many other similar solved problems, look into the lessons
- More complicated coin problems
in this site.