Question 437114
With coin problems, you have to keep track of the counts of each coin denomination as well as the value. For example, let d=number of dimes.  Then the value of the number of dimes = 10d.  Let n=number of nickels, 5n = value of nickels.  Similarly, q=number of quarters and 25q=value of the quarters.
.
We're told she has nickels, dimes, and quarters worth $2.05 or 205 cents.
5n + 10d + 25q = 205
.
We're also told she has 16 coins, so
n + d + q = 16
.
And we're told she has 3 times as many dimes as nickels:
3n = d
or
d = 3n
.
We can use simultaneous equations.
.
5n + 10d + 25q = 205
Divide by 5
n + 2d + 5q = 41
.
Substitute d = 3n in the equations
.
n + 2(3n) + 5q = 41
n + 6n + 5q = 41
7n + 5q = 41
.
n + 3n + q = 16
4n + q = 16
.
So we have two equations with two unknowns.
7n + 5q = 41
4n + q = 16
.
Multiply the second equation by 5
20n + 5q = 80
.
7n + 5q = 41
20n + 5q = 80
.
Subtract the second equation from the first.
-13n = -39
n = 3
.
Therefore d = 3n = 9
.
Therefore q = 16 - 3 -9 = 4
.
Our tentative answer is:
n = 3
d = 9
q = 4
.
Always check your answer.  In this case, we have to use the "value" equation since we used the "count" equation to figure out how many quarters.
What is the total value?
3*5 = 15 cents
9*10 = 90 cents
4*25 = 100 cents
45 + 90 + 100 = 205 cents
Correct.
.
Answer:  She has 3 nickels, 9 dimes, and 4 quarters.
.
Done.