To solve this problem, we can set up a system of equations based on the given information.
Let's denote the number of nickels, dimes, and quarters as n, d, and q respectively.
From the given information, we can create the following equations:
1) The total value of the coins is $9.50:
0.05n + 0.10d + 0.25q = 9.50
2) There are 130 coins in total:
n + d + q = 130
3) There are twice as many dimes as there are quarters:
d = 2q
Now, with these equations, we can solve for the number of nickels (n).
First, substitute equation 3 into equation 2:
n + 2q + q = 130
Simplifying, we get:
n + 3q = 130 (equation 4)
Next, we can multiply equation 4 by 0.05 (to represent nickels) and subtract it from equation 1 to eliminate the n variable:
0.05n + 0.10d + 0.25q - 0.05(n + 3q) = 9.50 - 0.05(130)
Simplifying, we get:
0.10d - 0.10q = 3.50 (equation 5)
Now we have two equations (4 and 5) with two variables (q and d).
Solving these equations simultaneously will give us the values of q and d, which we can then use to find n.
Let's solve equations 4 and 5:
From equation 5, we can rearrange it as:
0.10d = 0.10q + 3.50
Simplifying further:
d = q + 35
Now substitute this into equation 4:
n + 3q = 130
n + 3(q + 35) = 130
Simplifying:
n + 3q + 105 = 130
n + 3q = 25
Now, we have a new equation (n + 3q = 25) along with the equation (d = q + 35).
Using these equations, we can solve for q and d.
Substitute d = q + 35 into n + 3q = 25:
n + 3q = 25
n + 3(q + 35) = 25
Simplifying:
n + 3q + 105 = 25
n + 3q = -80
Now we have two equations:
d = q + 35
n + 3q = -80
To get the solution for q and d, we need to solve this system of equations.