from his paper route, andy collected $5.55 in nickels and dimes. The number of nickels was 6 more than the number of dimes. How many nickels and dimes were there?
n = d+6
5 n + 10 d = 555 cents
5(d+6) + 10 d = 555
5 d + 30 + 10 d = 555
etc
add up the values
If there were x dimes, then
10x + 5(x+6) = 555
To solve this problem, we can set up a system of equations based on the information given. Let's denote the number of nickels as "n" and the number of dimes as "d".
1. The first equation is based on the total value of the coins collected:
0.05n + 0.10d = 5.55
2. The second equation states that the number of nickels is 6 more than the number of dimes:
n = d + 6
We can now use these equations to solve for the values of n and d.
First, let's substitute the value of "n" from the second equation into the first equation:
0.05(d + 6) + 0.10d = 5.55
Simplifying the equation:
0.05d + 0.30 + 0.10d = 5.55
0.15d + 0.30 = 5.55
0.15d = 5.55 - 0.30
0.15d = 5.25
Next, divide both sides of the equation by 0.15 to isolate "d":
d = 5.25 / 0.15
d ≈ 35
So, the number of dimes is approximately 35.
Now, substitute this value of "d" back into the second equation to find the number of nickels:
n = 35 + 6
n = 41
Therefore, there were 41 nickels and 35 dimes.