SOLUTION: Please help solve this problem thanks- Solve each system by the addition method Nickels and Dimes. Winborne has 35 coins cinsisting of dimes and nickels. If the value of his

Click here to see ALL problems on Linear-equations

Question 7700: Please help solve this problem thanks-
Solve each system by the addition method
Nickels and Dimes. Winborne has 35 coins cinsisting of dimes and nickels. If the value of his coins is $3.30, then how many of each type does he have?
Answer by drglass(89) (Show Source):
You can put this solution on YOUR website!
We have two facts to start with

1) the number of nickels and the number of dimes add up to 35

2) the value of the nickels and dimes add up to $3.30

Let's call the number of nickels n and the number of dimes d. The number of coins add up to 35 and the value of the nickels is $0.05n and the value of the dimes is $0.10d. We can convert the two facts to a mathematical representation as follows:

1) n + d = 35

2) $0.05n + $0.10d = 3.30

Subtract n from both sides of the first equation to get

d + n - n = 35 - n

d = 35 - n

Now substitute d = 35 - n into the second equation:

0.05n + 0.10(35 - n) = 3.3

0.05n + 3.5 - 0.10n = 3.3

3.5 - 0.05n = 3.3

Subtract 3.3 from either side

3.5 - 0.05n - 3.3 = 3.3 - 3.3

0.2 - 0.05n = 0

Add 0.05n to either side to get

0.2 - 0.05n + 0.05n = 0.05n

0.2 = 0.05n

divide both sides by 0.05 to get:

=

So, n = 4. Since d + n = 35 and n = 4, we know d + 4 = 35. Subtract 4 from either side to get d = 31.

The answer is 4 nickels and 31 dimes.

To see that this is true, figure out how much 4 nickels and 31 dimes are worth. The 4 nickels add up to $0.20 and the 31 dimes add up to $3.10, the nickels and dimes add up to $3.30. This is exactly what the problem says.