rose counted her money and found that her 25 coins which were nickels, dimes, and quarters were worth $3.20. The number of dimes exceeded the number of nickels by 4. How many coins of each kind did she have?

I know that the expression for nickles would be 5x, for dimes it would be
10x + 40, but i don't understand how to do the expression for quarters.....can someone help me with the answer

This is a duplicate post. The answer can be found at the other site where it was posted

To solve this problem, we can use a system of equations to represent the given information. Let's use the following variables:

Let 'n' represent the number of nickels.
Let 'd' represent the number of dimes.
Let 'q' represent the number of quarters.

From the given information, we can write the following equations:

Equation 1: n + d + q = 25 (the total number of coins is 25)

Equation 2: 5n + 10d + 25q = 320 (the total value of the coins is $3.20)

From the problem statement, we know that the number of dimes exceeds the number of nickels by 4. This can be translated into another equation:

Equation 3: d = n + 4

Now we can solve the system of equations to find the values of 'n', 'd', and 'q'.

First, substitute Equation 3 into Equation 1:

n + (n + 4) + q = 25
2n + 4 + q = 25
2n + q = 21

Now we have a system of two equations:

Equation 4: 2n + q = 21
Equation 5: 5n + 10d + 25q = 320

To eliminate one variable, we can multiply Equation 4 by 5 and Equation 5 by 2:

10n + 5q = 105
10n + 20d + 50q = 640

Now we subtract Equation 4 from Equation 5:

(10n + 20d + 50q) - (10n + 5q) = 640 - 105
20d + 45q - 5q = 535
20d + 40q = 535

We can divide both sides of this equation by 5 to simplify it:

4d + 8q = 107

Now we have a simplified system of equations:

Equation 6: 4d + 8q = 107
Equation 7: 2n + q = 21

To solve this system, we can multiply Equation 7 by -4 and add it to Equation 6:

-8n - 4q = -84
4d + 8q = 107

-------------------------
0 - 4q = 23

Dividing both sides by -4, we get:

q = -23/4

Since the number of coins cannot be negative, there is no solution for this system of equations. It appears that there may be an error in the given information or equations.