if sue has 27 coins in all, totaling 2.10, of only dimes and nickels, and there are more dimes, how many more dimes than nickels does she have?
15 dimes = $1.50
12 nickles = $0.60
To solve this problem, you can set up a system of equations using the given information.
Let's assume that Sue has x dimes and y nickels.
From the problem statement, we know that Sue has a total of 27 coins, so we can write the first equation:
x + y = 27 ...(Equation 1)
Additionally, we are given that the total value of the coins is $2.10. Since each dime is worth 10 cents and each nickel is worth 5 cents, we can write the second equation:
10x + 5y = 210 ...(Equation 2)
Now, we have a system of two equations with two variables. We can solve this system to find the values of x and y, which represent the number of dimes and nickels, respectively.
One way to solve this system is by substitution. From Equation 1, we can express x in terms of y:
x = 27 - y
Substituting this value of x into Equation 2, we get:
10(27 - y) + 5y = 210
Simplifying the equation, we have:
270 - 10y + 5y = 210
Combining like terms, we get:
270 - 5y = 210
Now, we can solve for y:
-5y = 210 - 270
-5y = -60
y = -60 / -5
y = 12
So, Sue has 12 nickels.
To find the number of dimes, we substitute the value of y back into Equation 1:
x + 12 = 27
x = 27 - 12
x = 15
Therefore, Sue has 15 dimes.
Finally, to find the difference between the number of dimes and nickels, we subtract the number of nickels (y) from the number of dimes (x):
Difference = x - y
Difference = 15 - 12
Difference = 3
Hence, Sue has 3 more dimes than nickels.