Nadia has 32 coins made up of nickles, dimes, and quarters. The sum of the number of nickels and the number of quarters is three times the number of dimes. If the total value of the coins is $4.60, how many of each kind does she have?
I can't figure out how to make this word problem into 3 equations.
Wow, you waited all of 6 minutes before your reposted.
Check your previous post.
To solve this word problem, let's break it down into smaller steps and create equations for each step.
Step 1: Define variables
Let's define variables for the number of nickels, dimes, and quarters.
Let x be the number of nickels.
Let y be the number of dimes.
Let z be the number of quarters.
Step 2: Set up the equations
According to the problem, the sum of the number of nickels and the number of quarters is three times the number of dimes. This can be expressed as:
x + z = 3y (Equation 1)
Next, we know that the total value of the coins is $4.60. We can calculate the value of each coin, and then sum up the total value.
The value of a nickel is $0.05, the value of a dime is $0.10, and the value of a quarter is $0.25.
The value equation can be expressed as:
0.05x + 0.10y + 0.25z = 4.60 (Equation 2)
Step 3: Solve the system of equations
Now we have two equations: Equation 1 and Equation 2. We can solve them simultaneously to find the values of x, y, and z.
Equation 1: x + z = 3y
Equation 2: 0.05x + 0.10y + 0.25z = 4.60
We can solve this system of equations using any method, such as substitution, elimination, or matrices, but substitution is the simplest method in this case.
From Equation 1, we know that x + z = 3y. Thus, we can rearrange Equation 1 to solve for x in terms of y and z:
x = 3y - z
Substituting x = 3y - z into Equation 2, we get:
0.05(3y - z) + 0.10y + 0.25z = 4.60
Expanding and simplifying the equation, we have:
0.15y - 0.05z + 0.10y + 0.25z = 4.60
0.25y + 0.20z = 4.60
25y + 20z = 460 (Equation 3) [Multiplying both sides by 100 to get rid of decimals]
So now, we have two equations:
x + z = 3y (Equation 1)
25y + 20z = 460 (Equation 3)
Step 4: Solve for the variables
Solve the system of equations (Equation 1 and Equation 3) to find the values of x, y, and z.
From Equation 1, we can rewrite it as:
x = 3y - z
Substitute x in terms of y and z into Equation 3:
25y + 20z = 460
Now we have a system of two linear equations:
3y - z + z = 3y (from Equation 1)
25y + 20z = 460
Simplifying the system:
3y = 3y
25y + 20z = 460
By comparing the equations, we can see that the first equation is an identity, meaning there are infinitely many solutions. This is because the equation 3y = 3y holds true for any value of y.
To find a unique solution, we need another equation. However, the problem does not provide any other equations or information. As a result, we cannot determine the exact values of x, y, and z.
Therefore, there isn't a unique solution to this word problem without additional information.