a person has $4 in nickles, quarters, and dimes. there is a total of 36 coins altogether. the number of quarters is 4 less then the number of dimes. Algerbaicly determine the number of each type of coin using a system of 3 equations with 3 variables.
a).05N+.10D+.25Q=4
b)N+D+Q=36
Q=D-4
c) D-Q=4
To solve this problem algebraically, we need to define variables for each type of coin. Let's use the following variables:
Let n = number of nickels
Let q = number of quarters
Let d = number of dimes
We have three pieces of information given in the problem:
1. The person has a total of $4 in nickels, quarters, and dimes. We can express this information as an equation in terms of the values of the coins:
0.05n + 0.25q + 0.10d = 4
2. There is a total of 36 coins altogether. This can also be expressed as an equation:
n + q + d = 36
3. The number of quarters is 4 less than the number of dimes:
q = d - 4
We now have a system of three equations with three variables:
0.05n + 0.25q + 0.10d = 4
n + q + d = 36
q = d - 4
To solve this system, we can use substitution or elimination method. Let's solve it using the elimination method:
From equation (3), we can express q in terms of d:
q = d - 4
Substituting this value of q in the other two equations, we get:
0.05n + 0.25(d - 4) + 0.10d = 4 (Substituting q = d - 4)
n + (d - 4) + d = 36 (Substituting q = d - 4)
Simplifying these equations, we have:
0.05n + 0.25d - 1 + 0.10d = 4 (Expanding)
n + d - 4 + d = 36 (Expanding)
Now, let's rearrange these equations:
0.05n + 0.35d = 5 (Combining like terms)
n + 2d = 40 (Combining like terms)
We now have a reduced system of equations:
0.05n + 0.35d = 5 (Equation 1)
n + 2d = 40 (Equation 2)
We can solve this system of equations using substitution or elimination method to find the values of n and d. Once we have the values for n and d, we can substitute them back into any of the original equations to find the value of q.
This is the algebraic way to determine the number of each type of coin using a system of three equations with three variables.