Answer:
- Coat: $5
- Shirt: $7
- Dress: $4
Explanation:
You want the cost of cleaning for a coat, shirt, and dress, given the monthly cleaning costs for the numbers of items listed in the table.
Setup
Let c, s, d represent the costs of cleaning a coat, shirt, and dress, respectively. These are what the problem is asking for.
In order to find these costs, we need to make use of relations that we can write as equations. The problem statement tells us the cleaning costs for the three months are ...
7c +6s +3d = 89 . . . . . . cleaning cost for January
4c +6s +5d = 82 . . . . . . cleaning cost for February
8c +3s +4d = 77 . . . . . . cleaning cost for March
These three equations in three variables can be solved any number of ways. The first attachment shows a calculator's use of the matrix row-reduction method.
Solution
We recognize that the coefficients of s are related by a simple constant, so we can use this fact to eliminate the 's' variable from the equations.
Subtracting the second equation from the first gives ...
(7c +6s +3d) -(4c +6s +5d) = (89) -(82)
3c -2d = 7 . . . . . simplify
Subtracting the first equation from twice the third gives ...
2(8c +3s +4d) -(7c +6s +3d) = 2(77) -(89)
9c +5d = 65 . . . . . simplify
These two reduced equations have a coefficient of c that is related by a simple constant. This means we can eliminate the c variable by subtracting 3 times the first equation from the second:
(9c +5d) -3(3c -2d) = (65) -3(7)
11d = 44 . . . . . simplify
d = 4 . . . . . . . divide by 11
Substituting back into a convenient equation, we can find the values of the other variables:
3c -2(4) = 7
3c = 15 . . . . . . add 8
c = 5 . . . . . . divide by 3
and ...
8(5) +3s +4(4) = 77
3s = 21 . . . . . . subtract 56, simplify
s = 7 . . . . . divide by 3
The costs for cleaning each item are ...
- Coat: $5
- Shirt: $7
- Dress: $4
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Additional comment
An ad hoc solution method is detailed above. Its steps are based on eliminating a variable at each step, first reducing the system to 2 equations in 2 unknowns, then to 1 equation in 1 unknown. The choice of variable to eliminate and the way to do that is based on the coefficients we see. A different set of equations would result in different choices. One of the considerations in our choices is keeping the coefficients positive to help reduce errors.
The "reduced row-echelon form" of the augmented coefficient matrix shown in the attachment can be computed for any arbitrary equations. The calculator uses the same method all the time. This can also be done by hand, but tends to be somewhat tedious. Its convenience is that the calculator can do it simply after entering the coefficients of the equations.
Other algorithms can be used for solving the equations. Cramer's rule is one of those, making use of the matrix of coefficients.
We have also successfully used graphical methods for solving systems in more than 2 variables. The idea there is to use substitution to reduce the system to 2 variables.
For example, we could use the first equation to define d(c, s) = (89 -7c -6s)/3 and then graph the other two equations: 4c +6s +5d(c, s) = 82 and 8c +3s +4d(c, s) = 77. The solutions to these would be the point (c, s) = (5, 7), and the value of d would be found as d(5, 7) = 4. This is shown in the second attachment.
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