Complete Question:
Mike took clothes to the cleaners three times last month. First, brought 4 shirts and 1 pair of slacks and paid $11.45. Then brought 7 shirts, 2 pairs of slacks, and 2 sports and paid $35.39. Finally, brought 5 shirts and 2 sports and paid $21.43 . How much was charged for each shirt, each pair of slacks, and each sports coat?
Answer:
The charges
For each Shirt = $1.49
For each pair of slacks = $5.49
For each sports coat = $6.99
Step-by-step explanation:
Let Shirt = a
Let pair of slacks = b
Let sports coat = c
First, brought 4 shirts and 1 pair of slacks and paid $11.45
= 4a + b = 11.45 ..........Equation 1
b = 11.45 - 4a
Then brought 7 shirts, 2 pairs of slacks, and 2 sports and paid $35.39
= 7a + 2b + 2c = 35.39.........Equation 2
5 shirts and 2 sports and paid $21.43
5a + 2c = 21.43............Equation 3
Hence:
4a + b = 11.45 ..........Equation 1
7a + 2b + 2c = 35.39.........Equation 2
Using elimination method
Multiply Equation 2 by the coefficient of b = 1 in Equation 1
Multiply Equation 1 by the coefficient of b = 2 in Equation 2
8a + 2b = 22.9 .......Equation 4
7a + 2b + 2c = 35.39.........Equation 2
Subtracting Equation 2 from Equation 4
= a - 2c = -12.49 ........Equation 5
a = -12.49 + 2c
Subtituting -12.49 + 2c for a in Equation 3
5a + 2c = 21.43............Equation 3
5(-12.49 + 2c) + 2c = 21.43
= -62.45 + 10c + 2c = 21.43
Collecting like terms
10c + 2c = 21.43 + 62.45
12c = 83.88
c = 83.88/12
c = 6.99
5a + 2c = 21.43............Equation 3
Substituting 6.99 for c in Equation 3
5a + 2(6.99) = 21.43
5a + 13.98 = 21.43
5a = 21.43 - 13.98
5a = 7.45
a = 7.45/5
a = 1.49
4a + b = 11.45 ..........Equation 1
Substituting 1.49 for a in Equation 1
4(1.49) + b = 11.45
b = 11.45 - 4(1.49)
b = 11.45 - 5.96
b = 5.49
Therefore, since the charges
For each shirt = a
The charges for each Shirt = $1.49
For each pair of slacks = b
The charges for each pair of slacks = $5.49
For each sports coat = c
The charges for each sports coat = $6.99
. To make things simple, assume 
. The homogeneous part of the equation is
.
. Then
