Answer :
To determine the number of Delta (D), Beta (B), and Sigma (S) cars that can be produced in a month, we can set up and solve the system of linear equations based on the information provided about the hours required for painting, drying, and polishing.
We have the following constraints:
1. Painting: [tex]\(7D + 16B + 6S = 165\)[/tex]
2. Drying: [tex]\(2D + 5B + 5S = 94\)[/tex]
3. Polishing: [tex]\(5D + 6B + 3S = 89\)[/tex]
To solve this system of equations, we will use the method of elimination or substitution.
Let's use the elimination method:
1. Multiply the second equation by 2.5 (to match the drying term with the polishing equation):
[tex]\[ 2.5(2D + 5B + 5S) = 2.5 \times 94 \][/tex]
[tex]\[ 5D + 12.5B + 12.5S = 235 \quad \text{(Eq 4)} \][/tex]
2. Now, subtract the polishing equation (Eq 3) from this new equation (Eq 4):
[tex]\[ (5D + 12.5B + 12.5S) - (5D + 6B + 3S) = 235 - 89 \][/tex]
[tex]\[ 6.5B + 9.5S = 146 \quad \text{(Eq 5)} \][/tex]
[tex]\[ 13B + 19S = 292 \quad \text{(simplified by multiplying both sides by 2)} \][/tex]
3. Now, we can take the first equation and the new equation from step 2:
[tex]\[ 7D + 16B + 6S = 165 \quad \text{(Eq 1)} \][/tex]
[tex]\[ 13B + 19S = 292 \quad \text{(Eq 5)} \][/tex]
4. Let's express [tex]\(B\)[/tex] in terms of [tex]\(S\)[/tex] from Eq 5:
[tex]\[ 13B = 292 - 19S \][/tex]
[tex]\[ B = \frac{292 - 19S}{13} \][/tex]
5. Substitute this [tex]\(B\)[/tex] back into Eq 1 to find [tex]\(D\)[/tex] in terms of [tex]\(S\)[/tex]:
[tex]\[ 7D + 16\left(\frac{292 - 19S}{13}\right) + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{16 \times 292 - 16 \times 19S}{13} + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 304S}{13} + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 304S + 78S}{13} = 165 \][/tex] (Combining like terms)
[tex]\[ 7D + \frac{4672 - 226S}{13} = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 226S}{13} = 165 \][/tex]
[tex]\[ 7D \cdot 13 + 4672 - 226S = 165 \cdot 13 \][/tex]
[tex]\[ 91D + 4672 - 226S = 2145 \][/tex]
[tex]\[ 91D + 4672 - 226S = 2145 \][/tex]
[tex]\[ 91D = 2145 - 4672 + 226S \][/tex]
[tex]\[ 91D = -2527 + 226S \][/tex]
[tex]\[ D = -\frac{2527}{91} + \frac{226}{91} S \][/tex]
[tex]\[ D = -27.8 + 2.48S \][/tex]
Thus, we can express [tex]\(D\)[/tex] and [tex]\(B\)[/tex] in terms of [tex]\(S\)[/tex] as:
[tex]\[ D = -27.8 + 2.48S \][/tex]
[tex]\[ B = 22.4615 - 1.46S \][/tex]
6. These results indicate relationships dependent on [tex]\(S\)[/tex], which means that for any value of [tex]\(S\)[/tex], there will be corresponding values of [tex]\(D\)[/tex] and [tex]\(B\)[/tex] that satisfy all three equations. This implies that there are infinitely many solutions forming an infinite set of ordered pairs (D, B, S).
Hence, the correct answer is:
B. There are infinitely many combinations of each type of car that can be produced. Using ordered triplets, the solution can be written as [tex]\( \{(D, B, S) | D=-27.8 + 2.48S, B=22.4615 - 1.46S, S \text{ any real number}\} \)[/tex]
We have the following constraints:
1. Painting: [tex]\(7D + 16B + 6S = 165\)[/tex]
2. Drying: [tex]\(2D + 5B + 5S = 94\)[/tex]
3. Polishing: [tex]\(5D + 6B + 3S = 89\)[/tex]
To solve this system of equations, we will use the method of elimination or substitution.
Let's use the elimination method:
1. Multiply the second equation by 2.5 (to match the drying term with the polishing equation):
[tex]\[ 2.5(2D + 5B + 5S) = 2.5 \times 94 \][/tex]
[tex]\[ 5D + 12.5B + 12.5S = 235 \quad \text{(Eq 4)} \][/tex]
2. Now, subtract the polishing equation (Eq 3) from this new equation (Eq 4):
[tex]\[ (5D + 12.5B + 12.5S) - (5D + 6B + 3S) = 235 - 89 \][/tex]
[tex]\[ 6.5B + 9.5S = 146 \quad \text{(Eq 5)} \][/tex]
[tex]\[ 13B + 19S = 292 \quad \text{(simplified by multiplying both sides by 2)} \][/tex]
3. Now, we can take the first equation and the new equation from step 2:
[tex]\[ 7D + 16B + 6S = 165 \quad \text{(Eq 1)} \][/tex]
[tex]\[ 13B + 19S = 292 \quad \text{(Eq 5)} \][/tex]
4. Let's express [tex]\(B\)[/tex] in terms of [tex]\(S\)[/tex] from Eq 5:
[tex]\[ 13B = 292 - 19S \][/tex]
[tex]\[ B = \frac{292 - 19S}{13} \][/tex]
5. Substitute this [tex]\(B\)[/tex] back into Eq 1 to find [tex]\(D\)[/tex] in terms of [tex]\(S\)[/tex]:
[tex]\[ 7D + 16\left(\frac{292 - 19S}{13}\right) + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{16 \times 292 - 16 \times 19S}{13} + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 304S}{13} + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 304S + 78S}{13} = 165 \][/tex] (Combining like terms)
[tex]\[ 7D + \frac{4672 - 226S}{13} = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 226S}{13} = 165 \][/tex]
[tex]\[ 7D \cdot 13 + 4672 - 226S = 165 \cdot 13 \][/tex]
[tex]\[ 91D + 4672 - 226S = 2145 \][/tex]
[tex]\[ 91D + 4672 - 226S = 2145 \][/tex]
[tex]\[ 91D = 2145 - 4672 + 226S \][/tex]
[tex]\[ 91D = -2527 + 226S \][/tex]
[tex]\[ D = -\frac{2527}{91} + \frac{226}{91} S \][/tex]
[tex]\[ D = -27.8 + 2.48S \][/tex]
Thus, we can express [tex]\(D\)[/tex] and [tex]\(B\)[/tex] in terms of [tex]\(S\)[/tex] as:
[tex]\[ D = -27.8 + 2.48S \][/tex]
[tex]\[ B = 22.4615 - 1.46S \][/tex]
6. These results indicate relationships dependent on [tex]\(S\)[/tex], which means that for any value of [tex]\(S\)[/tex], there will be corresponding values of [tex]\(D\)[/tex] and [tex]\(B\)[/tex] that satisfy all three equations. This implies that there are infinitely many solutions forming an infinite set of ordered pairs (D, B, S).
Hence, the correct answer is:
B. There are infinitely many combinations of each type of car that can be produced. Using ordered triplets, the solution can be written as [tex]\( \{(D, B, S) | D=-27.8 + 2.48S, B=22.4615 - 1.46S, S \text{ any real number}\} \)[/tex]
