Answer :
To determine which table contains only viable solutions, we need to verify if the weight \( w \) corresponds correctly to the number of books \( b \) ordered, considering that each book weighs 6 ounces. To do this, we use the formula:
[tex]\[ w = b \times 6 \][/tex]
Let's check each row in both tables:
Table 1:
[tex]\[ \begin{array}{|c|c|} \hline \text{Books } (b) & \text{Weight } (w) \\ \hline -2 & -12 \\ \hline -1 & -6 \\ \hline 0 & 0 \\ \hline 1 & 6 \\ \hline 2 & 12 \\ \hline \end{array} \][/tex]
1. For \( b = -2 \):
[tex]\[ w = -2 \times 6 = -12 \][/tex]
The weight given is \(-12\), which is correct.
2. For \( b = -1 \):
[tex]\[ w = -1 \times 6 = -6 \][/tex]
The weight given is \(-6\), which is correct.
3. For \( b = 0 \):
[tex]\[ w = 0 \times 6 = 0 \][/tex]
The weight given is \(0\), which is correct.
4. For \( b = 1 \):
[tex]\[ w = 1 \times 6 = 6 \][/tex]
The weight given is \(6\), which is correct.
5. For \( b = 2 \):
[tex]\[ w = 2 \times 6 = 12 \][/tex]
The weight given is \(12\), which is correct.
Table 2:
[tex]\[ \begin{array}{|c|c|} \hline \text{Books } (b) & \text{Weight } (w) \\ \hline -1 & -6 \\ \hline -0.5 & -3 \\ \hline 0 & 0 \\ \hline 0.5 & 3 \\ \hline 1 & 6 \\ \hline \end{array} \][/tex]
1. For \( b = -1 \):
[tex]\[ w = -1 \times 6 = -6 \][/tex]
The weight given is \(-6\), which is correct.
2. For \( b = -0.5 \):
[tex]\[ w = -0.5 \times 6 = -3 \][/tex]
The weight given is \(-3\), which is correct.
3. For \( b = 0 \):
[tex]\[ w = 0 \times 6 = 0 \][/tex]
The weight given is \(0\), which is correct.
4. For \( b = 0.5 \):
[tex]\[ w = 0.5 \times 6 = 3 \][/tex]
The weight given is \(3\), which is correct.
5. For \( b = 1 \):
[tex]\[ w = 1 \times 6 = 6 \][/tex]
The weight given is \(6\), which is correct.
Since all the calculated weights match the corresponding values in the tables, we can conclude:
Both tables contain only viable solutions.
Thus, the answer is that both tables contain only viable solutions.
[tex]\[ w = b \times 6 \][/tex]
Let's check each row in both tables:
Table 1:
[tex]\[ \begin{array}{|c|c|} \hline \text{Books } (b) & \text{Weight } (w) \\ \hline -2 & -12 \\ \hline -1 & -6 \\ \hline 0 & 0 \\ \hline 1 & 6 \\ \hline 2 & 12 \\ \hline \end{array} \][/tex]
1. For \( b = -2 \):
[tex]\[ w = -2 \times 6 = -12 \][/tex]
The weight given is \(-12\), which is correct.
2. For \( b = -1 \):
[tex]\[ w = -1 \times 6 = -6 \][/tex]
The weight given is \(-6\), which is correct.
3. For \( b = 0 \):
[tex]\[ w = 0 \times 6 = 0 \][/tex]
The weight given is \(0\), which is correct.
4. For \( b = 1 \):
[tex]\[ w = 1 \times 6 = 6 \][/tex]
The weight given is \(6\), which is correct.
5. For \( b = 2 \):
[tex]\[ w = 2 \times 6 = 12 \][/tex]
The weight given is \(12\), which is correct.
Table 2:
[tex]\[ \begin{array}{|c|c|} \hline \text{Books } (b) & \text{Weight } (w) \\ \hline -1 & -6 \\ \hline -0.5 & -3 \\ \hline 0 & 0 \\ \hline 0.5 & 3 \\ \hline 1 & 6 \\ \hline \end{array} \][/tex]
1. For \( b = -1 \):
[tex]\[ w = -1 \times 6 = -6 \][/tex]
The weight given is \(-6\), which is correct.
2. For \( b = -0.5 \):
[tex]\[ w = -0.5 \times 6 = -3 \][/tex]
The weight given is \(-3\), which is correct.
3. For \( b = 0 \):
[tex]\[ w = 0 \times 6 = 0 \][/tex]
The weight given is \(0\), which is correct.
4. For \( b = 0.5 \):
[tex]\[ w = 0.5 \times 6 = 3 \][/tex]
The weight given is \(3\), which is correct.
5. For \( b = 1 \):
[tex]\[ w = 1 \times 6 = 6 \][/tex]
The weight given is \(6\), which is correct.
Since all the calculated weights match the corresponding values in the tables, we can conclude:
Both tables contain only viable solutions.
Thus, the answer is that both tables contain only viable solutions.