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.
