Let's fill in the missing pieces in your proof step-by-step.
Given: \(7(x-1)=2(3x+2)\)
[tex]\[
\begin{tabular}{|c|c|}
\hline
\text{Statement} & \text{Reason} \\
\hline
7(x-1)=2(3x+2) & \text{Given} \\
\hline
7(x - 1) = 2(3x + 2) & \text{Given} \\
\hline
7x - 7 = 6x + 4 & \text{Distributive Property (Expansion)} \\
\hline
7x - 7 = 6x + 4 & \text{4. Expand both sides} \\
\hline
7x - 6x - 7 = 6x - 6x + 4 & \text{Subtract } 6x \text{ from both sides} \\
\hline
x - 7 = 4 & \text{Combine like terms} \\
\hline
x - 7 + 7 = 4 + 7 & \text{Add 7 to both sides} \\
\hline
x = 11 & \text{Solve for } x \\
\hline
11 = x & \text{Reflexive Property} \\
\hline
\end{tabular}
\][/tex]
Thus, the completed proof table is as follows:
[tex]\[
\begin{tabular}{|c|c|}
\hline
\text{Statement} & \text{Reason} \\
\hline
7(x-1)=2(3x+2) & \text{Given} \\
\hline
7x - 7 = 6x + 4 & \text{Expand both sides} \\
\hline
x - 7 = 4 & \text{Subtract 6x from both sides} \\
\hline
x = 11 & \text{Add 7 to both sides} \\
\hline
11 = x & \text{Reflexive Property} \\
\hline
\end{tabular}
\][/tex]
By following these steps and completing the proof table, we demonstrate that [tex]\(x = 11\)[/tex].
