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Fill in the missing pieces in this proof.

Given: [tex]7(x-1)=2(3x+2)[/tex]

Prove: [tex]x = 11[/tex]

\begin{tabular}{|c|c|}
\hline
Statement & Reason \\
\hline
[tex]7(x-1)=2(3x+2)[/tex] & Given \\
\hline
[tex]7x-7=6x+4[/tex] & Distribute \\
\hline
[tex]7x-6x-7=4[/tex] & Subtraction Property of Equality \\
\hline
[tex]x-7=4[/tex] & Combine like terms \\
\hline
[tex]x=11[/tex] & Addition Property of Equality \\
\hline
& \\
\hline
\end{tabular}

Sagot :

GinnyAnswer
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].
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