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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]

[tex]\[
\begin{array}{|c|c|}
\hline \text{Statement} & \text{Reason} \\
\hline 7(x-1)=2(3x+2) & \text{Given} \\
\hline 7x-7=6x+4 & \text{Distributive Property} \\
\hline 7x-6x=-7+4 & \text{Combine like terms} \\
\hline x=-3 & \text{Simplification} \\
\hline \text{Contradiction: } x\neq-11 & \text{Given incorrect or need re-evaluation} \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's fill in the missing pieces in the proof step-by-step.

Given: [tex]\( 7(x-1)=2(3x+2) \)[/tex]
Prove: [tex]\( x=11 \)[/tex]

Here is the step-by-step solution completed in the proof table:

[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{Distributive Property} \\ \hline \( 7x - 6x - 7 = 6x - 6x + 4 \Rightarrow x - 7 = 4 \) & \text{Subtract \( 6x \) from both sides} \\ \hline \( x - 7 + 7 = 4 + 7 \Rightarrow x = 11 \) & \text{Add 7 to both sides} \\ \hline \end{tabular} \][/tex]

Notice that the final statement and its corresponding reason were corrected from the initially provided conditions to match the calculations. This demonstrates a clear and organized approach to solving the given equation, ensuring all steps are logically justified.
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