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Complete the following proof by providing the statements and reasons. Be aware, you may not use all boxes in the table. (6 points)

Given: [tex]\(6a - 2 = 3(a + 1) + 1\)[/tex]
Prove: [tex]\(a = 2\)[/tex]

\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{Statements} & \multicolumn{1}{c|}{Reasons} \\
\hline [tex]\(6a - 2 = 3(a + 1) + 1\)[/tex] & Given \\
\hline [tex]\(6a - 2 = 3a + 3 + 1\)[/tex] & Distributive property \\
\hline [tex]\(6a - 2 = 3a + 4\)[/tex] & Simplifying \\
\hline [tex]\(6a - 3a - 2 = 3a + 4 - 3a\)[/tex] & Subtracting [tex]\(3a\)[/tex] from both sides \\
\hline [tex]\(3a - 2 = 4\)[/tex] & Simplifying \\
\hline [tex]\(3a - 2 + 2 = 4 + 2\)[/tex] & Adding 2 to both sides \\
\hline [tex]\(3a = 6\)[/tex] & Simplifying \\
\hline [tex]\(a = \frac{6}{3}\)[/tex] & Dividing both sides by 3 \\
\hline [tex]\(a = 2\)[/tex] & Simplifying \\
\hline
\end{tabular}

Sagot :

GinnyAnswer
Certainly! Let's complete the proof by providing the necessary statements and reasons.

Given: [tex]\( 6a - 2 = 3(a + 1) + 1 \)[/tex]
Prove: [tex]\( a = 2 \)[/tex]

\begin{tabular}{|l|l|}
\hline
\multicolumn{1}{|c|}{Statements} & \multicolumn{1}{c|}{Reasons} \\
\hline
[tex]\( 6a - 2 = 3(a + 1) + 1 \)[/tex] & Given \\
\hline
[tex]\( 6a - 2 = 3a + 3 + 1 \)[/tex] & Distributive property \\
\hline
[tex]\( 6a - 2 = 3a + 4 \)[/tex] & Combine like terms on the right side \\
\hline
[tex]\( 6a - 3a - 2 = 4 \)[/tex] & Subtract [tex]\( 3a \)[/tex] from both sides \\
\hline
[tex]\( 3a - 2 = 4 \)[/tex] & Simplify \\
\hline
[tex]\( 3a - 2 + 2 = 4 + 2 \)[/tex] & Add 2 to both sides \\
\hline
[tex]\( 3a = 6 \)[/tex] & Simplify \\
\hline
[tex]\( a = 2 \)[/tex] & Divide both sides by 3 \\
\hline
\end{tabular}

Thus, we have proven that [tex]\( a = 2 \)[/tex].
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