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Complete the proof by selecting and dropping the correct reason in the spaces below.

[tex]\[
\begin{tabular}{|l|l|}
\hline
\multicolumn{1}{|c|}{\textbf{Statement}} & \textbf{Reason} \\
\hline
$8x - 5 = 2x + 1$ & \\
\hline
$6x - 5 = 1$ & Subtraction Property of Equality \\
\hline
$6x = 6$ & Addition Property of Equality \\
\hline
$x = 1$ & Division Property of Equality \\
\hline
\end{tabular}
\][/tex]

Reasons:
1. Subtraction Property of Equality
2. Addition Property of Equality
3. Division Property of Equality

Sagot :

GinnyAnswer
Let's go through the solution step-by-step and fill in the spaces with the correct reasons.

1. Original Equation:
- Statement: [tex]\(8x - 5 = 2x + 1\)[/tex]
- Reason: This is the given equation.

2. Applying the Subtraction Property of Equality:
- Statement: [tex]\(8x - 5 - 2x = 2x + 1 - 2x\)[/tex]
- Simplifies to: [tex]\(6x - 5 = 1\)[/tex]
- Reason: Subtraction Property of Equality

3. Applying the Addition Property of Equality:
- Statement: [tex]\(6x - 5 + 5 = 1 + 5\)[/tex]
- Simplifies to: [tex]\(6x = 6\)[/tex]
- Reason: Addition Property of Equality

4. Applying the Division Property of Equality:
- Statement: [tex]\(\frac{6x}{6} = \frac{6}{6}\)[/tex]
- Simplifies to: [tex]\(x = 1\)[/tex]
- Reason: Division Property of Equality

So the table should be filled as:

\begin{tabular}{|l|l|}
\hline
\multicolumn{1}{|c|}{Statement} & Reason \\
\hline
[tex]$8x - 5 = 2x + 1$[/tex] & Given Equation \\
\hline
[tex]$6x - 5 = 1$[/tex] & Subtraction Property of Equality \\
\hline
[tex]$6x = 6$[/tex] & Addition Property of Equality \\
\hline
[tex]$x = 1$[/tex] & Division Property of Equality \\
\hline
\end{tabular}

This completes the step-by-step solution showing how each statement follows from the previous one by applying the appropriate property of equality.
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