Complete the proof by dragging a reason to each box.

If [tex]$12=\frac{1}{3} x+5$[/tex], then [tex]$x=21$[/tex].

[tex]\[
\begin{tabular}{|l|l|}
\hline
Statement & Reason \\
\hline
$12=\frac{1}{3} x+5$ & Given \\
\hline
$7=\frac{1}{3} x$ & Subtraction Property of Equality \\
\hline
$21=x$ & Multiplication Property of Equality \\
\hline
$x=21$ & Symmetric Property of Equality \\
\hline
\end{tabular}
\][/tex]

Options:
1. Substitution Property of Equality
2. Symmetric Property of Equality
3. Multiplication Property of Equality
4. Distributive Property
5. Subtraction Property of Equality



Answer :

Let's complete the proof step-by-step and provide the reasons accordingly.

### Given Equation
The initial equation is:
[tex]\[ 12 = \frac{1}{3}x + 5 \][/tex]
Reason: Given

### Step 1: Subtract 5 from both sides
[tex]\[ 12 - 5 = \frac{1}{3}x \][/tex]
[tex]\[ 7 = \frac{1}{3}x \][/tex]
Reason: Subtraction Property of Equality

### Step 2: Multiply both sides by 3
[tex]\[ 7 \times 3 = \left(\frac{1}{3}x\right) \times 3 \][/tex]
[tex]\[ 21 = x \][/tex]
Reason: Multiplication Property of Equality

### Step 3: Write x on the left-hand side
[tex]\[ x = 21 \][/tex]
Reason: Symmetric Property of Equality

So, putting it all together in a tabular format, we have:

[tex]\[ \begin{tabular}{|l|l|} \hline \text{Statement} & \text{Reason} \\ \hline 12 = \frac{1}{3} x + 5 & \text{Given} \\ \hline 7 = \frac{1}{3} x & \text{Subtraction Property of Equality} \\ \hline 21 = x & \text{Multiplication Property of Equality} \\ \hline x = 21 & \text{Symmetric Property of Equality} \\ \hline \end{tabular} \][/tex]

This completes the proof that if [tex]\( 12 = \frac{1}{3}x + 5 \)[/tex], then [tex]\( x = 21 \)[/tex].