Sure, let's complete the table with the correct steps and justifications needed to solve the given equation [tex]\(\frac{2}{3} y + 15 = 9\)[/tex].
We will follow the already provided steps and justifications:
1. Start with the given equation: [tex]\(\frac{2}{3} y + 15 = 9\)[/tex]
2. Subtract 15 from both sides: [tex]\(\frac{2}{3} y + 15 - 15 = 9 - 15\)[/tex]
3. Simplify the equation: [tex]\(\frac{2}{3} y = -6\)[/tex]
4. Multiply both sides by the reciprocal of [tex]\(\frac{2}{3}\)[/tex], which is [tex]\(\frac{3}{2}\)[/tex]: [tex]\(\frac{2}{3} y \cdot \frac{3}{2} = -6 \cdot \frac{3}{2}\)[/tex]
5. Simplify: [tex]\(y = -9\)[/tex]
These steps and justifications can be placed in the table as follows:
\begin{tabular}{|c|c|}
\hline
Steps & Justifications \\
\hline
[tex]\(\frac{2}{3} y + 15 = 9\)[/tex] & Given \\
\hline
[tex]\(\frac{2}{3} y + 15 - 15 = 9 - 15\)[/tex] & Subtraction property of equality \\
\hline
[tex]\(\frac{2}{3} y = -6\)[/tex] & Simplification \\
\hline
[tex]\(\frac{2}{3} y \cdot \frac{3}{2} = -6 \cdot \frac{3}{2}\)[/tex] & Multiplication property of equality \\
\hline
[tex]\(y = -9\)[/tex] & Simplification \\
\hline
\end{tabular}
Let's put each part in context to ensure the solution path aligns correctly:
1. Step: [tex]\(\frac{2}{3} y + 15 = 9\)[/tex] Justification: Given
2. Step: [tex]\(\frac{2}{3} y + 15 - 15 = 9 - 15\)[/tex] Justification: Subtraction property of equality
3. Step: [tex]\(\frac{2}{3} y = -6\)[/tex] Justification: Simplification
4. Step: [tex]\(\frac{2}{3} y \cdot \frac{3}{2} = -6 \cdot \frac{3}{2}\)[/tex] Justification: Multiplication property of equality
5. Step: [tex]\(y = -9\)[/tex] Justification: Simplification
These are the correct steps and justifications needed to solve the equation [tex]\(\frac{2}{3} y + 15 = 9\)[/tex].
