ashleenbond
Answered

Drag each step and justification to the correct location on the table. Each step and justification can be used more than once, but not all steps and justifications will be used.

Order each step and justification that is needed to solve the equation below.

[tex]\[
\begin{array}{l}
\frac{2}{3} y + 15 = 9 \\
\qquad \frac{2}{3} y \cdot \frac{3}{2} = 6 \cdot \frac{3}{2} \quad \frac{2}{3} y \cdot \frac{3}{2} = -6 \cdot \frac{3}{2} \quad \frac{2}{3} y = 6 \quad y = -9
\end{array}
\][/tex]

Multiplication property of equality \quad [tex]\(\frac{2}{3} y = -6\)[/tex] \quad [tex]\(y = 9\)[/tex] \quad Subtraction property of equality

\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}



Answer :

GinnyAnswer
Certainly! Let's arrange each step and justification in the correct order to solve the given equation.

Given the steps:
[tex]\[ \begin{array}{l} \frac{2}{3} y+15=9 \\ \frac{2}{3} y+15-15=9-15 \\ \frac{2}{3} y=-6 \\ \frac{2}{3} y \cdot \frac{3}{2}=-6 \cdot \frac{3}{2} \\ y=-9 \end{array} \][/tex]

And the justifications that will be used:
\begin{itemize}
\item Given
\item Subtraction property of equality
\item Simplification
\item Multiplication property of equality
\end{itemize}

Now, we'll arrange them in the correct order in the table:

[tex]\[ \begin{tabular}{|c|c|} \hline Steps & Justifications \\ \hline \(\frac{2}{3} y+15=9\) & Given \\ \hline \(\frac{2}{3} y+15-15=9-15\) & Subtraction property of equality \\ \hline \(\frac{2}{3} y=-6\) & Simplification \\ \hline \(\frac{2}{3} y \cdot \frac{3}{2}=-6 \cdot \frac{3}{2}\) & Multiplication property of equality \\ \hline \(y=-9\) & Simplification \\ \hline \end{tabular} \][/tex]

This completes the step-by-step solution to solve the equation [tex]\(\frac{2}{3} y + 15 = 9\)[/tex].

Other Questions