Let's fill in the table based on the given domain and the function [tex]\( y = -\frac{2}{3}x + 7 \)[/tex].
1. For [tex]\( x = -6 \)[/tex]:
[tex]\[
y = -\frac{2}{3}(-6) + 7 = 4 + 7 = 11
\][/tex]
So, [tex]\( y \)[/tex] is 11 when [tex]\( x \)[/tex] is -6.
2. For [tex]\( y = 5 \)[/tex]:
We need to solve for [tex]\( x \)[/tex] in the equation [tex]\( 5 = -\frac{2}{3}x + 7 \)[/tex]:
[tex]\[
5 = -\frac{2}{3}x + 7 \implies -\frac{2}{3}x = 5 - 7 \implies -\frac{2}{3}x = -2 \implies x = -2 \cdot \frac{-3}{2} = 3
\][/tex]
So, [tex]\( x \)[/tex] is 3 when [tex]\( y \)[/tex] is 5.
3. For [tex]\( x = 15 \)[/tex]:
[tex]\[
y = -\frac{2}{3}(15) + 7 = -10 + 7 = -3
\][/tex]
So, [tex]\( y \)[/tex] is -3 when [tex]\( x \)[/tex] is 15.
4. For [tex]\( y = 15 \)[/tex]:
We need to solve for [tex]\( x \)[/tex] in the equation [tex]\( 15 = -\frac{2}{3}x + 7 \)[/tex]:
[tex]\[
15 = -\frac{2}{3}x + 7 \implies -\frac{2}{3}x = 15 - 7 \implies -\frac{2}{3}x = 8 \implies x = 8 \cdot \frac{-3}{2} = -12
\][/tex]
So, [tex]\( x \)[/tex] is -12 when [tex]\( y \)[/tex] is 15.
The completed table is:
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-6 & 11 \\
\hline
3 & 5 \\
\hline
15 & -3 \\
\hline
-12 & 15 \\
\hline
\end{tabular}
