Certainly! To create a table of values for the equation [tex]\(3x + 2y = 6\)[/tex] with [tex]\(x\)[/tex] values ranging from [tex]\(-1\)[/tex] to [tex]\(3\)[/tex], we need to determine the corresponding [tex]\(y\)[/tex] values. Let's solve for [tex]\(y\)[/tex] by isolating it on one side of the equation:
[tex]\[3x + 2y = 6\][/tex]
Subtract [tex]\(3x\)[/tex] from both sides:
[tex]\[2y = 6 - 3x\][/tex]
Now, divide both sides by 2 to solve for [tex]\(y\)[/tex]:
[tex]\[y = \frac{6 - 3x}{2}\][/tex]
We will substitute the given [tex]\(x\)[/tex] values [tex]\([-1, 0, 1, 2, 3]\)[/tex] into this equation to find the corresponding [tex]\(y\)[/tex] values.
1. For [tex]\(x = -1\)[/tex]:
[tex]\[
y = \frac{6 - 3(-1)}{2} = \frac{6 + 3}{2} = \frac{9}{2} = 4.5
\][/tex]
2. For [tex]\(x = 0\)[/tex]:
[tex]\[
y = \frac{6 - 3 \cdot 0}{2} = \frac{6}{2} = 3
\][/tex]
3. For [tex]\(x = 1\)[/tex]:
[tex]\[
y = \frac{6 - 3 \cdot 1}{2} = \frac{6 - 3}{2} = \frac{3}{2} = 1.5
\][/tex]
4. For [tex]\(x = 2\)[/tex]:
[tex]\[
y = \frac{6 - 3 \cdot 2}{2} = \frac{6 - 6}{2} = \frac{0}{2} = 0
\][/tex]
5. For [tex]\(x = 3\)[/tex]:
[tex]\[
y = \frac{6 - 3 \cdot 3}{2} = \frac{6 - 9}{2} = \frac{-3}{2} = -1.5
\][/tex]
Now, we can create the table of values:
[tex]\[
\begin{array}{c|c}
x & y \\
\hline
-1 & 4.5 \\
0 & 3 \\
1 & 1.5 \\
2 & 0 \\
3 & -1.5 \\
\end{array}
\][/tex]
So the table of values for the graph of [tex]\(3x + 2y = 6\)[/tex] with [tex]\(x\)[/tex] values ranging from [tex]\(-1\)[/tex] to [tex]\(3\)[/tex] is:
[tex]\[
\begin{array}{c|c}
x & y \\
\hline
-1 & 4.5 \\
0 & 3 \\
1 & 1.5 \\
2 & 0 \\
3 & -1.5 \\
\end{array}
\][/tex]
