To determine whether each ordered pair [tex]\((x, y)\)[/tex] satisfies the equation [tex]\(5x + 4y = -7\)[/tex], follow these steps:
1. For [tex]\((-3, 2)\)[/tex]:
- Substitute [tex]\(x = -3\)[/tex] and [tex]\(y = 2\)[/tex] into the equation:
[tex]\[
5(-3) + 4(2) = -15 + 8 = -7
\][/tex]
- Since [tex]\(-7 = -7\)[/tex] is true, [tex]\((-3, 2)\)[/tex] is a solution to the equation.
2. For [tex]\((1, -3)\)[/tex]:
- Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = -3\)[/tex] into the equation:
[tex]\[
5(1) + 4(-3) = 5 - 12 = -7
\][/tex]
- Since [tex]\(-7 = -7\)[/tex] is true, [tex]\((1, -3)\)[/tex] is a solution to the equation.
3. For [tex]\((2, 7)\)[/tex]:
- Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex] into the equation:
[tex]\[
5(2) + 4(7) = 10 + 28 = 38
\][/tex]
- Since [tex]\(38 \neq -7\)[/tex] is false, [tex]\((2, 7)\)[/tex] is not a solution to the equation.
4. For [tex]\((-4, 0)\)[/tex]:
- Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 0\)[/tex] into the equation:
[tex]\[
5(-4) + 0 = -20
\][/tex]
- Since [tex]\(-20 \neq -7\)[/tex] is false, [tex]\((-4, 0)\)[/tex] is not a solution to the equation.
Now, filling in the table based on the results we have:
\begin{tabular}{|c|c|c|}
\hline
[tex]$(x, y)$[/tex] & Is it a solution? \\
\cline{2-3}
& Yes & No \\
\hline
[tex]$(-3, 2)$[/tex] & X & 0 \\
\hline
[tex]$(1, -3)$[/tex] & X & 0 \\
\hline
[tex]$(2, 7)$[/tex] & 0 & X \\
\hline
[tex]$(-4, 0)$[/tex] & 0 & X \\
\hline
\end{tabular}
