Solve the following system of equations:

[tex]\[ \left\{\begin{array}{l}
y = 7x + 9 \\
2x - 3y = -8
\end{array}\right. \][/tex]

\begin{tabular}{|c|c|c|}
\hline
\multirow{2}{*}{[tex]$(x, y)$[/tex]} & \multicolumn{2}{|c|}{Is it a solution?} \\
\cline{2-3}
& Yes & No \\
\hline
[tex]$(2, 0)$[/tex] & & \\
\hline
[tex]$(-1, 2)$[/tex] & & \\
\hline
[tex]$(-3, -12)$[/tex] & [tex]$\bullet$[/tex] & \\
\hline
\end{tabular}



Answer :

Let's analyze each of the given coordinate pairs [tex]\((x, y)\)[/tex] to determine whether they satisfy the given system of equations:
[tex]\[ \left\{ \begin{array}{c} y = 7x + 9 \\ 2x - 3y = -8 \end{array} \right. \][/tex]

1. For the coordinate pair [tex]\((2, 0)\)[/tex]:
- Substitute [tex]\(x = 2\)[/tex] into the first equation:
[tex]\[ y = 7(2) + 9 = 14 + 9 = 23 \][/tex]
So, [tex]\( y \)[/tex] should be 23.
- However, the given [tex]\( y \)[/tex] is 0. Thus, [tex]\((2, 0)\)[/tex] does not satisfy [tex]\( y = 7x + 9 \)[/tex].
- As a result, [tex]\((2, 0)\)[/tex] is not a solution.

2. For the coordinate pair [tex]\((-1, 2)\)[/tex]:
- Substitute [tex]\(x = -1\)[/tex] into the first equation:
[tex]\[ y = 7(-1) + 9 = -7 + 9 = 2 \][/tex]
So, [tex]\( y \)[/tex] is indeed 2.
- Substitute [tex]\( x = -1 \)[/tex] and [tex]\( y = 2 \)[/tex] into the second equation:
[tex]\[ 2(-1) - 3(2) = -2 - 6 = -8 \][/tex]
This satisfies the second equation as well.
- Therefore, [tex]\((-1, 2)\)[/tex] is a solution.

3. For the coordinate pair [tex]\((-3, -12)\)[/tex]:
- Substitute [tex]\(x = -3\)[/tex] into the first equation:
[tex]\[ y = 7(-3) + 9 = -21 + 9 = -12 \][/tex]
So, [tex]\( y \)[/tex] is indeed -12.
- Substitute [tex]\( x = -3 \)[/tex] and [tex]\( y = -12 \)[/tex] into the second equation:
[tex]\[ 2(-3) - 3(-12) = -6 + 36 = 30 \][/tex]
This does not satisfy the second equation as it should equal [tex]\(-8\)[/tex].
- Therefore, [tex]\((-3, -12)\)[/tex] is not a solution.

Let's fill in the table accordingly:

[tex]\[ \begin{tabular}{|c|c|c|} \hline \multirow{2}{*}{$(2, y)$} & \multicolumn{2}{|c|}{Is it a solution?} \\ \cline{2-3} & Yes & No \\ \hline $(2,0)$ & & $\bullet$ \\ \hline $(-1,2)$ & $\bullet$ & \\ \hline $(-3,-12)$ & & $\bullet$ \\ \hline & 1 & 2 \\ \hline \end{tabular} \][/tex]

Thus, [tex]\((2, 0)\)[/tex] and [tex]\((-3, -12)\)[/tex] are not solutions, while [tex]\((-1, 2)\)[/tex] is a solution.