For each ordered pair, determine whether it is a sol
\[
\left\{\begin{array}{l}
12 x+3 y=9 \\
y=-4 x+3
\end{array}\right.
\]
\begin{tabular}{|c|c|c|}
\hline \multirow{2}{*}{$(x, y)$} & \multicolumn{2}{|c|}{ Is it a solution? } \\
\cline { 2 - 3 } & Yes & No \\
\hline$(0,-7)$ & 0 & 0 \\
\hline$(3,-9)$ & 0 & $\bigcirc$ \\
\hline$(-2,11)$ & 0 & 0 \\
\hline$(1,5)$ & 0 & 0 \\
\hline
\end{tabular}
So, the solutions to the system of equations are \(\boxed{(3,-9)}\) and \(\boxed{(-2,11)}\).
Step 1 :Substitute x = 0 and y = -7 into the first equation: \(12(0) + 3(-7) = 9\)
Step 2 :\(0 - 21 = 9\)
Step 3 :\(-21 \neq 9\)
Step 4 :Therefore, (0,-7) is not a solution to the system of equations.
Step 5 :Substitute x = 3 and y = -9 into the first equation: \(12(3) + 3(-9) = 9\)
Step 6 :\(36 - 27 = 9\)
Step 7 :\(9 = 9\)
Step 8 :Then substitute x = 3 and y = -9 into the second equation: \(-9 = -4(3) + 3\)
Step 9 :\(-9 = -12 + 3\)
Step 10 :\(-9 = -9\)
Step 11 :Therefore, (3,-9) is a solution to the system of equations.
Step 12 :Substitute x = -2 and y = 11 into the first equation: \(12(-2) + 3(11) = 9\)
Step 13 :\(-24 + 33 = 9\)
Step 14 :\(9 = 9\)
Step 15 :Then substitute x = -2 and y = 11 into the second equation: \(11 = -4(-2) + 3\)
Step 16 :\(11 = 8 + 3\)
Step 17 :\(11 = 11\)
Step 18 :Therefore, (-2,11) is a solution to the system of equations.
Step 19 :Substitute x = 1 and y = 5 into the first equation: \(12(1) + 3(5) = 9\)
Step 20 :\(12 + 15 = 9\)
Step 21 :\(27 \neq 9\)
Step 22 :Therefore, (1,5) is not a solution to the system of equations.
Step 23 :So, the solutions to the system of equations are \(\boxed{(3,-9)}\) and \(\boxed{(-2,11)}\).