Find each solution by graphing (e 3. $3 x+y=1$ and $y+2=0$ \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|} \hline & & & & & & & & & \\ \hline & & & & & & & & & \\ \hline & & & & & & & & & \\ \hline & & & & & & & & & \\ \hline & & & & & & & & & \\ \hline & & & & & & & & & \\ \hline & & & & & & & & & \\ \hline & & & & & & & & & \\ \hline & & & & & & & & & \\ \hline \end{tabular}

Step 1 ：Rewrite the first equation as \(y = 1 - 3x\) and the second equation as \(y = -2\).

Step 2 ：To graph the first equation, choose two points. When \(x = 0\), \(y = 1\). When \(x = 1\), \(y = -2\). So, we have the points (0,1) and (1,-2) for the first equation. The second equation is a horizontal line at \(y = -2\).

Step 3 ：The intersection point of the two lines is the solution to the system of equations. From the graph, we can see that the two lines intersect at the point (1,-2).

Step 4 ：Substitute \(x = 1\) and \(y = -2\) into the original equations to check if they are satisfied. For the first equation, \(3(1) + (-2) = 1\), which is true. For the second equation, \(-2 + 2 = 0\), which is also true.

Step 5 ：So, the solution to the system of equations is \(x = 1\) and \(y = -2\).

Step 6 ：\(\boxed{x = 1, y = -2}\)