Graph the system below and write its solution. \[ \begin{array}{l}y=\frac{1}{3} x-3 \\ 3 x+y=7\end{array} \]

To graph the system of equations, we can start by rearranging the second equation into slope-intercept form (y = mx + b). From the equation \(3x + y = 7\), we can solve for \(y\): \[ y = -3x + 7. \] Now we have two equations to graph: 1. \( y = \frac{1}{3}x - 3 \) 2. \( y = -3x + 7 \) **First Line:** To graph these equations, you can begin by plotting the y-intercepts. For the first equation, the y-intercept is at (0, -3), and for the second, it’s at (0, 7). Next, use the slope to determine another point; for the first equation \( \frac{1}{3} \) means rise 1 and run 3, leading to point (3, -2). For the second equation, with a slope of -3, go down 3 and right 1 to find point (1, 4). Connect the points to reveal the lines! **Second Line:** The intersection of these two lines will give us the solution to the system. By solving the equations simultaneously, we can substitute \(y\) from the first equation into the second equation: \[ 3x + \left(\frac{1}{3}x - 3\right) = 7 \] which simplifies to \(x = 3\). Plugging \(x = 3\) back into either equation gives \(y = -2\). Thus, the solution to the system is the point (3, -2).