 To solve a system by graphing, we need to first, well, graph the lines. So y equals negative x plus 3. Negative x plus 3, we know that 3 is the y-intercept, and the slope is negative 1. So if we graph that, we'd start at the y-intercept, which was 3, and then apply the slope of negative 1. So down 1, right 1, down 1, right 1, down 1, right 1. We could also go behind it, back 1, up 1. It would work just as well. And then graph the line, so that's one of them. Now the other one, y equals 2x minus 3. So again, the y-intercept is negative 3, and the slope is 2. So we'll go to the y-intercept, down 3, 1, 2, 3, and then apply the slope. The slope again was 2. So that means we rise to, run 1, up 2, over 1, up 2, over 1. And then if we wanted to, we could go back and then down 2 would work just as well. Graph the line, alright. And now since it does indeed ask us to solve the system, we do need to find where those two lines cross. It appears as though they cross at that point, and the coordinates would be 2, 1. So that would be the solution to our linear, our system of linear equations to...