Solving the equation for y puts it in slope-intercept form.
[tex]3x-5y=5\qquad\text{given equation}\\3x-5=5y\qquad\text{add 5y-5}\\y=\dfrac{3}{5}x-1\qquad\text{divide by 5}[/tex]
This tells you one point can be plotted on the y-axis at y=-1. Another point can be plotted 3 units up and 5 units to the right of that one, at (5, 2). The graph is the line through those points.
Since you're asked to plot at least 3 points, your third point could be 5 units left and 3 units down from the first one, at (-5, -4).
