From the graphs you can see that the graph of the line [tex]y= \frac{1}{3} x-3[/tex] lies under the graph of the line [tex]y= \frac{1}{3} x-1[/tex]. Then all solutions of unequality [tex]y\le \frac{1}{3} x-3[/tex] are solutions of unequality [tex]y\le \frac{1}{3} x-1[/tex], but not all solutions of unequality [tex]y\le \frac{1}{3} x-1[/tex] are solutions of unequality [tex]y\le \frac{1}{3} x-3[/tex].
For example, if x=1, y=-2
[tex]-2\le \frac{1}{3} \cdot 1-1 =- \frac{2}{3} \\ -2\ \textgreater \ \frac{1}{3} \cdot 1-3=- \frac{8}{3} [/tex].
Answer: Correct choice is B.
