Question

Solve this system of linear inequalities ( will give brainiest)

Answer 1

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