writing an inequality in two variables given a table of values\nwhich linear inequality could represent the…

Answer:

B. $y\leq - 2x + 3$

Explanation:

Step1: Test first point $(-4,-1)$

For $y<-2x + 3$, substitute $x=-4$ and $y = - 1$: $-1<-2\times(-4)+3$, $-1<8 + 3$, $-1<11$ (true). For $y\leq - 2x+3$, $-1\leq - 2\times(-4)+3$, $-1\leq8 + 3$, $-1\leq11$ (true). For $y>-\frac{1}{2}x - 3$, $-1>-\frac{1}{2}\times(-4)-3$, $-1>2 - 3$, $-1>-1$ (false). For $y\leq-\frac{1}{2}x - 3$, $-1\leq-\frac{1}{2}\times(-4)-3$, $-1\leq2 - 3$, $-1\leq - 1$ (true).

Step2: Test second point $(-2,4)$

For $y<-2x + 3$, substitute $x=-2$ and $y = 4$: $4<-2\times(-2)+3$, $4<4 + 3$, $4<7$ (true). For $y\leq - 2x+3$, $4\leq - 2\times(-2)+3$, $4\leq4 + 3$, $4\leq7$ (true). For $y>-\frac{1}{2}x - 3$, $4>-\frac{1}{2}\times(-2)-3$, $4>1 - 3$, $4>-2$ (true). For $y\leq-\frac{1}{2}x - 3$, $4\leq-\frac{1}{2}\times(-2)-3$, $4\leq1 - 3$, $4\leq - 2$ (false).

Step3: Test third point $(3,-3)$

For $y<-2x + 3$, substitute $x = 3$ and $y=-3$: $-3<-2\times3+3$, $-3<-6 + 3$, $-3<-3$ (false). For $y\leq - 2x+3$, $-3\leq - 2\times3+3$, $-3\leq-6 + 3$, $-3\leq - 3$ (true). For $y>-\frac{1}{2}x - 3$, $-3>-\frac{1}{2}\times3-3$, $-3>-\frac{3}{2}-3$, $-3>-\frac{3 + 6}{2}$, $-3>-\frac{9}{2}$ (true). For $y\leq-\frac{1}{2}x - 3$, $-3\leq-\frac{1}{2}\times3-3$, $-3\leq-\frac{3}{2}-3$, $-3\leq-\frac{9}{2}$ (false).

Step4: Conclusion

Since the points satisfy $y\leq - 2x + 3$ and not the other inequalities, the linear - inequality is $y\leq - 2x + 3$.