which system of linear inequalities has the point (3, -2) in its solution set?\n$y < - 3$\n$yleqslant\frac{2}…

Explanation:

Step1: Substitute x = 3 and y = - 2 into the first inequality

For (y<-3), substitute (y=-2). We get (-2<-3), which is false. For (y\leq\frac{2}{3}x - 4), substitute (x = 3) and (y=-2). First, calculate the right - hand side: (\frac{2}{3}\times3-4=2 - 4=-2). Since (y=-2) and (-2=-2), the inequality (y\leq\frac{2}{3}x - 4) is satisfied when (x = 3) and (y=-2). But since the first inequality (y<-3) is not satisfied, this is not the correct system. We need to check all the systems in a similar way. Let's assume we have the general form of checking a system (\left{\begin{array}{l}y < a_1x + b_1\y\leq a_2x + b_2\end{array}\right.) by substituting (x = 3) and (y=-2) into each inequality. Let's assume another system (\left{\begin{array}{l}y>-3\y\leq\frac{2}{3}x - 4\end{array}\right.)

Step2: Substitute into the first inequality

Substitute (y=-2) into (y > - 3). We have (-2>-3), which is true.

Step3: Substitute into the second inequality

Substitute (x = 3) and (y=-2) into (y\leq\frac{2}{3}x - 4). Calculate (\frac{2}{3}\times3-4=2 - 4=-2). Since (y=-2) and (-2=-2), the second inequality (y\leq\frac{2}{3}x - 4) is also true.

Answer:

The system (\left{\begin{array}{l}y>-3\y\leq\frac{2}{3}x - 4\end{array}\right.) has the point ((3,-2)) in its solution set.