Which system of linear inequalities has the point [tex]$(3, -2)$[/tex] in its solution set?

A. [tex]y \ \textless \ -3[/tex]

B. [tex]y \leq \frac{2}{3} x - 4[/tex]



Answer :

To determine if the point [tex]\((3, -2)\)[/tex] is in the solution set of the given system of linear inequalities, we need to check if it satisfies both inequalities:

1. [tex]\(y < -3\)[/tex]
2. [tex]\(y \leq \frac{2}{3}x - 4\)[/tex]

Let's evaluate the point [tex]\((3, -2)\)[/tex] for each inequality step by step.

### 1. Checking the inequality [tex]\(y < -3\)[/tex]:
Substitute the coordinates of the point [tex]\((3, -2)\)[/tex] into the inequality [tex]\(y < -3\)[/tex]:
- The [tex]\(y\)[/tex]-coordinate of the point is [tex]\(-2\)[/tex].

We need to check if:
[tex]\[ -2 < -3 \][/tex]

This statement is false since [tex]\(-2\)[/tex] is not less than [tex]\(-3\)[/tex].

### 2. Checking the inequality [tex]\(y \leq \frac{2}{3}x - 4\)[/tex]:
Substitute the coordinates of the point [tex]\((3, -2)\)[/tex] into the inequality [tex]\(y \leq \frac{2}{3}x - 4\)[/tex]:
- The [tex]\(x\)[/tex]-coordinate of the point is [tex]\(3\)[/tex], and the [tex]\(y\)[/tex]-coordinate is [tex]\(-2\)[/tex].

First, calculate the right-hand side of the inequality:
[tex]\[ \frac{2}{3} \cdot 3 - 4 \][/tex]
Perform the multiplication:
[tex]\[ 2 - 4 = -2 \][/tex]

So, the inequality becomes:
[tex]\[ -2 \leq -2 \][/tex]

This statement is true since [tex]\(-2\)[/tex] is equal to [tex]\(-2\)[/tex], and thus it satisfies the inequality [tex]\(y \leq -2\)[/tex].

### Conclusion:
The point [tex]\((3, -2)\)[/tex] satisfies the second inequality [tex]\(y \leq \frac{2}{3}x - 4\)[/tex] but does not satisfy the first inequality [tex]\(y < -3\)[/tex]. Therefore, the point [tex]\((3, -2)\)[/tex] is in the solution set of the inequality:
[tex]\[ y \leq \frac{2}{3} x - 4 \][/tex]
but not in the solution set of the inequality:
[tex]\[ y < -3 \][/tex]

Hence, the point [tex]\((3, -2)\)[/tex] is in the solution set of:
[tex]\[ y \leq \frac{2}{3} x - 4 \][/tex]