Given:
The system of inequalities:
![y\leq \dfrac{1}{2}x+2](https://tex.z-dn.net/?f=y%5Cleq%20%5Cdfrac%7B1%7D%7B2%7Dx%2B2)
![y](https://tex.z-dn.net/?f=y%3C-2x-3)
To find:
Whether the points (–3,–2) and (3,2) are in the solution set of the given system of inequalities.
Solution:
A point is in the solution set of the given system of inequalities if it satisfies both inequalities.
Check for the point (-3,-2).
![-2\leq \dfrac{1}{2}(-3)+2](https://tex.z-dn.net/?f=-2%5Cleq%20%5Cdfrac%7B1%7D%7B2%7D%28-3%29%2B2)
![-2\leq -1.5+2](https://tex.z-dn.net/?f=-2%5Cleq%20-1.5%2B2)
![-2\leq 0.5](https://tex.z-dn.net/?f=-2%5Cleq%200.5)
This statement is true.
![-2](https://tex.z-dn.net/?f=-2%3C-2%28-3%29-3)
![-2](https://tex.z-dn.net/?f=-2%3C6-3)
![-2](https://tex.z-dn.net/?f=-2%3C3)
This statement is also true.
Since the point (-3,-2) satisfies both inequalities, therefore (-3,-2) is in the solution set of the given system of inequalities.
Now, check for the point (3,2).
![2](https://tex.z-dn.net/?f=2%3C-2%283%29-3)
![2](https://tex.z-dn.net/?f=2%3C-6-3)
![2](https://tex.z-dn.net/?f=2%3C-9)
This statement is false because
.
Since the point (3,2) does not satisfy the second inequality, therefore (3,2) is not in the solution set of the given system of inequalities.