To determine which system of linear inequalities includes the point (3, 2) in its solution set, we need to check the inequalities against this point.
Start by considering a couple of example inequalities. Let’s say we have the following system:
- 1. y > 2x – 4
- 2. y < -x + 5
Now, we will substitute the coordinates of the point (3, 2) into each inequality:
For the first inequality:
2 > 2(3) - 4
2 > 6 - 4
2 > 2 (false)
This inequality is false when (3, 2) is plugged in, so this inequality does not hold for our point.
Next, for the second inequality:
2 < -3 + 5
2 < 2 (false)
This one is also false. So, neither of these inequalities includes the point (3, 2) in their solution set.
To find an appropriate system, let’s consider another example:
- 1. y > x – 1
- 2. y < -x + 6
Now checking (3, 2) for these:
For the first inequality:
2 > 3 - 1
2 > 2 (false)
This is still false. Now let’s try:
- 1. y > x – 2
- 2. y < -x + 8
Checking (3, 2):
For the first:
2 > 3 - 2
2 > 1 (true)
And for the second:
2 < -3 + 8
2 < 5 (true)
Both conditions are true for the point (3, 2). Therefore, this point lies in the solution set of the system of inequalities:
- y > x – 2
- y < -x + 8
In conclusion, the point (3, 2) is a solution to the system of inequalities: y > x – 2 and y < -x + 8.