To find an ordered pair (x, y) that satisfies both inequalities:
- y < x + 1: This means that for any value of x, y must be less than one unit above x.
- y < x + 3: Similarly, y must be less than three units above x.
Start by rewriting the inequalities for clarity:
- 1. y < x + 1
- 2. y < x + 3
Now let’s plot the lines represented by these inequalities:
- For the first inequality, the line y = x + 1 (has a slope of 1 and a y-intercept at 1).
- For the second inequality, the line y = x + 3 (also has a slope of 1 but a y-intercept at 3).
The region satisfying both inequalities will be below both lines in the Cartesian plane. Let’s analyze potential ordered pairs given:
- (2, 1): For x = 2, y = 1;
y < 2 + 1 (1 < 3) is true and
y < 2 + 3 (1 < 5) is true. - (3, 0): For x = 3, y = 0;
y < 3 + 1 (0 < 4) is true and
y < 3 + 3 (0 < 6) is also true. - (1, 3): For x = 1, y = 3;
y < 1 + 1 (3 < 2) is false. - (0, 1): For x = 0, y = 1;
y < 0 + 1 (1 < 1) is false.
Based on this analysis, both the ordered pairs (2, 1) and (3, 0) satisfy the inequalities. Therefore, the answer is:
- (2, 1)
- (3, 0)
In conclusion, the ordered pairs (2, 1) and (3, 0) are valid solutions that make both inequalities true.