Which ordered pair makes both inequalities true y x + 2?

To find an ordered pair that satisfies both inequalities y < 2x + 3 and y > x + 2, we can start by graphing the lines represented by the equations y = 2x + 3 and y = x + 2.

The first inequality y < 2x + 3 indicates that we are looking for values of y that lie below the line y = 2x + 3. The second inequality y > x + 2 indicates that we are looking for values of y that lie above the line y = x + 2.

By analyzing the graphs, we can identify the region where both conditions are satisfied. For instance, we can test the ordered pair (0, 3):

1. Substitute (0, 3) in the first inequality:

y < 2x + 3 → 3 < 2(0) + 3 → 3 < 3 (this is false)

2. Substitute (0, 3) in the second inequality:

y > x + 2 → 3 > 0 + 2 → 3 > 2 (this is true)

Since the pair (0, 3) does not satisfy the first inequality, we need to look for other pairs.

Let’s try the ordered pair (1, 4):

1. For the first inequality:

y < 2x + 3 → 4 < 2(1) + 3 → 4 < 5 (this is true)

2. For the second inequality:

y > x + 2 → 4 > 1 + 2 → 4 > 3 (this is true)

Thus, the ordered pair (1, 4) makes both inequalities true.

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