Which ordered pair makes both inequalities true y < x + 1 and y < x + 3?

To determine which ordered pair satisfies both inequalities, we need to analyze each pair given: (5, 2), (2, 1), (3, 0), and (1, 3).

The first inequality is y < x + 1. Rearranged, it predicts that for any x value, y must be less than that value plus one.

The second inequality is y < x + 3, indicating that y must also be less than x plus three.

Let’s test each ordered pair:

  • (5, 2):
    For the first inequality: 2 < 5 + 1 (2 < 6) is true.
    For the second inequality: 2 < 5 + 3 (2 < 8) is also true.
    –> Satisfies both.
  • (2, 1):
    For the first inequality: 1 < 2 + 1 (1 < 3) is true.
    For the second inequality: 1 < 2 + 3 (1 < 5) is also true.
    –> Satisfies both.
  • (3, 0):
    For the first inequality: 0 < 3 + 1 (0 < 4) is true.
    For the second inequality: 0 < 3 + 3 (0 < 6) is also true.
    –> Satisfies both.
  • (1, 3):
    For the first inequality: 3 < 1 + 1 (3 < 2) is false.
    –> Does not satisfy.

Therefore, the ordered pairs that satisfy both inequalities are (5, 2), (2, 1), and (3, 0).

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