Final answer:
To find the solution set of the inequality y² − 1y ≤ x+12, each ordered pair's values must be substituted into the inequality to check if they make a true statement. (5, -2), (4, 2), and (3, 1) satisfy the inequality, which means option (c) or (d) could be correct if no other differences exist between them.
Step-by-step explanation:
To determine which ordered pairs are in the solution set of the system of linear inequalities y² − 1y ≤ x+12, we need to plug in each x and y value from the ordered pairs into the inequality and see if it results in a true statement.
- For the ordered pair (5, -2), substituting x=5 and y=-2 into the inequality gives us (-2)² - (-2) ≤ 5 + 12, which simplifies to 4 + 2 ≤ 17. This is a true statement, so (5, -2) is part of the solution set.
- Next, for the ordered pair (3, 1), substituting x=3 and y=1 into the inequality gives us (1)² - 1 ≤ 3 + 12, which simplifies to 0 ≤ 15. This is also true, meaning (3, 1) is in the solution set.
- Lastly, for (4, 2), substituting x=4 and y=2 into the inequality gives us (2)² - 2 ≤ 4 + 12, which simplifies to 2 ≤ 16. This is true, indicating that (4, 2) is in the solution set.
Only the ordered pairs that make the inequality a true statement are in the solution set. Based on the test above, options (a), (c), and (d) all have (5, -2) and (4, 2) which are in the solution set, but (c) and (d) include (3, 1) which also satisfies the inequality, making option (c) or (d) correct if they are otherwise identical. Without additional context on the differences between option (c) and (d), we cannot determine which one is correct.