Substitute the point (1,1) into each inequality.
Check if the inequality holds true for the point (1,1).
2x+1"> y > 2 x + 1 becomes 3"> 1 > 3 , which is false.
y < 2 x − 1 becomes 1 < 1 , which is false.
y g e q 2 x + 1 becomes 1 g e q 3 , which is false.
y ≤ 2 x − 1 becomes 1 ≤ 1 , which is true.
The inequality that holds true is y ≤ 2 x − 1 .
Explanation
Understanding the Problem We are given that the point ( 1 , 1 ) lies on Han's graph of y ≤ 2 1 x + 2 . We need to find a second inequality from the given options such that ( 1 , 1 ) is also a solution to that inequality.
Solution Plan We will substitute the point ( 1 , 1 ) into each of the given inequalities and check which inequality holds true.
Checking the Inequalities
Consider the inequality 2 x+1"> y > 2 x + 1 . Substituting x = 1 and y = 1 , we get 2(1) + 1"> 1 > 2 ( 1 ) + 1 , which simplifies to 3"> 1 > 3 . This is false.
Consider the inequality y < 2 x − 1 . Substituting x = 1 and y = 1 , we get 1 < 2 ( 1 ) − 1 , which simplifies to 1 < 1 . This is false.
Consider the inequality y g e q 2 x + 1 . Substituting x = 1 and y = 1 , we get 1 g e q 2 ( 1 ) + 1 , which simplifies to 1 g e q 3 . This is false.
Consider the inequality y ≤ 2 x − 1 . Substituting x = 1 and y = 1 , we get 1 ≤ 2 ( 1 ) − 1 , which simplifies to 1 ≤ 1 . This is true.
Final Answer Therefore, the inequality that includes the point ( 1 , 1 ) as a solution is y ≤ 2 x − 1 .
Examples
Imagine you're designing a simple game where a character can move within certain boundaries. The first inequality, y ≤ 2 1 x + 2 , sets an upper limit to the character's movement. Now, you want to add another rule, y ≤ 2 x − 1 , to further restrict the character's position. This ensures the character stays within a specific, smaller area on the game map, making the game more challenging or guiding the player through a particular path. Understanding how inequalities work together helps define these boundaries precisely.