Which point lies on the boundary of (2x-y\le1) but not in the solution of (2x-y<1)?

Correct answer: ( (2,3) )

The direct answer is option B, (2,3). To solve this, first understand the word boundary. The boundary of the inequality 2x-y≤1 is the line obtained by replacing the inequality sign with equality: 2x-y=1. The strict inequality 2x-y<1 allows points only on one side of this line; it does not include the line itself. Check option B: put x=2 and y=3. Then 2x-y=2(2)-3=4-3=1. Therefore, (2,3) lies exactly on the boundary of 2x-y≤1, but it is not included in 2x-y<1 because 1<1 is false. Option A, (0,0), gives 2(0)-0=0, and 0<1 is true, so it is inside the strict solution, not merely a boundary point. Option B gives equality and is correct. Option C, (1,0), gives 2(1)-0=2, which is greater than 1, so it satisfies neither the required boundary equation nor the inequality. Option D, (3,4), gives 6-4=2, also greater than 1, so it is outside the solution. Remember: a line is included for ≤, but excluded for <. Equality is the exam clue for a boundary-only point.