Graphical solution of linear inequalities in two variables

Correct answer: C. Only on the boundary x+y=6

The governing concept is feasibility in a system of linear inequalities: a point belongs to the solution region only when it satisfies every inequality, and equality places it on the corresponding boundary. Substitute (x,y)=(4,2). We obtain x+2y=4+2(2)=8≤10, x+y=4+2=6≥6, and y=2≥1, so the point is feasible. Because x+y is exactly 6, the point lies on the boundary line x+y=6. It is not on x+2y=10, since the value is 8 rather than 10, and it is not on y=1, since y=2. Thus it lies on exactly one listed boundary, so option C is correct. Option A wrongly excludes equality, B ignores the satisfied inequalities, and D uses the wrong boundary.

Correct answer: C. (1, 0), (4, 0), (3, 2), (1, 3)

The governing concept is the graphical solution of simultaneous linear inequalities: the feasible region is the common part of all four half-planes, and its vertices occur at boundary intersections that satisfy every inequality. On y = 0, the restrictions give x ≥ 1, x ≤ 7, and x ≤ 4, so the relevant endpoints are (1, 0) and (4, 0). The two slanting boundaries intersect by solving x + 2y = 7 and 2x + y = 8, which gives (x, y) = (3, 2). On x = 1, the first slant boundary gives y = 3, while the second permits y = 6; hence (1, 3) is the limiting feasible point. Therefore option C is correct. Option A incorrectly uses (7, 0), which violates 2x + y ≤ 8; the other options contain points outside the feasible region.