Graphical solution of linear inequalities in two variables

Substituting \((2,k)\) into the first inequality gives \(2+3k<11\), so \(k<3\). The second inequality gives \(k-2\geq1\), so \(k\geq3\). No number can be both less than \(3\) and at least \(3\); hence, no such \(k\) exists. Although \(k=3\) is a close distractor, it fails the strict condition \(k<3\). Exam tip: substitute the point into each inequality separately and then check the intersection of the resulting conditions.

Substituting \((k,2)\) into the first inequality gives \(2k+2\leq10\), so \(k\leq4\). The second inequality gives \(k-2>1\), so \(k>3\). The common range is \(3<k\leq4\), hence option B is correct. At \(k=3\), the second strict inequality is not satisfied. Exam tip: take the intersection of the ranges obtained from all inequalities.

At the x-intercept, y=0. Substituting the given point (3,0) into the boundary line ax+2y=12 gives 3a+2(0)=12, so 3a=12. Hence, a=4. If a were 6, the x-intercept would be 12/6=2, not 3. Exam tip: for an x-intercept, always set y=0.

If (p>0), the (x)-intercept (\frac{6}{p}) is finite. If (p\leq 0), the region is not bounded in the (x)-direction in the first quadrant.

At ((2,3)), (x+2y=8) and (x-y=-1). Thus it lies on the boundary and also satisfies the second inequality.

The two inequalities describe the points lying between two parallel boundary lines. The first condition, \\(x+2y\\leq8\\), includes the line \\(x+2y=8\\), because equality is allowed. The second condition, \\(x+2y>4\\), excludes the line \\(x+2y=4\\), because equality is not allowed. Their common solution is therefore the region between these lines.

Combining the conditions gives \\(4<x+2y\\leq8\\). On a graph, the boundary for 8 is drawn as a solid line, while the boundary for 4 is drawn as a dashed line. The region between them is shaded, so it is a strip closed on one side and open on the other. Hence, option D is correct; it is neither empty nor only a boundary line.

The rectangle ([2,6]\times[1,5]) is cut by the line (x+y=8). The corners are ((2,1)), ((6,1)), ((6,2)), ((3,5)), and ((2,5)).

The direct answer is B, 6. The feasible region lies in the first quadrant and satisfies two upper bounds. To maximize the linear expression \(x-y\), inspect the corner points of the region. On the x-axis, y=0, the constraints give \(x\leq10\) and \(2x\leq12\), so the farthest feasible x-axis point is \((6,0)\). At this point, \(x-y=6-0=6\). Other relevant corners give smaller values: \((0,0)\) gives 0, \((0,5)\) gives -5, and the intersection of the two boundary lines gives \((14/3,8/3)\), where \(x-y=2\). Thus 6 is the maximum. Option A, 5, is less than the attainable value. Option B, 6, is correct. Option C, 4, and option D, 7, are not the maximum; 7 is not even feasible because x cannot exceed 6 when y is nonnegative. In linear programming, checking all vertices is the reliable method.

Among boundary corner-like points, (2x+y=6) at ((0,6)) is the smallest. Even in an unbounded region, a minimum often occurs at a boundary corner.

The direct answer is option C, 24 square units. The feasible region lies in the first quadrant. On the x-axis, y=0 gives 2x≤12, so the x-intercept is (6,0). On the y-axis, x=0 gives 2y≤12, so the y-intercept is (0,6). The two boundary lines intersect by solving x+2y=12 and 2x+y=12; subtracting gives x-y=0, so x=y, and then 3x=12 gives (4,4). The vertices are (0,0), (6,0), (4,4), and (0,6). Split the shape into two right triangles: the triangle with vertices (0,0),(6,0),(4,4) has area 12, and the triangle with vertices (0,0),(4,4),(0,6) also has area 12; total area is 24. Option A, 18, and option B, 20, are too small. Option C, 24, matches the calculated area. Option D, 30, overestimates the region. Include the boundary because the inequalities are non-strict.

At ((4,2)), (2x+y=10) and (x+3y=10). Choose the final option only after substituting the point in all inequalities.