Graphical solution of linear inequalities in two variables

Because the sign is (>), the boundary is excluded, and the origin does not satisfy the inequality. So shading is opposite to the origin.

Substituting \((2,k)\) into the first inequality gives \(2+k\leq 7\), so \(k\leq 5\). From the second inequality, \(2(2)-k<3\Rightarrow 4-k<3\Rightarrow -k<-1\), hence \(k>1\). Combining both conditions gives \(1<k\leq 5\). Option B incorrectly includes \(k=1\); at \(k=1\), \(2x-y=3\), which does not satisfy \(<3\). Exam tip: reverse the inequality sign when multiplying or dividing by a negative number.

Take pairwise intersections of the three boundary lines and keep valid points. The valid corners are ((3,1)), ((7,1)), and ((3,5)).

The direct answer is option A: a closed strip between two parallel lines. Rewrite the inequalities together: y≤x+4 and y≥x-2, so x-2≤y≤x+4. For every fixed x, y must lie between the lower value x-2 and the upper value x+4. The boundary lines y=x+4 and y=x-2 have the same slope, 1, but different intercepts, 4 and -2. Therefore they are distinct parallel lines and never meet. Because both signs include equality, both boundary lines are part of the solution. Hence the common region is the closed band or strip between them, extending infinitely in both directions. Option A is correct. Option B, the region outside both lines, is wrong because a point must lie below the upper line and above the lower line simultaneously, not outside both. Option C, only one line, is wrong because every point in the band between the lines is also included, not just the boundaries. Option D, empty region, is wrong because there are many solutions; for example, (0,0) satisfies 0≤4 and 0≥-2. The word closed is important: ≤ and ≥ include the boundary. If one sign were strict, the corresponding boundary would be excluded and the strip would not be closed on that side. Remember: same slope means parallel lines; between lower and upper bounds gives a strip.

The boundary line of the inequality is \(ax+y=4\). At the x-intercept, \(y=0\). Substituting the given point \((2,0)\) gives \(2a+0=4\), so \(2a=4\) and \(a=2\). If \(a=1\), the x-intercept would be \((4,0)\), not \((2,0)\). Exam tip: to find an x-intercept, always put \(y=0\).

From the rectangle (0\leq x\leq 4), (0\leq y\leq 3), the part above (x+y=5) remains. Its vertices are ((2,3)), ((4,1)), and ((4,3)).

Below or on the line means (\leq), and the first quadrant needs (x\geq 0), (y\geq 0). The boundary line is solid.

The condition (-3\leq x-y\leq 1) gives a strip between two parallel lines. Equality includes both boundaries.

The feasible region lies in the first quadrant and below both boundary lines. Its intercept-based vertices are \((0,0)\), \((6,0)\), and \((0,4)\). Solving \(x+4y=16\) and \(3x+2y=18\) gives their intersection as \((4,3)\). Therefore, the region has 4 corner points. Although \((5,1)\) satisfies both inequalities, it is not an intersection of active boundary lines, so it is not a corner. Exam tip: check axis intersections and intersections of all active boundary lines to identify vertices.

The point ((0,4)) lies on (2x+y=4), but checking the feasible polygon shows it is a valid corner, so this option set is inconsistent.

Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). While finding corners, keep only feasible intersections.

The vertices are ((2,1)), \(\left(\frac{15}{2},1\right)\), and \((2,\frac{14}{3})\). Check parallel distances carefully before using triangle area.

The same expression (2x-y) cannot be at most (4) and greater than (8) at the same time. Check contradictions before drawing the graph.

In a strict inequality, the boundary is excluded, so the line is dashed. The origin satisfies the condition, so shade its side.

The (\geq) inequalities give the upper-side region in the first quadrant. Boundaries are included and the region extends infinitely.

Substituting \((k,3)\) into the first inequality gives \(k+2(3)\leq 11\), so \(k\leq 5\). The second inequality gives \(3k-3>6\), so \(3k>9\) and hence \(k>3\). Combining both conditions gives \(3<k\leq 5\). Option B incorrectly includes \(k=3\), which does not satisfy the second inequality. Exam tip: solve each inequality separately and take their common interval.

The key corner is the intersection ((3,4)) of the two slant lines. Checking (x+y) at the corners gives the maximum (7).

At ((2,3)), (2x+5y=19) and (x-y=-1), so it cannot be rejected. For such questions, test all conditions carefully.

The region is a bounded triangle, but the boundary (x+y=6) is not included. In mixed inequalities, check each boundary separately.

The point ((2,4)) satisfies all conditions and is the intersection of two boundaries. To identify vertices, first test boundary intersections.

Together the two inequalities force equality (x+2y=12). Opposite inequalities with the same boundary often give the boundary line.

Inside the square (0\leq x\leq 5), (0\leq y\leq 5), the part above (x+y=7) remains. Its vertices are ((2,5)), ((5,2)), and ((5,5)).

The two boundary lines have the same slope, 3, but different intercepts, 2 and -4, so they are distinct parallel lines. Combining the inequalities gives 3x-4≤y≤3x+2. Thus y must be at least the lower line and at most the upper line. The points satisfying both conditions lie in the strip between the two parallel lines. Since both inequalities use ≤ or ≥, equality is permitted, so both boundary lines belong to the solution set; the strip is closed. Therefore option A is correct. The regions above both or below both satisfy only one side of the compound condition, and the lines do not conflict, so the solution is not empty.

For the same boundary, (\leq) and (>) give no common point. In such questions, first check logical contradiction.

Solving the two equations gives (x=\frac{44}{7}) and (y=\frac{24}{7}). Always test the intersection in all inequalities afterward.