Graphical solution of linear inequalities in two variables

At the origin, (2x-y=0), which is less than (4); to exclude it, we need (2x-y>4). In a strict region, the boundary is not included.

The intersection is \(\left(\frac{8}{3},\frac{8}{3}\right)\), where \(x+y=\frac{16}{3}\) is maximum. A linear expression attains its extreme value at a vertex.

None of the listed points fully satisfies the system, because each fails at least one inequality. Always verify all inequalities before choosing an axial point.

The inequality becomes (y\ge \frac{x-4}{2}), and ((2,3)) satisfies it. Decide above or below after rearranging the sign correctly.

Solving \(x+4y=20\) and \(3x+y=18\) gives \(\left(\frac{52}{11},\frac{42}{11}\right)\). Do not avoid fractions when finding exact vertices.

At the origin, (ax+by=0), and (0\le c) is true. A test point is the most reliable way to choose the sign.

Solving (x+y=7) and (x-y=3) gives ((5,2)). A common vertex comes from the intersection of two boundary lines.

The first-quadrant region (x+y\ge 5) extends outward infinitely. A closed polygon needs bounds in every direction.

Each inequality gives a half-plane; intersections of half-planes are convex. Thus, the segment joining any two solution points lies in the region. Exam tip: the feasible region need not be bounded.

The feasible vertices are ((0,0)), ((8,0)), ((6,4)), and ((0,7)); the maximum is (26) at ((6,4)). None of the listed options matches the correct value.

((3,0)) lies on the boundary (3x+y=9), and (>) does not include the boundary. Boundary points are not solutions for strict inequalities.

The vertices are ((2,2)), ((8,2)), and ((2,5)); the area is (\frac{1}{2}\times 6\times 3=9). So the correct value should be (9).

At ((2,2)), (4<5) and (6>4), so it is inside and not on a boundary. An interior point does not satisfy any boundary as equality.

At the (y)-intercept, (x=0), (y=3), so (3k=12) and (k=4). In intercept questions, set the other variable to zero.

At ((1,1)), (2<6), while at ((5,5)), (10>6), so the correct strict half-plane is (x+y<6). Use test points to decide direction.

From (x\ge 6) and (y\ge 3), (x+2y\ge 12), which contradicts (x+2y\le 10). Checking the minimum possible value quickly detects an empty region.

The intercepts are ((15,0)) and ((0,10)), so the area is (\frac{1}{2}\times 15\times 10=75). Finding intercepts first is the fastest method.

For ((4,2)), (10\le 9) is false, and no listed option satisfies all conditions. Verify every option carefully against all constraints.

At (x=0), the conditions give (y\le 10) and (3y\le 18), so the maximum is (y=6). To maximize a variable, look for the tightest bound in that direction.

Solving \(x+2y=16\) and \(2x+y=14\) together gives \(\left(\frac{22}{3},\frac{13}{3}\right)\). In a graph, always check key vertices from boundary intersections.

The three half-planes form a closed triangle. In exams, first find the intersection points of boundary lines.

Both inequalities select the upper sides of the lines, and in the first quadrant the region extends infinitely. Since (\geq) is used, boundaries are included.

For parallel lines with the same slope, (2x+1) is always greater than (2x-3). Hence no (y) can satisfy both conditions.

Direct answer: Option B, the empty set. The first inequality says \(x+y\geq4\), so the sum must be 4 or more. The second says \(x+y<4\), so the same sum must be strictly less than 4. No number can be both at least 4 and less than 4. Therefore no point \((x,y)\), even one satisfying \(x\geq0\), can satisfy all the inequalities. Option B is correct. Option A, the line \(x+y=4\), is not included because the strict inequality \(x+y<4\) excludes the boundary, and the first inequality alone cannot rescue it. Option C, the first quadrant, contains many points with sums below 4 and many with sums above 4, but it does not satisfy both contradictory conditions. Option D, the region above the line, is also not a solution because one inequality requires one side of the line while the other requires the opposite side, with no overlap. The condition \(x\geq0\) becomes irrelevant after the contradiction is found. Memory cue: \(S\geq4\) and \(S<4\) together always give no solution.

Solving the two equations gives \(x=\frac{18}{5}\) and \(y=\frac{14}{5}\). This is the inner corner on the graph.