Graphical solution of linear inequalities in two variables

Since the sign is (\geq), the boundary is included, and (y) is greater than or equal to the line. So the region above the line is shaded.

At ((1,4)), (x+2y=9) and (2x+y=6). It satisfies all conditions and lies in the region including the boundary.

At ((6,4)), both (x+y=10) and (x-y=2) hold as equalities. So it is the valid intersection of both boundary lines.

Inside the rectangle (0\leq x\leq 6), (0\leq y\leq 4), the part above (x+2y=10) remains. Hence the vertices are ((2,4)), ((6,2)), and ((6,4)).

Above the (x)-axis means (y\geq 0), and below or on the line means (y\leq 2x-5). Convert the words directly into inequalities.

The direct answer is option C, 7. A linear expression over a bounded polygon reaches its maximum at a vertex, so first identify the important corner points. On the x-axis, y=0; the inequality 3x+y≤21 gives x≤7, so (7,0) is feasible. At this point x-y=7-0=7. The other relevant corners are (0,0), (0,7), and the intersection of 3x+y=21 with x+3y=21, which is (5.25,5.25); their values of x-y are 0, -7, and 0. Thus 7 is the maximum. Option A, 0, occurs at several balanced points but is not largest. Option B, 5, is feasible in value but can be improved to 7. Option C, 7, is achieved at (7,0). Option D, 21, incorrectly treats the right side of an inequality as the value of x-y. Exam cue: to maximize x-y, prefer large x and small y, while still checking all constraints.

The first two lines are parallel and strict inequalities create an open strip. The condition (x>0) restricts it to the right half-plane, but the region remains unbounded.

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

Solving the two boundary equations gives \(x=\frac{16}{5}\) and \(y=\frac{27}{5}\). It is a good method to test the intersection in all inequalities.

The intersection of the two slant lines is \(\left(\frac{21}{5},\frac{18}{5}\right)\). Use valid intercepts on the axes to choose all polygon corners.

Since \(x\geq4\) and \(y\geq3\), the minimum possible value of \(x+y\) is \(7\), attained at \((4,3)\). Therefore, \(x+y\leq m\) can be satisfied only when \(m\) is at least \(7\), i.e. \(m\geq7\). Although \(m>12\) also gives a non-empty region, it is not the required condition because \(m=7\) already gives the solution point \((4,3)\). Exam tip: For parameter-based inequalities, first find the extreme possible value of the relevant expression.

Checking the corners gives (10) at ((0,5)), (12) at ((4,0)), and (16) at \(\left(\frac{14}{5},\frac{18}{5}\right)\). A linear expression attains its maximum at a corner.

The vertices are ((2,0)), ((6,0)), ((3,6)), and ((2,7)). Using the shoelace method or splitting into parts gives area (20).

The slant boundaries intersect at \(\left(\frac{14}{5},\frac{18}{5}\right)\). Use valid intercepts on the axes to list all corners.

Direct answer: Option C, \(\frac{25}{2}\) square units. The inequalities \(x\geq2\) and \(y\geq1\) place the region to the right of \(x=2\) and above \(y=1\). The inequality \(x+y\leq8\) places it below the line \(x+y=8\). The boundary intersections are: \(x=2,y=1\), giving \((2,1)\); with \(y=1\), \(x+1=8\), giving \((7,1)\); with \(x=2\), \(2+y=8\), giving \((2,6)\). These three points form a right triangle. Its horizontal base is \(7-2=5\), and its vertical height is \(6-1=5\). Therefore area \(=\frac12\times5\times5=\frac{25}{2}\) square units. Option C is correct. Option A, 10, would require a smaller triangle. Option B, 12, is not obtained from the base-height calculation. Option D, 15, is also too large; it ignores the one-half factor used for a triangle or uses an incorrect dimension. The boundary lines are included because all inequalities are non-strict, but boundaries do not change area. Exam cue: shift the origin to \((2,1)\); the remaining triangle has legs 5 and 5.

The direct answer is option B, (3,3). A point is a solution only if it satisfies every inequality, including non-negativity. Test option A, (1,2): 2x+y=4, which is less than 7, so it fails the first condition. Test option B, (3,3): 2x+y=6+3=9≥7, and x+2y=3+6=9≤12; both coordinates are non-negative, so it works. Test option C, (5,4): 2x+y=14≥7, but x+2y=13>12, so it fails the second condition. Test option D, (0,6): 2x+y=6<7, so it fails the first condition, even though x+2y=12. Therefore only option B is a solution. The safe method is substitution: calculate both left sides separately and then check the signs and coordinate restrictions. Memory cue: satisfying one inequality is not enough; a solution must pass all four tests.

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

Direct answer: Option C, no solution. The first inequality requires \(y\leq x+1\), while the second requires \(y>x+4\). For every value of x, \(x+4\) is exactly 3 greater than \(x+1\). Thus the second condition demands that y be above a higher line, while the first demands that y be at or below a lower line. These requirements cannot overlap. For example, if \(y>x+4\), then automatically \(y>x+1\), which contradicts \(y\leq x+1\). Option C is correct. Option A, a closed strip, is wrong because the two lines are parallel but the required sides are separated with no common strip. Option B, an unbounded region, is wrong because unboundedness matters only if at least one common point exists; here there are none. Option D, only the line \(y=x+1\), is wrong because points on that line fail the strict condition \(y>x+4\). The strict sign also excludes boundaries, but even replacing it with a non-strict sign would still leave no overlap because the lines differ by 3. Memory cue: compare the two right sides before drawing; the higher lower-bound line makes the intersection impossible.

Since \(x\geq 3\) and \(y\geq 4\), the minimum possible value of \(x+y\) occurs at \((3,4)\) and is \(3+4=7\). This point satisfies \(x+y\leq m\) only when \(m\geq 7\). For \(m=7\), the solution region consists of the single point \((3,4)\), so \(m>7\) is not necessary. Exam tip: for parameter-based inequalities, first find the minimum or maximum value of the relevant expression under the given bounds.

The direct answer is option D, 18. The feasible region is in the first quadrant and is bounded by x+y≤6 and 2x+y≤8. Its vertices are (0,0), (4,0), the intersection (2,4), and (0,6). At these points, 2x+3y equals 0, 8, 16, and 18 respectively. Therefore the maximum is 18 at (0,6). Option A, 12, is not the largest value; it can be exceeded by the feasible point (0,6). Option B, 14, is likewise smaller than 18. Option C, 16, is the value at (2,4), but another vertex gives 18. Option D, 18, is correct. The corner-point method works because a linear objective over a closed polygon reaches its extreme value at a corner. Do not choose a value merely because it resembles a boundary number; calculate the objective at every vertex.

The two slant boundaries meet at ((2,2)). Checking corner-like boundary points gives the minimum (x+y=4).

The condition (-4\leq x-y\leq 2) gives a strip between two parallel lines. Equality includes both boundaries.

The two lines are parallel and give ( -x+2<y<-x+6). Because the inequalities are strict, neither boundary is included.

Together the two inequalities force equality (x+3y=9). Opposite inequalities with the same boundary often give only the line.

A point satisfies the system only when all three conditions are true. For (2,3), the first condition gives \(x+y=2+3=5\), so \(x+y\leq5\) is true and the point lies exactly on the boundary \(x+y=5\). The second condition is \(x=2\geq1\), which is true. The third is \(y=3\leq4\), also true.

Thus (2,3) is inside the solution set, but it is on one of its boundary lines. It is not on the boundaries \(x=1\) or \(y=4\), because its coordinates are 2 and 3. Therefore the precise description is “in the solution region and on the boundary \(x+y=5\),” which is choice A. The equality is allowed because the first inequality uses \(\leq\).