Graphical solution of linear inequalities in two variables

Substituting ((1,1)) gives (1+2=3<8), and the dashed boundary requires (<). Exam tip: Check both the test point and boundary type.

Together, the two inequalities force (x+2y=4). Exam tip: Opposite inequalities with the same boundary may reduce to a line.

The vertices are ((1,2)), ((1,9)), and ((8,2)), giving perpendicular lengths (7) and (7). Exam tip: For shifted axis bounds, read base and height carefully.

Dividing (-3y<6-2x) by a negative number gives (y>\frac{2}{3}x-2). Exam tip: Keep a dashed boundary for strict inequalities.

At ((2,4)), (4\cdot2+4=12), so it lies on the boundary. Exam tip: For boundary checking, replace the inequality by its equation.

The second inequality is (x+y>4), which cannot overlap with (x+y\leq 3). Exam tip: Simplify proportional boundary lines first.

From (x-1\leq 3), we get (x\leq 4), and all boundaries are included. Exam tip: A hidden upper bound may come from comparing inequalities.

The intercepts are ((16,0)) and ((0,4)), whose distances from the origin are (16) and (4). Exam tip: You can also compare squared distances.

The two boundary lines meet at ((2,2)), which is the nearest corner of the outer unbounded region. Exam tip: In two (\geq) regions, the intersection point is often crucial.

The inequality (x-y\geq 2) gives (y\leq x-2), and ((0,0)) makes it false. Exam tip: Pay close attention to the sign of (y).

The vertices are ((-1,-2)), ((-1,6)), and ((7,-2)), with perpendicular sides (8) and (8). Exam tip: Find intercepts carefully when bounds are negative.

Both have the same side, but (\leq) includes the boundary line. Exam tip: The solid versus dashed line difference earns marks in graph questions.

Counting possible (y)-values for (x=0) to (8) gives a total of (33) points. Exam tip: For such questions, write the maximum (y) for each (x).

Solving the two equations gives (x=2), (y=1). Exam tip: Find boundary intersections before drawing the common region.

Inside the square (0\leq x\leq 4), (0\leq y\leq 4), the condition (x+y\geq 6) leaves the upper-right triangle. Exam tip: Draw the larger region first and then cut it.

On the (x)-axis, (y=0), and below it requires (y<0). Exam tip: Treat axis lines as boundary lines too.

The intercepts are ((9,0)) and ((0,6)), so the area is (\frac{1}{2}\cdot9\cdot6=27). Exam tip: Wrong intercepts can make the whole graph wrong.

For every point, (x-2y) is either at most (4) or greater than (4). Exam tip: In or questions, think of the union of regions.

For ((2,3)), both (3>2) and (2+3<6) are true. Exam tip: Immediately reject options with equality on open boundaries.

From (y=0), we get ((0,0)) and ((6,0)), and (2x=-x+6) gives ((2,4)). Exam tip: Find axis and slant-line intersections separately.

Between the two parallel lines and the axes, a bounded quadrilateral is formed in the first quadrant. Exam tip: View the strip between parallel lines after cutting it by the axes.

Dividing by a negative gives (y\leq 2x+3), and the boundary is included because the inequality is non-strict. Exam tip: After reversing the sign, check boundary type separately.

The given triangle lies in the first quadrant below the line (x+y=5). Exam tip: Identify axis conditions from vertices first.

Both inequalities are strict, so both boundary lines are excluded. Exam tip: Decide boundary type separately for each inequality.

The inequality x ≥ 1 keeps the region to the right of the vertical line x = 1, while y ≥ 2 keeps it above the horizontal line y = 2. The condition x + y ≤ 7 keeps it below the line x + y = 7. The boundary intersections are (1, 2), (5, 2), and (1, 6): substituting y = 2 gives x = 5, and substituting x = 1 gives y = 6. These points form a right triangle. Its horizontal base is 5 − 1 = 4 units and its vertical height is 6 − 2 = 4 units. Therefore area = 1/2 × 4 × 4 = 8 square units. Hence option B is correct; the other choices do not equal the triangle’s area.