Graphical solution of linear inequalities in two variables

For \((4,3)\), \(x+2y=4+2(3)=10\), so the point lies on the line \(x+2y=10\). Since the inequality uses \(\le\), every point on the boundary line is included in the solution region. For \((1,4)\), the value is \(9\), so it satisfies the inequality but does not lie on the given line. Exam tip: Include the boundary line whenever the sign is \(\le\) or \(\ge\).

The direct answer is option D: below the line, including the line itself. The boundary line is \(y=-3x+9\). The inequality says that the chosen point must have a y-value less than or equal to the y-value on the line at the same x. Less than means below; the symbol \(\le\) also includes equality, so the boundary line must be included. Because equality is included, draw the boundary as a solid line, then shade the lower side. Option A is wrong because above the line gives larger y-values, corresponding to \(y>-3x+9\). Option B is wrong because the boundary alone is only the equality \(y=-3x+9\), not all points satisfying the inequality. Option C is wrong because points on both sides do not all satisfy the condition. Option D correctly says below including the boundary. Test point cue: \((0,0)\) gives \(0\le9\), so the side containing the origin is shaded.

Subtracting \(x\) from both sides of \(2y+x\ge 14\) gives \(2y\ge 14-x\). Dividing both sides by the positive number \(2\) does not reverse the inequality sign, so \(y\ge \frac{14-x}{2}\). In option C, the numerator has been written with the sign reversed. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

The governing concept is finding the intercepts of the boundary line and combining them with the first-quadrant restrictions. Replace the inequality by its boundary equation 2x+y=12. On the x-axis, y=0, so 2x=12 and x=6; the x-intercept is (6,0). On the y-axis, x=0, so y=12; the y-intercept is (0,12). The first quadrant imposes x≥0 and y≥0, and the inequality selects the side containing the origin. Consequently, the feasible region is the triangle bounded by both axes and the line, with vertices (0,0), (6,0), and (0,12). Thus option C is correct. Option A swaps the intercept values, while B and D either include non-boundary points or use incorrect intercepts.

The point ((3,2)) satisfies all conditions because (3\ge 2), (2\ge 1), and (3+2\le 7). Check each condition separately.

The direct answer is option B: the line \(x+y=6\). A point satisfies both inequalities only if it is at most 6 and at least 6 at the same time. Therefore its value must be exactly 6: \(x+y=6\). Geometrically, \(x+y\le6\) is the half-plane on one side of the line, while \(x+y\ge6\) is the half-plane on the other side. Their common part is only the boundary line, and both include that boundary because the signs contain equality. Option A is wrong because points with \(x+y<6\) fail the second inequality. Option B is correct because every point on the line satisfies both. Option C is wrong because many points do satisfy both, for example \((0,6)\). Option D is wrong because the origin has sum 0, not 6. Memory cue: a quantity that is both no greater than and no smaller than a number must equal that number.

Solving (x+2y=8) and (3x-y=3) gives (x=2) and (y=3). Verify the intersection in both equations.

From (-2y\le 10-5x), reversing the sign gives (y\ge \frac{5x-10}{2}). The inequality sign changes when dividing by a negative.

To the right, (x)-values are greater than (3), and a solid line includes equality. Therefore (x\ge 3) is correct.

For every point above the line \(y=-2\), the value of \(y\) is greater than \(-2\). A dashed boundary means that points on the line are not included, so equality is excluded. Therefore, the inequality is \(y>-2\). The option \(y\geq -2\) would include the boundary line and would be shown with a solid line. Exam tip: shading above a horizontal line means larger \(y\)-values; shading below means smaller \(y\)-values.

The governing idea is to interpret the inequalities as half-planes. The conditions x≥0 and y≥0 restrict points to the first quadrant. The inequality x+y≥9 selects the side of the line x+y=9 that lies away from the origin, including the line itself. This region contains points such as (9,0), (0,9), (10,10), and infinitely many points farther northeast. There is no upper bound on x or y: if a point satisfies the conditions, sufficiently increasing either coordinate continues to satisfy them. Therefore the region extends indefinitely and is unbounded. Option D is correct. It is not a bounded triangle; the triangle near the origin corresponds to x+y≤9. It is not empty because (9,0) is a solution, and it is certainly more than a single line segment.

The constraints give \(0\le x\le 4\) and \(0\le y\le 3\). Therefore, the region is a rectangle with length 4 units parallel to the \(x\)-axis and width 3 units parallel to the \(y\)-axis. Its area is \(4\times 3=12\) square units. A value such as 10 may result from identifying the shape incorrectly. Exam tip: first write the minimum and maximum bounds of \(x\) and \(y\) to identify the region.

The governing concept is using the axis intercepts of a linear boundary to find the area of the feasible triangle. The boundary is 2x+3y=18. Setting y=0 gives 2x=18, so x=9 and the x-intercept is (9,0). Setting x=0 gives 3y=18, so y=6 and the y-intercept is (0,6). Together with the origin, the first-quadrant region is a right triangle with perpendicular legs 9 and 6. Its area is 1/2×base×height=1/2×9×6=27 square units. Therefore option A is correct. The values 18 and 36 result from missing or doubling the one-half factor, while 54 does not follow from the intercepts. The non-strict inequalities include the boundary but do not change the area.

The direct answer is option D: the vertex (1,2) is formed by x = 1 and y = 2. A vertex is commonly obtained where two boundary lines intersect. Check the point in x = 1: its x-coordinate is 1, so it lies on that boundary. Check it in y = 2: its y-coordinate is 2, so it also lies on that boundary. The third boundary gives x + y = 1 + 2 = 3, not 8, so the point is not on x + y = 8. Option A is wrong because (1,2) does not satisfy x + y = 8. Option B is wrong for the same reason. Option C is wrong because one line alone does not determine an ordinary intersection vertex. Option D is correct. Memory cue: test a vertex by substituting its coordinates into each boundary equation.

\(ax+by=c\) is the boundary line. Since the sign is \(<\), points on this line do not satisfy the inequality, so the solution is an open half-plane on one side. Exam tip: draw a dashed boundary for \(<\) or \(>\).

Dividing both sides of \(4x+2y\le 16\) by 2 gives \(2x+y\le 8\). To draw the graph, replace the inequality sign with an equality sign to obtain the boundary line. Therefore, the boundary line is \(2x+y=8\). Option B is the simplified inequality, not its boundary line. Exam tip: For a boundary line, always replace the inequality sign with \(=\).

The intersection of (x=2) and (y=3) is ((2,3)), and it satisfies (x+y\le 10). Therefore it is a vertex.

The symbol \(<\) represents a strict inequality, so points satisfying \(x+2y=6\) are excluded and the boundary is dashed. A solid line is used for \(\le\) or \(\ge\). Exam tip: decide the line style from the inequality sign first.

An intersection point must satisfy both line equations. Substituting \(y=2\) into \(2x+5y=20\) gives \(2x+5(2)=20\), or \(2x+10=20\). Thus \(2x=10\) and \(x=5\). Therefore, the intersection point is \((5,2)\). The point \((2,5)\) has \(y=5\), so it cannot lie on the line \(y=2\). Exam tip: Substitute the fixed variable from one equation into the other equation.

The direct answer is option C, the point \((2,4)\). A common region means that both inequalities must be true at the same point. For \((2,4)\), substitute \(x=2\) and \(y=4\): the first inequality becomes \(4\ge2+1=3\), which is true, and the second becomes \(4\le5\), which is also true. Therefore the point belongs to the intersection. Option A, \((5,5)\), fails because \(5\ge5+1\) means \(5\ge6\), which is false. Option B, \((3,3)\), satisfies \(3\ge4\) falsely, although it satisfies \(y\le5\); one failed condition is enough to reject it. Option C satisfies both conditions, so it is correct. Option D, \((4,6)\), satisfies \(6\ge5\), but fails \(6\le5\). Always test both inequalities, not just one. Memory cue: common region means AND, so every condition must pass.

The direct answer is option A: the point is in the solution region and on its boundary. Substitute \((x,y)=(1,4)\) into \(2x+y\ge6\): \(2(1)+4=2+4=6\). Since the result is exactly 6, the point lies on the boundary line \(2x+y=6\). The inequality uses \(\ge\), so equality is included; hence the point is also part of the solution region. Option A is correct. Option B is wrong because the point is not outside: it satisfies the inequality exactly. Option C is wrong because “but not on the boundary” contradicts the equality just calculated. Option D is wrong because the position can be determined directly by substitution. Remember that \(\ge\) and \(\le\) include the boundary, whereas strict signs \(>\) and \(<\) exclude it.

The intersection of (x=6) and (y=0) is ((6,0)), and it satisfies (x+2y\le 8). Hence it is a vertex of the region.

No (x) can be both less than or equal to (1) and greater than or equal to (4). Therefore there is no common solution.

The inequality \(2\le y\le 6\) means that the \(y\)-coordinate of every solution point lies from 2 to 6, including both boundaries. The lines \(y=2\) and \(y=6\) are horizontal, so the closed strip between them is the solution region. The line \(y=2\) alone is only the lower boundary, not the complete solution set. Exam tip: inequalities involving only \(y\) have horizontal boundary lines.

The sign \(<\) is strict, so the boundary \(x-y=3\) is not included and must be dashed. Testing \((0,0)\) gives \(0<3\), which is true; therefore, shade the side containing that point. Exam tip: use a dashed line for \(<\) or \(>\).