Graphical solution of linear inequalities in two variables

Adding \(y\) to both sides of \(3x-y\le 0\) gives \(3x\le y\), or \(y\ge 3x\). Hence, the solution is the region on and above the line \(y=3x\), including the line. The region \(y\le 3x\) lies below the line, so it is the closest but incorrect distractor. Exam tip: Adding or subtracting the same quantity does not reverse an inequality sign.

The region is a (2\times 3) rectangle, so its area is (6). Recognize bounded rectangles quickly.

The intercepts are (4) and (4), so area is (\frac{1}{2}\times 4\times 4=8). Remember the intercept triangle formula.

The half-plane containing the true test point is the solution side. This is the quickest graphical check.

At origin the value is (0), so the origin side is (\le 6) and the opposite side is (\ge 6). Use equality signs when boundary is included.

The boundary is (x-2y=6), giving (x=6) and (y=-3). Draw the equality line before the inequality.

On the x-axis, the y-coordinate is zero, so set y=0 in the main inequality. Then 4x+0≤8 gives 4x≤8, and division by positive 4 gives x≤2. The non-negativity condition x≥0 restricts the possible values to 0≤x≤2, so the maximum x-coordinate on the x-axis is 2. It is attained at the boundary point (2,0), which satisfies 4(2)+0=8. Therefore option A is correct. Option B would result from incorrectly using 2x≤8. Option C treats the coefficient 4 as if it were absent. Option D does not reach the x-intercept. The y≥0 condition is automatically satisfied on the x-axis at y=0, while the boundary equation determines the intercept precisely.

At every point on the \(y\)-axis, \(x=0\). Substituting \(x=0\) in \(x+4y\le 12\) gives \(4y\le 12\), so \(y\le 3\). Since \(y\ge 0\), the highest point of the feasible region on the \(y\)-axis is \((0,3)\). Hence, the maximum value of \(y\) is 3. For option 4, \(4(4)=16\), which exceeds 12. Exam tip: to find a value on an axis, set the other variable equal to 0.

A feasible point must satisfy all four inequalities simultaneously. Test option A, (5,0): x+y=5≤6 is true, x-y=5-0=5≤2 is false, and although non-negativity holds, one failed condition is enough to make the point infeasible. Thus option A is correct. For (2,2), the values are 4≤6, 0≤2, and both coordinates are non-negative, so it is feasible. For (1,4), the values are 5≤6, -3≤2, with x and y non-negative, so it is feasible. For (3,2), the values are 5≤6, 1≤2, and both coordinates are non-negative, so it is also feasible. The key concept is intersection: the feasible region is the common part of all four half-planes, not the region satisfying only the first inequality.

((2,1)) is on the boundary because (2(2)+1=5), but a strict inequality excludes the boundary. Do not include equal points for (<).

Pairwise intersections of (x=2), (y=1), and (x+y=6) give these vertices. Solve boundary lines pairwise in exams.

Below the line gives (y\le), and including the boundary includes equality. Remember (y\le) for below the line.

The region lies inside the square (0\le x\le 4), (0\le y\le 4) and above (x+y=4). Draw the larger rectangle first in exams.

Substituting ((6,0)) gives (12\le 10), which is false, so the point is outside. Use substitution for point location.

The boundary line of an inequality is obtained by replacing the inequality sign with equality. Thus the boundary of x+y≥3 is the line x+y=3. Test the options in this equation: for (1,2), 1+2=3, so it lies exactly on the boundary. For (1,1), the sum is 2, so it is not on the line. For (3,1), the sum is 4, and for (0,0), the sum is 0; neither equals 3. Notice that the inequality describes a half-plane, while the boundary is only the line separating the accepted and rejected regions. Because the inequality is inclusive, the boundary belongs to the solution region, but the question asks specifically for a point on that line. Therefore option A is the only correct answer.

A point lies in the common region only if it satisfies both inequalities at the same time. For option A, (x,y)=(2,3): the first condition gives 3≥0, which is true, and the second gives 3≤2(2)=4, also true. For option B, y≤2x becomes 3≤2, which is false. For option C, it becomes 1≤2(−1)=−2, which is false, even though its y-value is positive. For option D, y≥0 is true, but y≤2x becomes 2≤0, which is false. Therefore only option A belongs to the intersection of the two half-planes. The governing concept is simultaneous satisfaction: passing one inequality is insufficient when a common region is requested.

The graph of a system is the common shaded region. In exams, intersect the half-planes of all inequalities.

A solution region is empty when the inequalities cannot be satisfied simultaneously by even one point. In option C, a point would have to satisfy y > 4 and y < 1 at the same time. No real number can be greater than 4 and less than 1, so the two boundary half-planes do not overlap and the intersection is empty. Hence option C is correct. Options A and B describe nonempty horizontal or vertical bands, since values such as y = 3 and x = 2 satisfy them. Option D also has solutions, for example (x, y) = (0, 0), which satisfies both conditions.

The governing idea is graphical solution of simultaneous linear inequalities. The conditions x ≥ 0 and y ≥ 0 restrict the solution to the first quadrant, including both coordinate axes. The boundary x + y = 8 has intercepts (8, 0) and (0, 8). Because the sign is ≤, the region on the side containing the origin is selected, and the boundary line is included. Consequently, the feasible region has vertices (0, 0), (8, 0), and (0, 8), forming a closed triangle. It is bounded, so it cannot be an unbounded strip. Although the axes meet at right angles, the figure has only three sides and is not a square or rectangle. Therefore option A is correct.

A point is a solution only if it satisfies every inequality at the same time. Test option B, (3, 2): x + 2y = 3 + 2(2) = 7 ≤ 10, and 2x − y = 2(3) − 2 = 4 ≥ 3. Both coordinates are non-negative, so all four conditions hold. The distractors fail as follows: (1, 4) gives 2x − y = −2, which is too small; (0, 5) gives −5, also too small; and (2, 5) gives x + 2y = 12, which exceeds 10. Thus (3, 2) is the only feasible point, so option B is correct. Direct substitution is sufficient and avoids any ambiguity from a sketch.

The standard origin test determines which half-plane satisfies a linear inequality. Substitute the origin (0, 0) into y > 2x − 1: 0 > 2(0) − 1, which becomes 0 > −1. This statement is true, so the origin lies in the solution region. It is not on the boundary, because the boundary equation is y = 2x − 1; when x = 0, the boundary point has y = −1, not y = 0. Also, the strict sign > excludes the boundary itself. Therefore option A is correct. The test is useful because a linear boundary divides the plane into two half-planes, and checking one point identifies the required side without needing to shade the entire graph.

To graph a linear inequality in two variables, first draw its boundary line and then test a point that is not on the line. The origin is convenient here. Substituting (0, 0) into 3x − 2y ≤ 6 gives 0 − 0 ≤ 6, or 0 ≤ 6, which is true. Therefore the entire half-plane containing the origin satisfies the inequality and must be selected. Because the sign is ≤ rather than <, the boundary line 3x − 2y = 6 is included and should be drawn as a solid line. The opposite half-plane contains points that fail the inequality; selecting only the boundary or both sides would not describe the solution set.

Answer: (2, 4), so option C is correct. A solution of a system of inequalities must satisfy every inequality at the same time. Test each point directly. For A, (0,2), x - 2y = 0 - 4 = -4, so the first inequality is true, but x + y = 2, and 2 ≥ 3 is false. For B, (4,1), x - 2y = 4 - 2 = 2, and 2 ≤ -4 is false. For C, (2,4), x - 2y = 2 - 8 = -6, so -6 ≤ -4; x + y = 6, so 6 ≥ 3; and y = 4 ≤ 5. All three conditions hold. For D, (-3,6), the value y=6 violates y≤5, even though checking another inequality is unnecessary once one condition fails. Therefore C is the only solution. Memory cue: for simultaneous inequalities, one failed condition rejects the point; every condition must pass.

To classify the point, substitute x = 9 and y = 0 into every inequality. We obtain x + y = 9, so x + y ≤ 9 is true with equality; equality means the point lies on the boundary line x + y = 9. Also, x + 2y = 9 + 0 = 9 ≥ 8, and the non-negativity conditions are satisfied because x = 9 ≥ 0 and y = 0 ≥ 0. In fact, y = 0 places the point on the x-axis as well, which is another boundary of the feasible region. Thus (9, 0) belongs to the solution region and lies on its boundary, rather than in its interior. Therefore option A is correct; options C and D incorrectly claim that valid inequalities fail.

A point in the open solution region must satisfy both strict inequalities. Strict signs also mean that a point on either boundary line is excluded. For option B, (2, 2), the first inequality gives 2 < −2 + 5 = 3, which is true, and the second gives 2 > 2 − 1 = 1, also true. Hence (2, 2) lies in the required intersection. Option A, (3, 2), fails because 2 > −3 + 5 = 2 is false; equality would not satisfy the strict condition. Option C, (1, 4), lies on the first boundary because 4 = −1 + 5. Option D, (4, 0), fails the second inequality since 0 is not greater than 3. Therefore option B is the only valid answer.