Graphical solution of linear inequalities in two variables

For a boundary line, replace the inequality sign by equality. In exams, draw the line first and then choose the region.

The boundary line for \(3x-2y\ge 12\) is \(3x-2y=12\). Since \(\ge\) includes equality, every point on the boundary is part of the solution region, so the line is drawn solid. A dashed line is used only for strict inequalities such as \(>\) or \(<\). Exam tip: use a solid line for \(\le\) and \(\ge\), and a dashed line for \(<\) and \(>\).

Since the sign is \(>\), the boundary \(2x+3y=6\) is not included, so it is dashed. Testing the origin gives \(0>6\), which is false; hence shade the opposite half-plane. Exam tip: decide the boundary style before choosing the region.

Substituting ((1,4)) gives (1+8=9), which is greater than (8). In point testing, place both coordinates carefully.

The x-intercept is the point where the line meets the x-axis, so \(y=0\) there. Substituting \(y=0\) in \(x+3y=6\) gives \(x=6\). Therefore, the x-intercept is \(6\). The value \(3\) is related to the y-intercept, not the x-intercept. Exam tip: To find an x-intercept, always put \(y=0\).

To find the \(y\)-intercept, set \(x=0\). Substituting in \(4x+y=8\) gives \(4(0)+y=8\), so \(y=8\). Hence, the line meets the \(y\)-axis at \((0,8)\), and \(8\) is correct. \(4\) is only the coefficient of \(x\), not an intercept. Exam tip: always put \(x=0\) to find the \(y\)-intercept.

To graph y ≥ 2x + 1, first draw the boundary line y = 2x + 1. The inequality says that y is greater than or equal to the value given by the line for the same x-coordinate. Therefore the solution consists of the half-plane above the line, and because equality is allowed, the boundary itself is included and is drawn as a solid line. A point test also confirms this: (0, 2) satisfies 2 ≥ 1 and lies above the line, whereas (0, 0) does not satisfy 0 ≥ 1. Thus option B is correct; option A represents the opposite inequality, option C omits valid points, and option D includes points that do not satisfy the condition.

Both coordinates are non-negative, so the region is the first quadrant including axes. Always identify non-negativity conditions first.

The inequality \(x\le 4\) means that the \(x\)-coordinate of every point is 4 or less. The boundary \(x=4\) is a vertical line, and it is included because the inequality contains \(\le\). Therefore, the required region is the half-plane to the left of the line, including the line. The region on the right represents \(x>4\). Exam tip: an inequality involving \(x\) has a vertical boundary line.

In (y>3), (y) is greater than (3) and the boundary is excluded. Shade above the dashed horizontal line.

Adding the equations gives (2x=8), so (x=4) and (y=2). Intersection points form corners of a feasible region.

From the first line, y=10-2x. Substituting this in x+2y=8 gives x+2(10-2x)=8, so -3x=-12 and x=4. Then y=10-2(4)=2. Hence, the intersection point is (4, 2). For example, (2, 4) does not satisfy the first equation because 2(2)+4=8, not 10. Exam tip: Verify an intersection point by substituting it into both linear equations.

The axes and the line (x+2y=10) form a closed triangle. First-quadrant conditions can make the region bounded.

The intersection of (x=1) and (y=2) is ((1,2)), and it satisfies (x+y\le 7). To find vertices, intersect boundary lines pairwise.

The conditions x ≥ 0 and y ≥ 0 restrict the solution to the first quadrant. The inequality x + y ≥ 4 selects the side on or above the line x + y = 4. This region contains points such as (4, 0), (0, 4), (5, 0), and (0, 10), and it continues indefinitely as x or y increases. For example, every point of the form (t, 4) with t ≥ 0 satisfies all three inequalities, so there is no finite boundary enclosing the region. Hence the solution region is unbounded, making option B correct. It is not empty and is not a single point; it is also not bounded because it extends without limit.

The governing concept is the intersection of solution sets of simultaneous linear inequalities. A point is in the feasible region only if it satisfies every condition: both coordinates must be non-negative and their sum must not exceed 3. Point A, (0,0), satisfies x ≥ 0, y ≥ 0, and 0 + 0 = 0 ≤ 3. Point B, (1,1), also satisfies the coordinate conditions and has sum 2 ≤ 3. Point C, (2,2), has non-negative coordinates, but its sum is 2 + 2 = 4, which is greater than 3; it violates the third inequality. Point D, (3,0), has sum 3 and is allowed because the boundary is included by ≤. Therefore option C is outside the region.

If the test point gives false, the solution region is not on that side. Hence choose the opposite half-plane.

For the origin test, put \(x=0\) and \(y=0\). The inequality \(x-2y\ge 6\) becomes \(0\ge 6\), which is false. Therefore, the origin does not lie in the solution half-plane, so the region on the side of the line \(x-2y=6\) opposite to the origin is selected. Option A uses the same inequality but incorrectly calls it true. Exam tip: after substituting a test point, always check whether the resulting inequality is true or false.

The x-intercept is found by putting y=0 in the equation: 2x=20, so x=10 and the point is (10,0). The y-intercept is found by putting x=0: 5y=20, so y=4 and the point is (0,4). Hence option A gives both intercepts. The other options interchange the values or fail to satisfy the original equation, so they cannot represent the line’s axis intercepts.

The line intercepts are ((8,0)) and ((0,6)), and the axes meet at ((0,0)). These are the triangle vertices in the first quadrant.

(x\le 5) gives the left side and (y\le 3) gives the lower side. The common region is the overlap of both conditions.

The governing distinction is between strict and non-strict inequalities. The solution x + 2y > 8 contains only points whose left-hand side is strictly greater than 8. Every point on the boundary line x + 2y = 8 gives equality, not a value greater than 8, so no boundary point satisfies the strict inequality. Consequently the boundary is excluded and would be drawn as a dashed line in a graph. Both listed points, (8, 0) and (0, 4), lie on that boundary because 8 + 2(0) = 8 and 0 + 2(4) = 8; neither is included. Therefore option B is correct, while option A and the two point-specific options incorrectly include equality cases.

The symbol \(\ge\) means “greater than or equal to.” Hence, every point satisfying \(2x+3y=18\) also satisfies \(2x+3y\ge 18\). Therefore, the boundary line belongs to the solution region and is drawn as a solid line. With \(>\), the boundary would not be included and would be drawn dashed. Exam tip: use a solid boundary line for \(\ge\) or \(\le\).

At the point of intersection, both equations must be satisfied. Substituting x=0 into x+4y=12 gives 4y=12, so y=3. Hence, the intersection is (0, 3). The point (3, 0) is the x-intercept, whereas x=0 gives a point on the y-axis. Exam tip: To find the y-intercept of a line, put x=0.

The line y=0 represents the x-axis. Substituting y=0 in 5x+2y=20 gives 5x=20, so x=4. Hence, the intersection point is (4, 0). The point (0, 10) is the y-intercept of the line, not a point on the x-axis. Exam tip: To find an x-intercept, put y=0.