Graphical solution of linear inequalities in two variables

Substituting \((0,1)\) gives \(2(0)-1=-1\), and \(-1\leq 0\) is true. Therefore, the half-plane containing \((0,1)\) is shaded. The boundary line \(2x-y=0\) is also included because the inequality contains \(\leq\). Exam tip: if a test point satisfies the inequality, shade the side containing that point.

The direct answer is option A, a solid line. The inequality is \(x+y\le6\). Its boundary is \(x+y=6\). Because the sign is \(\le\), points on the boundary satisfy the inequality exactly and must be included in the solution. An included boundary is drawn with a solid line. Option A is therefore correct. Option B, a dashed line, is used for strict inequalities such as \(<\) or \(>\), where boundary points are excluded. Option C, points only, does not show the complete boundary line. Option D, no line, is wrong because the boundary is needed to separate the solution region. The region would be selected using a test point, but the line itself remains solid. Memory cue: less-than-or-equal or greater-than-or-equal means solid; strict signs mean dashed.

The boundary line \(x=5\) is vertical. The inequality \(x\leq 5\) requires the x-coordinate to be 5 or less, so the solution region lies to the left of the line and includes the boundary. The right side represents \(x>5\). Exam tip: for a vertical line \(x=a\), shade left for \(x\leq a\) and right for \(x\geq a\).

The boundary line of a linear inequality is obtained by replacing its inequality sign with an equality sign. Therefore, changing x + 2y ≤ 10 to equality gives x + 2y = 10. This line separates the plane into the two half-planes x + 2y < 10 and x + 2y > 10. Since the original symbol is ≤, the points on the boundary itself satisfy the inequality and belong to the solution region; consequently, the line is normally drawn as a solid line. Option B describes only the strict side below the boundary, option C describes the opposite side, and option D excludes rather than identifies the boundary. Hence option A is correct.

Substituting the origin \((0,0)\) gives \(0+0\ge 9\), or \(0\ge 9\), which is false. Therefore, the half-plane containing the origin does not satisfy the inequality, so the opposite half-plane is shaded. Exam tip: For an inequality with \(\ge\) or \(\le\), the boundary line is included and should be drawn as a solid line.

For \((1,2)\), \(x+y=1+2=3\), and \(3\leq5\) is true. Hence, this point lies in the solution region. In contrast, the closest distractor \((3,3)\) gives \(x+y=6\), which is greater than 5. Exam tip: Substitute the coordinates of a point into the inequality to test whether it belongs to the solution region.

Substitution must make the expression strictly greater than (8). Here no equality is allowed in a strict inequality.

The governing concept is testing points in a two-variable linear inequality. A point belongs to the solution region only when its coordinates make the inequality true. Because the sign is strict, the expression must be greater than 8; equality is not enough. For A, 0 + 2(4) = 8, so it lies on the boundary and is excluded. For B, 5 + 2(1) = 7, and for C, 3 + 2(1) = 5; both values are less than 8. For D, 2 + 2(4) = 10, and 10 > 8. Thus D satisfies the inequality and is the only valid point. The strict sign also explains why the boundary point A is not accepted.

Substituting \((3,1)\) into the inequality gives \(2(3)+1=7\). Since \(7\leq 7\) is true, the point belongs to the solution region. It lies on the boundary line \(2x+y=7\), and the boundary is included because the inequality contains \(\leq\). Option C is incorrect because the point does not lie on either axis. Exam tip: Test a point by directly substituting its coordinates into the inequality.

With \(\leq\) or \(\geq\), equality is allowed. Hence the boundary line \(2x+y=9\) is included in the solution region and is drawn as a solid line. For \(<\) and \(>\), the boundary line is excluded and is usually drawn dashed. Exam tip: if the inequality sign contains an equality bar, draw a solid boundary line.

Option D uses the strict inequality \(<\). The points on the boundary line \(y=x+2\) are not part of its solution region, so the boundary is drawn as a dashed line. In contrast, \(\leq\) and \(\geq\) include the boundary and require a solid line. Exam tip: use a dashed boundary for \(<\) and \(>\).

For the origin, substitute \(x=0\) and \(y=0\). Then \(x+y\leq 1\) becomes \(0+0\leq 1\), i.e., \(0\leq 1\), which is true. Therefore, the origin lies in the solution region of the inequality. \(1\leq 0\) is false and does not result from the substitution. Exam tip: To check whether a point belongs to a solution region, substitute its coordinates in the inequality.

Substituting the origin \((0,0)\) gives \(x+y=0+0=0\). Since \(0\geq -3\) is true, the origin belongs to the solution region of the inequality. It is not only on the boundary, because the boundary line is \(x+y=-3\), whereas the sum at the origin is 0. Exam tip: To test whether a point lies in a solution region, substitute its coordinates into the inequality and check whether it is true.

Substituting the origin \((0,0)\) into \(x-y\leq 2\) gives \(0-0\leq 2\), i.e. \(0\leq 2\). This is true, so the origin lies in the half-plane that satisfies the inequality. Option B incorrectly calls this true statement false. Exam tip: shade the side of the boundary line for which a test point satisfies the inequality.

Substituting the origin \((0,0)\) gives \(0-0\geq 2\), i.e. \(0\geq 2\), which is false. Therefore, the origin is not in the solution region, so the side containing the origin is not shaded. The boundary line \(x-y=2\) still exists. Exam tip: if the test point makes the inequality true, choose its side; if false, choose the opposite side.

The inequality \(x+2y\leq 6\) includes equality. Hence, the points on the boundary line \(x+2y=6\) belong to the solution region, so the boundary is drawn as a solid line. A dashed line is used only for strict inequalities such as \(<\) or \(>\). Exam tip: draw a solid boundary whenever \(\leq\) or \(\geq\) appears.

The boundary line for x+2y>6 is x+2y=6. Since the inequality sign is >, points on this line are not included in the solution region. Therefore, the boundary line is drawn dashed. A solid line is used only for inequalities containing ≥ or ≤. Exam tip: First replace the inequality sign with = to obtain the boundary line, then use the sign to decide whether the line is solid or dashed.

The line \(y=4\) is horizontal. For every point on or below it, the \(y\)-coordinate is less than or equal to 4, so \(y \leq 4\) is correct. Although \(y<4\) would represent only the region strictly below the line, it would not include the line itself; \(\leq\) includes the boundary line. Exam tip: For the region below a horizontal line \(y=a\), use \(y\leq a\).

The line \(x=-2\) is vertical. Every point to its right has an \(x\)-coordinate greater than \(-2\), so the required solution region is \(x>-2\). The inequality \(x<-2\) represents the region to the left, whereas inequalities involving \(y\) represent horizontal regions. Exam tip: for the region to the right of \(x=a\), write \(x>a\).

For the point \((2,1)\), \(x+2y=2+2(1)=4\), so it satisfies the equation \(x+2y=4\) and lies on the line. For \((0,1)\), the left-hand side is \(2\), not \(4\), so it does not lie on the line. Exam tip: Substitute the coordinates in the line equation; a point lies on the line only when the equality is satisfied.

Since ‘>’ is strict, points on \(x+2y=6\) are excluded, so the boundary is dashed. In exams, test \((0,0)\) to choose the side; a solid line is used for \(\ge\) or \(\le\).

Substituting \((8,0)\) gives \(x+y=8+0=8\), so \(8\leq 8\) is true. Therefore, the point belongs to the solution region. Since equality holds, it lies on the boundary line \(x+y=8\). Option B is incorrect because the inequality is satisfied. Exam tip: For \(\leq\) or \(\geq\), points giving equality are included in the solution set.

Substituting the origin \((0,0)\) gives \(0\leq4\), so the required region lies on the origin side of \(x+y=4\). Since the boundary is included, the sign must be \(\leq\); \(<\) excludes the line. Exam tip: test the given point first.

To obtain the boundary line, write the inequality as \(2x+3y=12\). At the \(y\)-intercept, \(x=0\). Thus, \(3y=12\), so \(y=4\). Therefore, option 4 is correct. Setting \(y=0\) gives the \(x\)-intercept, not the \(y\)-intercept. Exam tip: always set \(x=0\) to find a line's \(y\)-intercept.

The inequality \(x+y\leq10\) describes all points whose coordinate sum is at most 10. To test a listed point, substitute its x- and y-coordinates and check whether the resulting inequality is true. For \((2,3)\\), the sum is 5, so it is inside the region. For \((5,5)\\), the sum is 10, so it lies on the boundary and is included. For \((0,9)\\), the sum is 9, so it is also inside.

For \((6,7)\\), the sum is 13, and \(13\leq10\) is false. Therefore this point is outside the solution region, making option C correct. The boundary line is \(x+y=10\), and because the sign includes equality, points on that line belong to the solution set.