Graphical solution of linear inequalities in two variables

A strict inequality (>) excludes the boundary, so its line is dashed. The inequality (\leq) includes the boundary, so its line is solid.

Substituting ((0,0)) gives (0\geq 6), which is false, so the half-plane containing the origin is rejected. In exams, choose a test point not lying on the boundary.

Boundary intersections give ((2,3)), ((4,1)), and ((4,3)). On a graph, keep only intersections satisfying all inequalities.

For ((3,1)), (2x+y=7), so it fails the strict inequality. This item needs a point strictly inside the region.

Since (0 \le 12) is true, the half-plane containing the origin is taken. In exams, check the boundary line first and then use a test point.

A point belongs to the solution region only if its coordinates make the inequality true. For each option, substitute its x- and y-values into \(x+2y>8\). At (2,3), the left side is 2+6=8, so it fails because the inequality is strict. At (5,1), the value is 7, so it fails. At (3,3), the value is 3+6=9, which is greater than 8. Thus choice C is correct.

The point (0,4) gives 0+8=8, which is also only on the boundary and is not included in a strict inequality. The boundary line is \(x+2y=8\); points satisfying the equality lie on that line, while the required region contains only points giving a value greater than 8. Therefore (3,3) is the only listed point in the solution region.

From (3x-2y \ge 6), we get (y \le \frac{3x-6}{2}). When dividing by a negative number, the inequality sign reverses.

The sign (<) is strict, so the boundary line is not included. Such a boundary is drawn dotted in the graph.

Solving (x+y=6) and (x+2y=8) gives ( (4,2) ). Vertices usually come from intersections of boundary lines.

At ( (3,2) ), (8<10) and (9<12), so it is strictly inside. A boundary point must make at least one inequality an equality.

Since (0\ge-5) is true, the side containing the origin is the solution region. A simple test point quickly decides the shading.

At ( (2,2) ), both left sides equal (6), which is less than (10). In a combined solution, all inequalities must be true together.

Solving the two equations gives (x=3) and (y=2). To find a vertex, write the related boundary lines as equalities.

In the first quadrant, the region above (x+y=5) extends without bound. In such questions, check both direction and axis restrictions.

The governing concept is redundancy in a system of linear inequalities. An inequality is redundant when every point satisfying the other restrictions automatically satisfies it, so removing it does not enlarge or change the feasible region. From x ≤ 4 and y ≤ 3, adding the upper bounds gives x + y ≤ 4 + 3 = 7. Thus every feasible point already satisfies x + y ≤ 7, which is stronger than x + y ≤ 8. The condition x + y ≤ 8 therefore eliminates no additional point and is redundant, making option C correct. The non-negativity conditions and the separate upper bounds still impose independent restrictions. The conclusion follows from the two upper bounds alone and does not require a complete graph.

The governing idea is that a linear inequality in two variables describes a half-plane whose boundary is obtained by replacing the inequality sign with equality. Starting with 5x − 2y < 10, subtract 5x from both sides to get −2y < 10 − 5x. Dividing by −2 reverses the inequality sign, so y > (10 − 5x)/(−2) = (5x − 10)/2. Therefore option A is correct. The boundary line is y = (5x − 10)/2, and because the original sign is strict, the boundary is not included and would be drawn dashed. Option B forgets to reverse the sign; C has the wrong direction and expression; D changes the constant incorrectly.

At ( (2,3) ), (-1\le2) and (5\ge4) are both true. In a common region, every given condition must be checked.

The boundary line is (2x+5y=20), giving intercepts (x=10) and (y=4). Drawing the line first is the base of graphical solution.

At the origin, (0>24) is false, so the side containing the origin is not the solution. If the test point fails, shade the other side.

Subtracting the two lines gives (2x=4), so (x=2) and (y=3). For vertices, solve the pair of linear equations.

The conditions (x\le5) and (y\le4) with the axes form a bounded rectangular region. A bounded region needs closing boundaries in all directions.

We need (x-1< -x+7), so (2x<8) and (x<4). For a region between two lines, compare the upper and lower bounds.

The direct answer is option B, (2,3). To solve this, first understand the word boundary. The boundary of the inequality 2x-y≤1 is the line obtained by replacing the inequality sign with equality: 2x-y=1. The strict inequality 2x-y<1 allows points only on one side of this line; it does not include the line itself. Check option B: put x=2 and y=3. Then 2x-y=2(2)-3=4-3=1. Therefore, (2,3) lies exactly on the boundary of 2x-y≤1, but it is not included in 2x-y<1 because 1<1 is false. Option A, (0,0), gives 2(0)-0=0, and 0<1 is true, so it is inside the strict solution, not merely a boundary point. Option B gives equality and is correct. Option C, (1,0), gives 2(1)-0=2, which is greater than 1, so it satisfies neither the required boundary equation nor the inequality. Option D, (3,4), gives 6-4=2, also greater than 1, so it is outside the solution. Remember: a line is included for ≤, but excluded for <. Equality is the exam clue for a boundary-only point.

Putting (x=0) gives (4y\le16), so (y\le4). To find an axis intercept limit, put the other variable equal to zero.

Solving (x+y=7) and (x-y=1) gives ( (4,3) ). The intersection of active boundaries gives a vertex.