Graphical solution of linear inequalities in two variables

Correct answer: D. (3x+y=6) dashed and (x-y=2) solid

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

Correct answer: A. The half-plane containing ((0,0))

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.

Correct answer: B. ((2,3)), ((4,1)), ((4,3))

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

Correct answer: B. Point ((3,1))

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

Correct answer: A. Boundary is solid and the side containing origin is shaded

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.

Correct answer: C. ( (3,3) )

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.

Correct answer: B. (y \le \frac{3x-6}{2})

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

Correct answer: B. The line (4x+y=10) is dotted

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

Correct answer: C. ( (4,2) )

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

Correct answer: C. ( (3,2) )

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

Correct answer: B. The side containing origin is accepted

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

Correct answer: D. It satisfies neither inequality

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

Correct answer: A. ( (3,2) )

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

Correct answer: C. Unbounded region

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

Correct answer: C. x + y ≤ 8

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.

Correct answer: A. y > (5x − 10)/2

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.

Correct answer: D. ( (2,3) )

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

Correct answer: A. ( (10,0) ) and ( (0,4) )

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

Correct answer: C. On the opposite side of the solution region

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.

Correct answer: B. ( (2,3) )

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

Correct answer: C. (x\ge0), (y\ge0), (x\le5), (y\le4)

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

Correct answer: A. (x<4)

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

Correct answer: B. ( (2,3) )

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.

Correct answer: C. ( (0,4) )

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

Correct answer: B. ( (4,3) )

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