Graphical solution of linear inequalities in two variables

For the point \((2,5)\), substitute \(x=2\) and \(y=5\). Since \(5\geq2\) is true, the point belongs to the solution region. As \(5>2\), it is not on the boundary line \(y=x\); it lies above it. Exam tip: To test a point, substitute its \(x\)- and \(y\)-coordinates directly into the inequality.

In (-y\leq 4-2x), dividing by (-1) reverses the sign. In exams, always reverse the sign when multiplying or dividing by a negative number.

The graphical rule for a linear inequality is to replace its inequality sign by an equality sign to obtain the boundary line. Therefore, the boundary of x − 2y < 6 is x − 2y = 6, making option A correct. Because the original sign is the strict inequality <, points lying on the boundary are not included in the solution region, so the line is usually drawn dashed. That drawing convention does not change the boundary equation. Option B is still an inequality, option C reverses the comparison and describes another region, and option D represents a different condition altogether. Hence only option A gives the required boundary line.

Substituting \((2,1)\) gives \(1\leq -2+3=1\), which is true. Hence, the point lies in the solution region and is on the boundary line. For \((1,4)\), \(4\leq 2\) is false, so it is not a solution. Exam tip: Substitute the coordinates of a point into the inequality to test whether it belongs to the region.

Check the inequality at \((0,0)\):
\[x+y>1\Rightarrow 0+0>1\Rightarrow 0>1,\]
which is false. So \((0,0)\) is not included in the solution region of \(x+y>1\).

The governing concept is the point-testing method for a linear inequality. To determine the position of a point, substitute its coordinates into the left-hand expression and compare the result with the right-hand side. For (2,2), x+2y=2+2(2)=2+4=6. Thus the inequality becomes 6≤6, which is true with equality. The point therefore belongs to the solution set and lies exactly on the boundary line x+2y=6, rather than strictly inside either half-plane. Because the inequality is non-strict, the boundary is included in the solution region. Hence option B is correct. An interior point would make x+2y<6, whereas an exterior point would give x+2y>6. Option D is unnecessary because direct substitution determines the position.

The governing rule is that a point belongs to the solution region of 4x+y≤16 exactly when substitution makes the inequality true. Test every option: for (5,0), 4x+y=20, so 20≤16 is false. For (0,20), the value is 20, again false. For (3,3), the value is 4(3)+3=12+3=15, and 15≤16 is true; therefore this point lies in the required half-plane. For (4,2), the value is 16+2=18, so it is also outside. Since only one option satisfies the inequality, option C is the unambiguous answer. The boundary itself would be represented by 4x+y=16, but the selected point is strictly inside the solution side because its value is less than 16.

The relevant concept is checking a point against both the inequality and its boundary equation. Substituting the origin into x+2y gives 0+2(0)=0. Hence the inequality becomes 0≤0, which is true, so the origin is included in the solution set. At the same time, equality holds, and the corresponding boundary line is x+2y=0. Therefore (0,0) lies on that boundary, not strictly in the interior of the selected half-plane. Option A correctly states both facts. Option B is wrong because the inequality is satisfied. Option C is wrong because an interior point would produce a strict result, x+2y<0, rather than equality. Option D is also wrong because substitution completely determines the position.

A point belongs to the common solution region only when it satisfies every inequality in the system. Test option A, (3,1): its coordinate sum is 3+1=4, and 4≤5 is true; also 3≥1 is true. Thus (3,1) satisfies both conditions. For option B, (4,3), the first inequality fails because 4+3=7>5. For option C, (1,4), the sum condition is true, but x≥y is false because 1<4. For option D, (2,4), both the sum condition 6≤5 and the comparison 2≥4 fail. Therefore option A is the only unambiguous answer. Graphically, it lies in the intersection of the half-planes below or on x+y=5 and on the side x≥y.

The governing method is substitution of a test point into the inequality. For the origin, x=0 and y=0, so x-y=0-0=0. The required condition is 0≥2, which is false because zero is less than two. Consequently, the origin is outside the half-plane represented by x-y≥2, so option A is correct. Option B is false because 0-0 equals 0, not 2. Option C is an invalid general claim; the origin is included only when it actually satisfies the given inequality. Option D is irrelevant and also incorrect: the boundary equation x-y=2 can be written as y=x-2, which is a sloping line, not a vertical line. Since the inequality is non-strict, points on the boundary would be included, but the origin is not on or in the correct side.

To determine whether a point is in a linear-inequality region, substitute its coordinates into the complete inequality, not merely into part of the expression. For (x,y)=(2,3), x+2y=2+2(3)=2+6=8. Since 8≤8 is true, the point is included. It lies exactly on the boundary line x+2y=8, and the symbol ≤ includes equality; therefore option A is correct. Option B reverses the numerical result and incorrectly claims that 8 is greater than 8. Option C uses x+y instead of the required weighted expression x+2y. Option D compares the coordinates, which is unrelated to the stated inequality. A graph would show the point on the solid boundary, not outside the feasible region.

The governing algebraic rule is to isolate y while preserving the inequality direction. Starting with x+3y<9, subtract x from both sides: 3y<9-x. Then divide by the positive number 3, so the inequality sign does not reverse: y<(9-x)/3=3-x/3. Therefore option A is correct. Option B reverses the sign without justification; a reversal would occur only when dividing or multiplying by a negative number. Option C fails to divide both terms on the right by 3 and also changes the coefficient incorrectly. Option D has both an incorrect sign pattern and an incorrect constant. This rearranged form is useful for graphing: the boundary is y=3-x/3, and because the original sign is strict, the boundary is drawn dashed and the region below it is selected.

For a linear boundary written in slope-intercept form y=mx+c, the coefficient m of x is the slope and c is the y-intercept. Replacing the strict inequality with equality gives the boundary line y=-(1/2)x+4. Comparing this with y=mx+c shows m=-1/2, so option A is correct. The inequality sign tells us which side of the line is shaded and, because it is strict, the boundary is dashed; it does not alter the slope. Option B, 4, is the y-intercept, not the slope. Option C loses the negative sign, which would change the line’s direction. Option D incorrectly treats the constant term as a negative slope. Thus the line decreases as x increases: an increase of 2 in x produces a decrease of 1 in y, confirming the slope -1/2.

The line x = 3 is vertical, and every point on this line has x-coordinate exactly 3. Any point to its right has a larger x-coordinate, so the required half-plane is represented by x > 3. The strict inequality is used because the phrase “to the right” excludes the boundary line itself. If the boundary were included, the correct inequality would be x ≥ 3. Thus option C is correct. Option A, x < 3, describes the half-plane to the left of the line. Option B describes only the boundary, not a two-dimensional half-plane. Option D, y > 3, describes points above a horizontal line y = 3 and does not determine a region relative to the vertical line x = 3. Therefore only option C matches both the direction and the boundary.

The boundary y = −2 is a horizontal line. Every point on it has y-coordinate equal to −2, while points above it have y-coordinates greater than −2. Therefore the required half-plane is y > −2, so option C is correct. The inequality is strict because “above the line” excludes the boundary itself; if the boundary were included, the expression would be y ≥ −2. Option B, y < −2, represents the region below the line. Options A and D concern the vertical boundary x = −2, describing regions to its right and left respectively, so they cannot represent a region above y = −2. The key distinction is that vertical position is determined by the y-coordinate, not by x.

The equations x = 2 and x = 7 are two distinct vertical, parallel lines. The equations y = 1 and y = 5 are two distinct horizontal, parallel lines. Their intersections produce the four vertices (2,1), (2,5), (7,1), and (7,5). The horizontal distance is 7 - 2 = 5 units and the vertical distance is 5 - 1 = 4 units. Thus the bounded region has four right angles and opposite sides equal and parallel: it is a rectangle. It is not a square because its side lengths are 5 and 4, not equal. Therefore option B is correct.