Graphical solution of linear inequalities in two variables

The direct answer is option A, \(x+y\le5\). Start with \(2x+2y\le10\). Every term and both sides can be divided by the positive number 2, so the inequality sign does not change: \((2x+2y)/2\le10/2\), giving \(x+y\le5\). Option A is correct. Option B reverses the inequality without reason; division by a positive number never reverses it. Option C changes addition into subtraction, which is not a valid simplification. Option D leaves an incorrect coefficient on x and is not obtained by dividing every term by 2. If the divisor were negative, the sign would reverse, but 2 is positive. Memory cue: divide every term, and reverse an inequality only when multiplying or dividing by a negative number.

To find the \(y\)-intercept, put \(x=0\). Then \(3(0)+y=15\), so \(y=15\). Hence, the line meets the \(y\)-axis at \((0,15)\), and its \(y\)-intercept is \(15\). The value \(5\) is the \(x\)-intercept, obtained by putting \(y=0\). Exam tip: for a \(y\)-intercept, always set \(x=0\).

At every point on the x-axis, y=0. Substituting this in 5x+y=20 gives 5x=20, so x=4; hence the line intersects the x-axis at (4, 0). Therefore, option C is correct. Exam tip: to find the x-intercept, set y equal to zero; 5 is only the coefficient of x, not the intercept.

Substituting \((0,0)\) gives \(0+4(0)\geq 12\), i.e. \(0\geq 12\), which is false. Hence, the half-plane containing the origin is not shaded; the half-plane opposite to the origin is shaded. The boundary line \(x+4y=12\) is included because the inequality has \(\geq\). Exam tip: if a test point makes the inequality false, shade the side opposite to that point.

Substituting the origin \((0,0)\) gives \(2(0)+3(0)\geq 6\), or \(0\geq 6\), which is false. Hence, points on the side containing the origin do not satisfy the inequality, so that side is not the solution region. The solution lies on the opposite side of the line \(2x+3y=6\). Exam tip: Use the origin as a test point to choose the shaded side when the boundary line does not pass through it.

The region satisfying both inequalities is their intersection. In exams, identify the common shaded part.

In the first quadrant, the line (x+y=4) forms a triangle with the axes. This is a very common exam pattern.

Substituting \((0,3)\) gives \(0+2(3)=6\). This equals the right-hand side, so the point lies on the boundary line \(x+2y=6\). Since the inequality uses \(\leq\), points on the boundary are included in the solution set. Exam tip: Boundary points are included for \(\leq\) and \(\geq\), but not for \(<\) and \(>\).

Substituting (0,3) gives x+2y=0+2(3)=6. Hence, the point lies on the boundary line x+2y=6. Since the inequality uses <, a point giving equality is not included in the solution region. Option C is incorrect because it would require the value to be less than 6. Exam tip: Test a point by substituting its coordinates into the inequality and comparing the result.

To identify the slope, first replace the inequality by its boundary equation: y = 5 − x. Comparing it with slope-intercept form y = mx + c gives m = −1. Since the coefficient of x is negative, the line has a negative slope; as x increases by 1, y decreases by 1. Thus option B is correct. The inequality sign determines which side of the boundary is the solution region, but it does not change the slope of the boundary line. A positive slope would rise to the right, a zero slope would produce a horizontal line, and an undefined slope would produce a vertical line. This boundary instead falls from left to right.

In \(y\geq 2x-3\), the equality sign is included. Hence, the points on the boundary line \(y=2x-3\) are part of the solution region, so it is drawn as a solid line. A dotted line is used only for strict inequalities such as \(>\) or \(<\). Exam tip: use a solid line for \(\leq\) and \(\geq\).

From (x-y\leq 2), (-y\leq 2-x), so (y\geq x-2). The sign changes when dividing by a negative number.

Substituting \((1,2)\) gives \(3(1)+2=5\). Since \(5\leq 9\) is true, \((1,2)\) belongs to the solution set of the inequality. It is not on the boundary line, because points on the boundary must satisfy \(3x+y=9\). Exam tip: To test a point, substitute its coordinates into the inequality and check whether the result is true.

Substituting \((3,2)\) gives \(3(3)+2=11\). Since \(11\leq 9\) is false, the point is not in the solution set of the inequality. It is not on the boundary either, because a boundary point must satisfy \(3x+y=9\). Exam tip: Test a point by substituting its \(x\)- and \(y\)-coordinates into the inequality and checking whether the statement is true.

The direct answer is option B: the left half-plane including the vertical axis. The inequality \(x\leq0\) talks only about the x-coordinate. A point has x-coordinate 0 on the y-axis, x-coordinate less than 0 to the left of the y-axis, and x-coordinate greater than 0 to the right. Therefore all points on the left side and all points on the y-axis satisfy the inequality. The boundary is the vertical line \(x=0\), which is the y-axis. Because the sign is \(\leq\), the boundary is included, so it is drawn as a solid line. Option A, the right half-plane, corresponds mainly to \(x\geq0\), not \(x\leq0\). Option B is correct because it includes points with negative x and points with x equal to zero. Option C, the upper half-plane, is controlled by y-values and would relate to an inequality such as \(y\geq0\), not x. Option D, the lower half-plane, relates to \(y\leq0\), not x. For example, \((-2,5)\) satisfies the inequality because \(-2\leq0\), while \((3,-4)\) does not because \(3\leq0\) is false. The y-coordinate can be anything; only x matters. Memory cue: x controls left-right, y controls up-down; the equality sign includes the boundary.

In the Cartesian plane, y is the vertical coordinate, measured from the horizontal x-axis. The inequality y ≥ 0 means that the y-coordinate may be positive or exactly zero. Consequently, every point above the x-axis is included, and every point on the x-axis is also included because it has y = 0. The boundary is therefore included, so the region is the upper half-plane together with the horizontal axis. It is not only the origin, because infinitely many points satisfy the condition. The lower half-plane has y < 0, while left and right regions are described by conditions on x. Thus option A is correct.

The boundary line of the given inequality is \(x+5y=10\). For the \(x\)-intercept, put \(y=0\); this gives \(x=10\), so the point is \((10,0)\). For the \(y\)-intercept, put \(x=0\); then \(5y=10\), so \(y=2\) and the point is \((0,2)\). Option B incorrectly interchanges the intercept values. Exam tip: To find intercepts of a line, set one variable to zero at a time.

Dividing both sides of the inequality by the positive number \(2\) gives \(6x/2+2y/2\geq 12/2\), i.e., \(3x+y\geq 6\). The inequality sign does not change when dividing by a positive number. Option B incorrectly reverses the sign. Exam tip: Check whether the number used for multiplication or division is negative; the inequality sign reverses only for a negative number.

The boundary line of 2x + y ≤ 4 is 2x + y = 4. Its intercepts are (2, 0), obtained by putting y = 0, and (0, 4), obtained by putting x = 0. Testing the origin gives 2(0) + 0 = 0 ≤ 4, so the required side is the side containing the origin, below the line. The restrictions x ≥ 0 and y ≥ 0 limit the points to the first quadrant and include the coordinate axes. Hence the common solution is the closed triangular region with vertices (0, 0), (2, 0), and (0, 4). Therefore option A is correct. The other options either use a wrong quadrant, the wrong side of the line, or an unnecessarily large region.

A point lies in the solution region if substituting its coordinates into x+2y>4 gives a value greater than 4. For option D, 1+2(3)=7, and 7>4, so (1,3) lies in the solution region. For option B, 1+2(1)=3, which is not greater than 4. Exam tip: For the strict inequality >, points on the boundary line are not included.

The boundary line of the inequality is \(4x+3y=12\). For the x-intercept, put \(y=0\): \(4x=12\), so \(x=3\), giving \((3,0)\). For the y-intercept, put \(x=0\): \(3y=12\), so \(y=4\), giving \((0,4)\). In option B, the intercept values have been interchanged. Exam tip: to find an intercept on one axis, set the other variable equal to zero.

From (-3y\geq 6-x), we get (y\leq \frac{x-6}{3}). The inequality sign changes when dividing by a negative number.

Substituting the origin gives \(0\leq \frac{0}{2}+2\), i.e., \(0\leq 2\), which is true. Therefore, the half-plane on the side containing the origin is the solution region. Since the sign is \(\leq\), the boundary line is also included; it is not removed. Exam tip: if a test point makes an inequality true, shade the side containing that point.

Substituting
(4,2) gives
\(2(4)+2=10\). Hence, the point lies on the boundary line \(2x+y=10\). Since the inequality uses \(>\), equality is not included, so this point is not part of the solution region. Option B is incorrect because it includes equality. Exam tip: For strict inequalities, \(>\) and \(<\), points on the boundary line are excluded.

The direct answer is option B: the region to the right of \(x=1\) and above \(y=2\), including both boundary lines. The inequality \(x\geq1\) means that a point's x-coordinate is 1 or greater, so the point lies on or to the right of the vertical line \(x=1\). The inequality \(y\geq2\) means that its y-coordinate is 2 or greater, so the point lies on or above the horizontal line \(y=2\). To satisfy both inequalities together, take the common overlapping region: right and above. Since both signs include equality, the lines \(x=1\) and \(y=2\) are included and should be solid boundaries. Option A says left and below, which would represent \(x\leq1\) and \(y\leq2\), the opposite directions. Option B is correct because every point there has both required coordinates. Option C, only \((1,2)\), is too small; many points such as \((2,3)\), \((5,2)\), and \((1,6)\) satisfy both inequalities. Option D, the whole plane, includes points such as \((0,0)\), which fail both conditions. A quick test point in the correct region is \((2,3)\): \(2\geq1\) and \(3\geq2\). Memory cue: x inequalities move left-right, y inequalities move up-down; combine them by taking the overlap.