Graphical solution of linear inequalities in two variables

The governing concept is interval notation. Square brackets indicate that the corresponding endpoint is included in the set. Thus [1,4] contains every real number from 1 through 4, including both 1 and 4. In inequality form this is 1≤x≤4, so option A is correct. Option B represents the open interval (1,4), excluding both endpoints. Option C includes 1 but excludes 4, corresponding to [1,4). Option D excludes 1 but includes 4, corresponding to (1,4]. A number-line check gives closed dots at both 1 and 4, which confirms the two “less than or equal to” signs. Therefore A precisely translates the given interval.

Substituting ((0,0)) gives (0\leq 4), so the point lies in the solution region. In exams, test a convenient point first.

Since (>) does not include equality, the boundary line is dashed. Remember dashed lines for (<) and (>) in exams.

In (y\leq 3), the values of (y) are (3) or less. In exams, identify above or below for horizontal lines.

In (x> -2), all (x)-values are greater than (-2), so shading is to the right. For inequalities in (x), check left or right in exams.

Substituting ((0,0)) gives (0\geq 6), which is false, so the origin side is not the solution. In exams, shade the opposite side when the test is false.

The boundary line is made by replacing the inequality with equality, so it is (x+y=5), and it is dashed because of (<). In exams, first convert the inequality into an equation.

To identify the solution region of an inequality, substitute a convenient test point into the inequality. The point belongs to the region if the resulting statement is true. The boundary line is \(x+2y=8\), and the inequality includes all points for which \(x+2y\) is at most 8. Because the symbol is \(\leq\), points on the boundary are included as well.

Test the options. For \((0,0)\), we get \(0+2(0)=0\), and \(0\leq8\) is true. Thus option B lies in the solution region. The point \((5,3)\) gives 11, \((9,1)\) gives 11, and \((4,4)\) gives 12; all are greater than 8. Hence those points are outside the required region.

The symbol (\geq) includes equality, so the boundary line is part of the solution. Draw a solid line for (\leq) and (\geq) in exams.

Substituting ((0,0)) gives (0\leq 9), so the side containing the origin is selected. Use a test point to decide the shading direction in exams.

Equality is not included in (x-y<2), so the boundary line (x-y=2) is dashed. In exams, do not include the line for (<).

Substituting ((2,1)) gives (2+1=3), and (3\leq 3) is true. Boundary points are included for (\leq) in exams.

The governing concept is testing whether an ordered pair belongs to the solution region of a two-variable inequality. Substitute x=1 and y=3 into y<2x. The left side becomes 3, while the right side becomes 2(1)=2, so the required statement is 3<2. This statement is false. Therefore (1,3) does not satisfy the inequality and cannot lie in the solution region. It is actually on the side of the boundary line y=2x where y is greater than 2x. The boundary itself would be y=2x, and the strict symbol < excludes the boundary from the solution set. Hence option B is correct. Option A reverses the comparison, option C wrongly claims equality, and option D is false because the point’s x-coordinate is 1 rather than 0.

In (x\leq 4), (x)-values are (4) or less, so the region is to the left. In exams, read (x\leq a) as the left side.

Substituting ((0,0)) gives (0\geq -2), which is true, so shade the origin side. If the test point is true, take that side in exams.

Substituting ((0,0)) gives (0\geq 7), which is false, so the solution is the side opposite the origin. Shade the other side when the test is false in exams.

The boundary line is (2x+3y=12), giving (x=6) when (y=0) and (y=4) when (x=0). The intercept method draws the line quickly in exams.

In the boundary line (4x+y=8), putting (y=0) gives (x=2). For the (x)-intercept, put (y=0) in exams.

In the boundary line (x+5y=10), putting (x=0) gives (y=2). For the (y)-intercept, put (x=0) in exams.

Substituting ((1,1)) gives (1\leq 2), which is true, so the point is in the solution region. Calculate the right side carefully in exams.

In (y> -2), (y)-values are greater than (-2), so shade above the line. Do not include the boundary line for (>) in exams.

In (x\geq 0), (x) is non-negative, so the (y)-axis and the right side are included. Remember (x=0) is the (y)-axis in exams.

In (y\leq 0), (y) is zero or negative, so the (x)-axis and the lower side are included. Remember (y=0) is the (x)-axis in exams.

Substituting ((3,0)) gives (3>2), which is true, so it lies in the solution region. In exams, check the candidate point directly.

Substituting ((2,0)) gives (4\geq 4), true with equality. Boundary points are included when the symbol is (\geq).