Graphical solution of linear inequalities in two variables

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

The direct answer is option A, \(x+y=6\). A boundary line separates the shaded solution region from the non-solution region. To obtain it, replace the inequality sign by equality: \(x+y\le 6\) becomes \(x+y=6\). The line passes through \((6,0)\) and \((0,6)\). Option A is exactly this line. Option B, \(x-y=6\), has a different slope and represents another line. Option C, \(x+y<6\), is still an inequality and describes a region rather than the boundary equation itself. Option D, \(x=6\), is a vertical line and ignores y. The symbol \(\le\) later tells us to include the boundary and shade the side containing a test point such as \((0,0)\). Memory cue: boundary means change < or \(\le\) into = first.

In \(x+2y\geq 10\), the symbol \(\geq\) means that the points on the boundary line \(x+2y=10\) are included in the solution set. Therefore, the boundary is drawn as a solid line. A dotted line is used only for \(>\) or \(<\), where the boundary is not included. Exam tip: Use a solid line for \(\leq\) and \(\geq\).

For \(<\) or \(>\), points on the boundary are excluded from the solution region, so the boundary is dotted. With \(\leq\) or \(\geq\), the line is included and must be solid. Exam tip: an equality bar means a solid boundary.

Substituting the test point
(0,0) gives
0+0>3, i.e.
0>3, which is false. Hence, the side containing the origin is not the solution region; the opposite side must be shaded. Since the inequality is strict (
>), the boundary line
x+y=3 is drawn dashed and is not included. Exam tip: shade the side containing a test point only when that point satisfies the inequality.

The governing concept is the intercept method for graphing a linear equation. The x-intercept is found by putting y=0, so x+0=4 and the point is (4,0). The y-intercept is found by putting x=0, so 0+y=4 and the point is (0,4). Therefore option A gives both correct intercepts. Option B would correspond to x+y=2, while (0,0) does not satisfy the equation and (4,4) gives 8 rather than 4. The points in option D also give sums of 5, not 4. Although the question uses a line rather than an inequality region, it belongs to the graphical treatment of linear inequalities in two variables because intercepts are a standard graphing tool for the boundary line.

To graph y ≤ 2x + 1, first draw the boundary line y = 2x + 1. Because the inequality contains ≤ rather than <, the boundary line is included and should be drawn as a solid line. The symbol y ≤ means that, for each fixed x, the allowed y-values are less than or equal to the line’s y-value; therefore the solution region lies below the line and includes the line itself. A test point such as (0,0) satisfies 0 ≤ 1, confirming that the side containing the origin is shaded. Thus option B is correct. The region above the line corresponds to y ≥ 2x + 1, while left/right descriptions do not represent this inequality.

In the inequality \(y>-x+2\), the value of \(y\) must be greater than the corresponding value on the line \(y=-x+2\), so the region above the line is shaded. Since ‘>’ is a strict inequality, the boundary line is not included and is usually drawn as a dashed line. The region below the line represents \(y<-x+2\). Exam tip: Test a point, such as \((0,0)\), to confirm the correct region.

The inequality \(x\geq 2\) means that the \(x\)-coordinate of every point is 2 or greater than 2. The boundary \(x=2\) is a vertical line, and it is included because the sign is \(\geq\). Therefore, the region to the right of the line is correct. Points to the left have \(x<2\). Exam tip: inequalities involving \(x\) shade left or right, whereas inequalities involving \(y\) shade above or below.

The boundary line of the inequality is \(3x+2y=12\). Substituting \((2,3)\) gives \(3(2)+2(3)=6+6=12\), so this point lies on the boundary line. For \((3,2)\), the value is \(13\), so it is not on the line. Exam tip: To test a boundary point, replace the inequality sign with an equality sign.

Substituting \((0,0)\) gives the left-hand side as \(0+3(0)=0\). Thus, the inequality becomes \(0\leq 9\), which is true. Hence, the half-plane containing the origin is shaded in the graph. ‘False’ is incorrect because \(0\) is less than \(9\). Exam tip: After substituting a test point in a linear inequality, choose the side for which the inequality is true.

To obtain the boundary line of an inequality, replace the inequality sign \(\geq\) with the equality sign \(=\). Hence, the boundary line is \(2x-y=4\). The expression \(2x-y>4\) is still an inequality, not an equation of a line. Exam tip: For \(\geq\) or \(\leq\), the boundary line is drawn as a solid line.

The symbol \(\leq\) includes equality. At every point on the line \(x+y=7\), the value of \(x+y\) is 7, so all such points satisfy \(x+y\leq 7\). Therefore, the boundary line is included in the solution region and is drawn as a solid line. It is not limited to one point or to the origin. Exam tip: A boundary is included whenever the inequality has \(\leq\) or \(\geq\).

The inequality (x-y<1) is strict, so equality is not included. Its boundary line is x-y=1; every point on this line gives x-y=1, which is not less than 1. Therefore, the boundary line is excluded from the solution and is usually drawn as a dashed line. Exam tip: use a dashed line for < or >, and a solid line for ≤ or ≥.

An x-intercept is found by setting y = 0, because a point on the x-axis has zero y-coordinate. Substituting gives 2x + 3(0) = 6, so 2x = 6 and x = 3; the x-intercept is (3,0). A y-intercept is found by setting x = 0, because a point on the y-axis has zero x-coordinate. Then 2(0) + 3y = 6, so y = 2; the y-intercept is (0,2). Therefore option A is correct. Option B reverses the coefficients and intercept values, option C treats the constant 6 as both coordinates, and option D gives points that do not lie on the required axes or line.

Substituting \((5,0)\) gives \(4x+y=4(5)+0=20\). Since \(20>16\), the point does not satisfy \(4x+y\leq16\), so it is not in the solution region. It is not on the boundary either, because boundary points satisfy \(4x+y=16\). Exam tip: Test a point by substituting its \(x\)- and \(y\)-coordinates into the inequality.

To test whether a point belongs to the solution set, substitute its coordinates into the inequality. For (x,y) = (2,3), the left-hand side is x + 2y = 2 + 2(3) = 2 + 6 = 8. The inequality requires this value to be at least 8, and equality is allowed because the symbol is ≥. Therefore the point lies exactly on the boundary line x + 2y = 8 and is included in the solution region, so option A is correct. It is not excluded merely because it is on the boundary. Option B would apply if the inequality were false, option C is not the required classification, and option D is unrelated because the boundary line can certainly be drawn.

In \(y\geq x-1\), the value of \(y\) must be equal to or greater than the value on the line \(y=x-1\). Hence, the half-plane above the line is shaded. Since \(\geq\) includes equality, the boundary line is drawn solid. The region below the line represents \(y<x-1\), so it is not correct. Exam tip: for \(y\geq\) or \(y>\), generally choose the region above the line.

For the point \((1,3)\), substitute \(x=1\) and \(y=3\): \(3\leq -2(1)+6\), i.e. \(3\leq 4\), which is true. Hence, the point lies in the solution region of the inequality. It is not on the boundary line because \(3\neq 4\). Exam tip: To test a point, substitute its \(x\)- and \(y\)-coordinates in the inequality and check whether the statement is true.

Substituting the origin \((0,0)\) gives \(x+y=0+0=0\). Hence, it lies on the boundary line \(x+y=0\). Since the inequality uses \(\geq\), the boundary line where equality holds is included in the solution region. Therefore, the origin is on the boundary and is also part of the solution. Exam tip: For \(\geq\) or \(\leq\), the boundary line is included.

For the equation \(x-2y=4\), putting \(y=0\) gives \(x=4\), so \((4,0)\) lies on the line. Putting \(x=0\) gives \(-2y=4\), hence \(y=-2\), so \((0,-2)\) also lies on it. Therefore, option A is correct. The points in the other options do not satisfy the equation. Exam tip: To obtain two convenient points for a line, set \(x=0\) and \(y=0\) successively to find the intercepts.

Substituting \((0,5)\) gives \(2x+5y=2(0)+5(5)=25\). Since \(25\leq20\) is false, the point is not a solution of the inequality. A point lies on the boundary line only when the value is exactly \(20\), but here it is \(25\). Exam tip: test a point by directly substituting its \(x\)- and \(y\)-coordinates into the inequality.

The inequalities describe coordinate restrictions. The condition x ≥ 0 permits points on the y-axis and to its right, while y ≥ 0 permits points on the x-axis and above it. Their common part is therefore the first quadrant together with both positive coordinate axes and the origin. Equality is important: because the signs are ≥ rather than >, points satisfying x = 0 or y = 0 are included. The second quadrant has x < 0, the third has both coordinates negative, and the fourth has y < 0, so none of those can be the complete solution region. Hence option C is correct.

To find the boundary line of \(x\leq 5\), replace the inequality sign with an equality sign. Hence, the boundary line is \(x=5\), which is a vertical line. All points satisfying \(x\leq 5\) lie on this line or to its left. \(y=5\) is a horizontal line, so it is not correct. Exam tip: Replace \(\leq\) or \(\geq\) with \(=\) to obtain the boundary line.

The boundary of an inequality is obtained by replacing the inequality sign with an equality sign. For \(y\geq-4\), the boundary equation is \(y=-4\). This is a horizontal line because the y-coordinate remains fixed at -4 while x can have any value. The boundary divides the coordinate plane into points above it and points below it.

Since the original sign is \(\geq\), the boundary itself is included, and the solution region consists of the points on or above the line \(y=-4\). For example, \((0,0)\) satisfies \(0\geq-4\), while \((0,-5)\) does not. Thus option B gives the boundary line. The equations involving x or both variables describe different lines and cannot be the boundary of this inequality.