Graphical solution of linear inequalities in two variables

For the boundary line (x+3y=6), (y=0) gives (x=6) and (x=0) gives (y=2). Find both intercepts separately in exams.

Substituting ((3,0)) gives (9\leq 6), which is false, so it is not in the solution region. In exams, find the option that makes the inequality false.

From (x-2y\leq 0), we get (x\leq 2y), so (y\geq \frac{x}{2}). Remember sign changes when handling negative coefficients in exams.

The governing concept is checking a point in a linear inequality by substitution. For the point (0,2), use x=0 and y=2 in y≤2x+1. The left side is 2, while the right side is 2(0)+1=1, so the inequality becomes 2≤1, which is false. Therefore the point is not in the solution region, making option C correct. It is also not on the boundary line, because the boundary equation y=2x+1 gives y=1 when x=0; the point on the boundary at x=0 would be (0,1), not (0,2). The symbol ≤ includes the boundary whenever equality holds, but equality does not hold here. Option A is rejected by the failed substitution, option B confuses the point with the boundary point, and option D is false because the origin is (0,0).

The symbol (\geq) includes equality, so (y=2x+1) is drawn as a solid line. Use solid lines for equality-including signs in exams.

The boundary line is (x+y=1), and ((0,1)) gives (0+1=1). Check equality to identify boundary points in exams.

Dividing the whole inequality by (2) gives (x+y>2). The sign does not change when dividing by a positive number in exams.

The symbol \(\leq\) includes equality. Hence the boundary \(x+2y=6\) is drawn as a solid line, and its points belong to the solution region. Exam tip: use a dashed line only for \(<\) or \(>\).

The direct answer is option D: \((0,3)\). To test a point in the solution region of \(x+2y\geq4\), substitute its x- and y-coordinates and check whether the inequality is true. For option A, \((0,0)\): \(0+2(0)=0\), and \(0\geq4\) is false. For option B, \((1,1)\): \(1+2(1)=3\), and \(3\geq4\) is false. For option C, \((2,0)\): \(2+2(0)=2\), and \(2\geq4\) is false. For option D, \((0,3)\): \(0+2(3)=6\), and \(6\geq4\) is true. Therefore D is the only point in the solution region. The boundary line is \(x+2y=4\), and because the sign is \(\geq\), the boundary is included and the side giving values at least 4 is shaded. Option A is wrong because its left side is 0, option B is wrong because it gives 3, and option C is wrong because it gives 2; all are less than 4. Option D works because its value is 6, which is greater than 4. A common mistake is to look only at positive coordinates; positive coordinates do not guarantee a solution. Always substitute. Memory cue: for a point question, calculate the left side first and compare it directly with the right side.

Substituting ((1,0)) gives (0>-1), which is true, so the point is in the solution region. Be careful with negative signs in exams.

From (x-y\geq 0), (-y\geq -x), so (y\leq x). Change the inequality sign when multiplying or dividing by (-1) in exams.

Substituting ((2,1)) gives (5<5), which is false, so it is not a solution. Equality is not accepted in strict inequalities in exams.

In the boundary line (x+4y=8), putting (x=0) gives (y=2). Write intercepts as ordered pairs in exams.

Substituting ((0,0)) gives (0\geq 10), false, so the origin side is not the solution. Use the origin as a test point when it is not on the line.

The symbol (>) does not include equality, so (x+2y=6) is a dashed line. Think of boundary and shading separately in exams.

Substituting ((0,2)) gives (10\leq 10), true, so it is on the boundary and in the solution. Equality is included for (\leq) in exams.

Substituting ((0,0)) gives (0<6), true, so the origin side is the solution. A true test means shade that same side in exams.

In (y\geq 5), the line (y=5) is included and (y) is greater than or equal to it. Choose the above region for a horizontal line in exams.

For the boundary line (x+6y=12), (y=0) gives (x=12) and (x=0) gives (y=2). Draw the boundary first, then choose the region.

Substituting ((2,2)) gives (6\leq 5), which is false, so the point is not a solution. Avoid addition mistakes in exams.

Dividing by (2) gives (y\leq \frac{x+4}{2}). The inequality sign does not change when dividing by a positive number in exams.

Substituting ((0,0)) gives (0\geq 0), true with equality. Include boundary points for (\geq) in exams.

At the origin the value is (0), which is less than (6), so the opposite side is (x+y\geq 6). If the line is included, choose (\leq) or (\geq) in exams.

Substituting ((0,0)) gives (0\geq 2), which is false, so the origin side is not the solution. In exams, shade the opposite side when the test is false.

Since (<) does not include equality, the boundary line will be dashed. In exams, do not include the boundary line for (<) and (>).