Replacing the equals sign in a linear equation with an inequality turns a line into a half-plane. The line itself is the boundary, solid or dashed according to the symbol, and a single test point decides which side gets shaded.
Core ideas
A linear equation in two unknowns had a line for its solution set. An inequality has half the plane, and finding which half is one substitution.
A solution is still an ordered pair that makes the statement true, exactly as in “The Coordinate Plane”. There are simply infinitely many of them filling an area rather than lying along a line, and shading is how an infinite set gets drawn.
≤ and ≥ include the line, drawn solid. < and > exclude it, drawn dashed. The dashed line is not decoration: (0, −4) satisfies 4x − 2y ≥ 8 and fails 4x − 2y > 8, and the drawing has to record that difference.
Substitute one point that is not on the line. A true statement means that point's side is the solution set; a false one means the other side. (0, 0) is the easiest test point whenever the boundary misses the origin, and when the boundary passes through it, any other point will do.
The material
How to draw the region, why the test point is more reliable than the symbol, and the one algebraic move that reverses an inequality.
A linear inequality in two variables is a statement of the form Ax + By < C, with ≤, > or ≥ equally allowed and A and B not both zero. Its solution set is the set of ordered pairs making it true, which is a half-plane, including its edge or not.
The line Ax + By = C splits the plane into two open regions. Every point in one region makes Ax + By larger than C, every point in the other makes it smaller, and points on the line make it equal. That is why one test point settles the whole region: the expression cannot change which side of C it falls on without crossing the line.
For 2x + 3y ≤ 12 the boundary meets the axes at (6, 0) and (0, 4), and it is solid. Testing (0, 0) gives 0 ≤ 12, which is true, so the shaded half-plane is the one containing the origin: the side below and to the left of the line.
It is tempting to read > as "above" and < as "below". That works only when the inequality has already been solved for y with a positive coefficient, and it fails as soon as the coefficient of y is negative.
The written symbol is > and the region is below. Both facts are correct, and they are consistent because solving for y required dividing by −2, which reversed the inequality into y < 2x − 4. A test point reaches the same conclusion without needing the rearrangement at all, which is why it is the more reliable method.
If a < b and c is negative then ac > bc. This is the single rule from “Linear Inequalities” in Linear Equations and Inequalities that survives into two variables, and it is where a correct boundary line ends up with the wrong side shaded. Adding or subtracting anything, and multiplying by a positive, leave the direction alone.
y < 3x has the origin on its boundary, so (0, 0) cannot be the test point: it satisfies the equation, not the inequality, and it lies on the edge rather than in either region. Any other point works. Testing (1, 0) gives 0 < 3, which is true, so the shaded region is the one containing (1, 0), below and to the right of the line.
For 4x − 2y > 8: (0, 0) gives 0, false. (3, 1) gives 10, true. (0, −4) gives 8, and 8 > 8 is false, so this point sits exactly on the dashed boundary and is not a solution. (1, 2) gives 0, false. Only (3, 1) is in the solution set, and (0, −4) is the one worth pausing on: it would be a solution if the symbol were ≥.
The reverse direction appears in modelling and in Systems and Matrices. Find the boundary line's equation from two of its points, choose ≤ or < from whether the line is solid or dashed, and then fix the direction by testing a point taken from inside the shaded region. The last step is the same substitution as before, run with the answer known and the symbol unknown.
Interactive
The feasible region, its corners, and whether it is empty
Method
Boundary, style, test, shade. Doing the test before choosing a side keeps the symbol from misleading you.
Swap the inequality for an equals sign and draw that line. Intercepts are usually quickest: for 4x − 2y = 8 they are (2, 0) and (0, −4).
Solid when the symbol includes equality, dashed when it does not. Make this decision from the original inequality, before any rearranging, so a flipped sign cannot change it. Rearranging never changes whether the boundary is included.
Use (0, 0) unless the boundary passes through it, in which case pick any convenient point off the line. Substitute into the original inequality and evaluate to a true or false statement.
True means shade the test point's side; false means shade the other. Then test one point from inside the shaded region: it must satisfy the inequality. That second substitution is the check that catches a reversed sign.
The first is straightforward. The second has a negative coefficient on y, which is where the shading goes wrong.
The second example is the whole lesson in one picture. The symbol is >, the region is below, and there is no contradiction: solving for y divided by −2 and flipped the sign to <. Anyone shading above has applied a rule that is only valid after the rearrangement has been done.
The point (0, −4) is on the boundary and gives 8 > 8, which is false. With ≥ in place of > it would be a solution and the line would be solid. One symbol changes the status of every point on the line and no other point in the plane.
Both confirmations tested a point from inside the shaded region rather than re-reading the picture. Shading is easy to draw on the wrong side of a correctly drawn line, and only a substitution catches it.
For a faded graph, use 3x − 2y ≤ 6. The boundary 3x − 2y = 6 is supplied. Decide its line style, find two points on it, test the origin in the original inequality, shade the winning side, and confirm with a second point. Finally solve for y and use the reversed sign as an independent check on the shading.
Given 5x − 2y ≥ 10 you should draw the solid boundary from two checked points, choose and substitute a test point, shade the correct half-plane, and confirm the result after solving for y. You should also handle a boundary through the origin by choosing a different test point, and read an inequality back off a shaded picture.
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