Graphing InequalitiesEdexcel IGCSE Maths: Revision notes

Every linear inequality (e.g. $y < 2x + 1$) has a boundary line given by the equation you get when you replace the inequality sign with an equals sign. So $y < 2x + 1$ has boundary line $y = 2x + 1$.

To draw it:

1. Rearrange into a form you can plot (e.g. $y = mx + c$, or find two points that satisfy the equation).
2. Draw the line across the grid.
3. Decide whether the line itself is included (solid) or excluded (dashed) — this depends on the inequality symbol.

Vertical and horizontal lines work the same way: $x = 3$ is a vertical line through $x = 3$; $y = -2$ is a horizontal line through $y = -2$.

The type of line tells you whether points on the line count as solutions:

• Solid line — used for $\leq$ or $\geq$. The line IS part of the region (points on it satisfy the inequality).
• Dashed line — used for $<$ or $>$. The line is NOT part of the region (points on it do not satisfy the inequality); it just shows the boundary.

A useful memory trick: $\leq$ and $\geq$ both have a line under the symbol — think 'line under, line on the graph' (solid). Strict inequalities $<$ and $>$ have no line under them, so the boundary is dashed.

Once the boundary line is drawn, you need to identify the region that satisfies the inequality.

Method — the test point method:

1. Pick a point that is clearly NOT on the line (the origin $(0,0)$ is easiest, if the line doesn't pass through it).
2. Substitute its coordinates into the original inequality.
3. If the inequality is true, the region containing that point is the solution — shade/label that side.
4. If it's false, the solution is the other side of the line.

Shortcut without a test point: for $y > \ldots$ or $y \geq \ldots$, the region is above the line; for $y < \ldots$ or $y \leq \ldots$, the region is below the line. For $x > \ldots$, the region is to the right; for $x < \ldots$, to the left.

When a question gives two or more inequalities together (e.g. $x \geq 1$, $y \geq 0$, $x + y \leq 6$), the solution is the region satisfying all of them simultaneously.

Algebraically, without drawing:

1. Find the boundary line for each inequality.
2. Determine each region separately (above/below/left/right, using the test point method).
3. The overall solution is the intersection — every point must be on the correct side of every line.

On an exam paper you are usually asked to shade the unwanted regions and leave the required region blank (labelled $R$), or shade the wanted region directly — always check which convention the question specifies.

The feasible region is the area (or, in linear programming problems, the enclosed polygon) satisfying every inequality in a system. Its corners are called vertices.

To find a vertex algebraically:

1. Identify the two boundary lines that meet at that corner.
2. Solve their equations simultaneously (substitution or elimination) to find the $(x, y)$ point where they cross.
3. Check the point satisfies all the other inequalities too — a crossing point is only a genuine vertex of the feasible region if it lies within all the constraints.

Vertices matter because, in optimisation problems, the maximum or minimum value of an expression like $3x + 2y$ over the feasible region always occurs at one of the vertices — so you evaluate the expression at each vertex and compare.