Graphical Inequalities and Shaded Regions

Drawing the Lines

To sketch an inequality, first treat it as an equation to find the boundary line. This is the line formed by replacing the inequality sign with an equals sign.

• Vertical lines take the form $x = k$ and pass through $k$ on the $x$-axis.
• Horizontal lines take the form $y = k$ and pass through $k$ on the $y$-axis.
• Diagonal lines take the form $y = mx + c$ and may require rearranging before plotting.

If you need to rearrange an equation and multiply or divide the inequality by a negative number, you must flip the inequality sign.

Finding the Feasible Region

Once your lines are drawn, you must determine which side of each line satisfies the inequality. The area where all given inequalities are satisfied simultaneously is called the feasible region.

• AQA's official guidance is to create a shaded region over the unwanted (non-required) areas.
• This leaves the wanted feasible region clear and unshaded, making it clearly legible.
• In an exam, you must almost always label this clear area with a capital 'R'.

To check which side to shade, use the Test Point method. Choose a coordinate not on the line, such as $(0, 0)$. Substitute these values into the inequality; if the mathematical statement is true, that side is the "wanted" region, so you should shade the opposite side. If the statement is false, that side is the "unwanted" region, so you should shade it.

Worked Example: Sketching Multiple Inequalities

Represent the region R satisfied by $x \leq 4$, $y > 2$, and $y \leq x + 1$.

Step 1: Plot the boundary line $x = 4$.

• Because the inequality is $\leq$ (inclusive), use a solid line.
• The wanted area is $x \leq 4$ (left), so shade the unwanted area to the right ($x > 4$).

Step 2: Plot the boundary line $y = 2$.

• Because the inequality is $>$ (strict), use a dashed line.
• The wanted area is $y > 2$ (above), so shade the unwanted area below ($y < 2$).

Step 3: Plot the boundary line $y = x + 1$.

• Because the inequality is $\leq$ (inclusive), use a solid line (gradient 1, $y$-intercept 1).
• Test the point $(0, 0)$: $0 \leq 0 + 1$ is true.
• Since $(0, 0)$ is below the line and is "wanted", shade the unwanted area above the line.

Step 4: Label the final region.

• Mark the remaining clear triangular area with a capital 'R'.

[Insert diagram: A coordinate grid showing three boundary lines. A solid vertical line at $x = 4$, a dashed horizontal line at $y = 2$, and a solid diagonal line $y = x + 1$. The area to the right of $x = 4$, the area below $y = 2$, and the area above $y = x + 1$ are all shaded out to represent the unwanted regions. This leaves a clear, unshaded triangular feasible region bounded by vertices at $(1, 2)$, $(4, 2)$, and $(4, 5)$. This clear region is labelled with a large capital 'R'.]

Integer Solutions

Higher Tier questions may ask you to "mark with a cross all points with integer coordinates in region R."

• Points that sit exactly on solid lines are included in the region and should be marked.
• Points that sit exactly on dashed lines are excluded from the region and must NOT be marked.