Representing Inequalities as Regions

Exam code: 4024

Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Finding Regions using Inequalitiesvideo

Finding Regions using Inequalities

Finding regions using inequalities

What are 2D inequalities?

A 2D inequality involves two variables
- E.g. $y < x$ or $x + y \geq 8$
The solution to a 2D inequality is a region in the $x y$ plane
The solution to a 1D inequality can also be represented as a region in the $x y$ plane
- E.g. $y \geq 2$ represents the region of points that lie on or above the line $y = 2$

How do we draw inequalities on a graph?

A set of 2D inequalities can be shown graphically using straight lines and shaded regions
To draw the correct lines:
- Replace the inequality sign with “=” and draw that line
  - Use a solid line for ≤ or ≥ (to indicate the line is included)
  - Use dotted line for < or > (to indicate the line is not included)
To decide which side of the line is the wanted side:
- if "y ≤ ..." or "y < ..." then the wanted region is below the line
- if "y ≥ ..." or "y > ..." then the wanted region is above the line
- If you are unsure
  - substitute the coordinates from a point on one side of the line into the inequality
  - determine whether or not the inequality holds true on that side
- For vertical lines:
  - the wanted region for $x < k$ is to the left of $x = k$
  - the wanted region for $x > k$ is to the right of $x = k$
To do the shading:
- Shade the unwanted sides of each line (unless the question says otherwise)
  - You are shading away any parts you don't want
  - This will leave behind a clear region that is the wanted region (rather than trying to look for the wanted region under multiple shades)
  - Label the wanted region R (unless the question says otherwise)

Worked example

Show, graphically, the region that is satisfied by all three inequalities below:

$3 x + 2 y \geq 12$ $y < 2 x$ $x < 3$

Label this region R.

Answer:

First draw the three straight lines: $3 x + 2 y = 12$ , $y = 2 x$ and $x = 3$

Use your knowledge of Straight Line Graphs, $y = m x + c$
You may wish to rearrange $3 x + 2 y = 12$ to the form $y = m x + c$ first

$\begin{array}{rcl} 2 y & = & - 3 x + 12 \\ y & = & - \frac{3}{2} x + 6 \end{array}$

The line $3 x + 2 y \geq 12$ is a solid line because of the "≥"

The lines $y < 2 x$ and $x < 3$ are dotted lines because of the "<"

Now we need to shade the unwanted regions

For $3 x + 2 y \geq 12$ (or $y \geq - \frac{3}{2} x + 6$ ), the unwanted region is below the line
We can check this with the point (0, 0)

$\begin{array}{rcl} " 3 (0) + 2 (0) & \geq & 12 " \end{array}$ is false therefore (0, 0) does lie in the unwanted region for $\begin{array}{rcl} 3 x + 2 y & \geq & 12 \end{array}$

For $y < 2 x$ , the unwanted region is above the line
Check with another point, for example (1, 0)

$\begin{array}{rcl} " 0 & < & 2 (1) " \end{array}$ is true, so (1, 0) lies in the wanted (i.e. unshaded) region for $\begin{array}{rcl} y & < & 2 x \end{array}$

For $x < 3$ , shade the unwanted region to the right of $x = 3$
If unsure, check with a point

Finally, don't forget to label the region R

Finding Regions using Inequalitiesvideo

Finding regions using inequalities

What are 2D inequalities?

How do we draw inequalities on a graph?

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