Representing Inequalities as Regions (Cambridge International IGCSE Mathematics)

Finding Regions using Inequalities

What are 2D inequalities?

• A 2D inequality involves two variables

  • E.g. $y<x$ or $x+y\geq8$

• The solution to a 2D inequality is a region in the $xy$ plane

• The solution to a 1D inequality can also be represented as a region in the $xy$ plane

  • E.g. $y\geq2$ 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:

$3x+2y\geq12$ $y<2x$ $x<3$

Label this region R.

Answer:

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

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

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

The line $3x+2y\geq12$ is a solid line because of the "≥"

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

Now we need to shade the unwanted regions

For $3x+2y\geq12$ (or $y\geq-\frac{3}{2}x+6$), the unwanted region is below the line We can check this with the point (0, 0)

$\begin{gathered} " 3 ( 0 ) + 2 ( 0 ) & \text{greater} \text{or} \text{equal} \text{than} 12 " \end{gathered}$ is false therefore (0, 0) does lie in the unwanted region for $\begin{gathered} 3 x + 2 y & \text{greater} \text{or} \text{equal} \text{than} 12 \end{gathered}$

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

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

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