# Representing Inequalities Graphically – A Level Maths Revision

It is possible to solve linear and quadratic inequalities graphically, using graphs to show the regions that satisfy single or multiple inequalities, as well as algebraically.

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• To practise some of the prior knowledge needed to represent inequalities graphically, try these multiple-choice questions.

Consider the two shaded regions below:

When drawing inequalities graphically, you can either shade the regions as you eliminate them or shade the final region. Unless stated in the question, you can do either, however with longer questions it is usually easier to eliminate areas as you go, leaving the final region unshaded. That is what we will do in these examples.

The shaded region on both of the above graphs is the same, it is below the line π¦ = 2. However, graph 2 is drawn with a dashed line while graph 1 is a solid line.

This is used to differentiate between π¦ > 2 (dashed line) and π¦ β₯ 2 (solid line); if the line is dashed, or broken, then the values on the line are not included.

Now, consider the next two graphs. The line is π¦ = π₯ and it is solid, so each must represent either π¦ β₯ π₯ or π¦ β€ π₯.

To decide which area represents which inequality, use a point to test it; in this case, (-2, 2). At this coordinate, π¦ β₯ π₯.

Remember that you are shading to eliminate the area that does not satisfy the inequality. On graph 3, the point lies in the shaded region. This means graph 3 must represent π¦ β€ π₯.

On graph 4, the point (-2, 2) lies in the unshaded region. This means graph 4 must represent π¦ β₯ π₯.

Example Question 1

By plotting the graphs, show the region that satisfies the inequalities π¦ + π₯ β€ 5, π¦ > -1 and π₯ β₯ 1. Label the region with an R.

Start by plotting π¦ + π₯ β€ 5 (which rearranges to π¦ β€ 5 β π₯). This uses the inequality β€ so draw a solid line. Now consider the point (4, 4) β this point is above the line, and it doesnβt satisfy the inequality: 4 + 4 = 8, and 8 is not less than 5.

You can therefore eliminate the values above the line by shading.

Next, plot π¦ > -1. This time, use a dashed line as the line π¦ = -1 is not included. The region of interest is above the line, where values of π¦ are greater than -1, so shade the region below the line to eliminate it.

Finally, plot π₯ β₯ 1. The line should be solid as the line π₯ = 1 is included. You are interested in the region to the right of the line, where the values of π₯ are greater than 1, so shade the area to the left of the line to eliminate it.

You have represented all three inequalities graphically, leaving the unshaded region. Before you finish, it is important to label that area with an R, as directed in the question.

To check the answer, choose a point in the region R and make sure the coordinates satisfy all 3 inequalities.

Letβs test the point (2, 1). In this case, x = 2 and π¦ = 1:

π¦ + x = 3, so π¦ + π₯ β€ 5
π¦ = 1, so π¦ > -1
π₯ = 2, so π₯ β₯ 1

All three inequalities are satisfied, so you have chosen the correct region.

Example Question 2

By plotting the graphs, show the region that satisfies the inequalities $\boldsymbol{y<\frac{1}{3}x+10}$, π¦ > 5 and π¦ β₯ π₯2. Mark the region with an R.

Begin by plotting the graph $y<\frac{1}{3}x+10$. As the inequality sign is <, use a broken line and shade above it to show that you’re eliminating that region.

Next, plot π¦ > 5. Again, this is a broken line. This time, shade below it: the area where the values of π¦ are not greater than 5.

Finally, plot π¦ β₯ π₯2 as a solid line.

Here, it is harder to decide which region to shade by eye, so chose a point to check. Consider the point (-4, 4), which is below the line of the graph.

For (-4, 4), π₯2 = 16 and π¦ = 4. This means (-4, 4) is in the region where the inequality is not satisfied, so shade that region to eliminate it.

Putting this together gives:

Again, you can use a point to check all three inequalities are satisfied; this time, use (0, 8):

$\frac{1}{3}x+10=10$ and π¦ = 8, so $y<\frac{1}{3}x+10$

π¦ = 8, so π¦ > 5

π₯2 = 0 and π¦ = 8, so π¦ β₯ π₯2

All three inequalities are satisfied, so this is the correct region.

Example Question 3

By plotting the graphs, find all the integer values of π₯ and π¦ that satisfy the inequalities π¦ < π₯3, -1 < π¦ β€ 1 and π¦ β₯ π₯ β 3.

Start this question as before, by plotting each of the three graphs.

First, plot π¦ < π₯3. This is a broken line with the shading above (or to the left) of the line. You can check this by considering the point (0, 1): in this case π₯3 = 0 and π¦ = 1. This does not satisfy the inequality, so we shade this region.

Next, plot -1 < π¦ β€ 1. Use a dashed line to represent -1 < π¦ and a solid line to represent π¦ β€ 1. You are interested in the region between these two lines, so shade the regions outside them.

Finally, plot π¦ β₯ π₯ β 3. This is a solid line. Shade below it: the area where the values of π¦ are not greater than π₯ β 3.

Unlike in previous questions which just ask you to represent the inequalities graphically, here you are asked to state the integer solutions that satisfy the inequalities. This is just each point in the unshaded region where π₯ and π¦ are whole numbers.

Some are relatively easy to spot, such as (1, 0). However, some take more thought. Consider the points (2, 1) and (0, -1). These both lie on lines, but (2, 1) lies on a solid line while (0, -1) lies on a dashed line. This means that you should include (2, 1) as a pair of values, but not (0, -1).

Next, consider the points (-1, -1), (1, 1), (2, -1) and (4, 1). Each of these lie where two lines meet. (-1, -1) is the point where two broken lines meet, so this canβt be included as the values on a dashed line do not satisfy the inequalities.

(1, 1) and (2, -1) both lie on points where a broken line meets a solid line. Again, you donβt include the values; while (1, 1) satisfies 1 < π¦ β€ 1, it doesnβt satisfy π¦ < π₯3. The same logic applies to (2, -1).

Finally, consider (4, 1). This is the point where two solid lines meet. The values on a solid line satisfy the inequalities so you can include (4, 1).

The graph above shows the final values of π₯ and π¦ included.

These are the coordinates (1, 0), (2, 0), (3, 0), (2, 1), (3, 1) and (4, 1).

Note that the values of π₯ and π¦ are paired – the values will only satisfy the inequalities if they are in this pair.

Remember, these are just the integer values. There are an infinite number of non-integer solutions to the three inequalities.

Representing Inequalities Graphically – Practice Questions

1) Show each region that satisfies the following inequalities. Use axes where -10 β€ π₯ β€ 10 and -10 β€ π¦ β€ 10. Mark the region with an R.

a. π¦ > 1; π₯ + π¦ < 10; π₯ β₯ 4 and π¦ β₯ 2π₯ β 10.

b. π¦ β₯ π₯2 β 5; π¦ β π₯ < 5; π¦ β₯ 0 and π₯ β₯ 0.

c. π¦ β₯ π₯2 + 2π₯ β 3; and π¦ < -2π₯ β 1.

d. π¦ β€ 7 β π₯2; y > π₯2.

2) Identify the inequalities that describe the area marked with an R.

a.

b.

3) By plotting the graphs, find the integer solutions to the inequalities π¦ + π₯ β€ 1, π¦ β₯ π₯3 β 2, π¦ > -2 and π¦ < 1 + 3π₯ β π₯2. Use axes where -3 β€ π₯ β€ 4 and -3 β€ π¦ β€ 4.

Answers

1)

a. π¦ > 1; π₯ + π¦ < 10; π₯ β₯ 4 and π¦ β₯ 2π₯ β 10.

b. π¦ β₯ π₯2 β 5; π¦ β π₯ < 5; π¦ β₯ 0 and π₯ β₯ 0.

c. π¦ β₯ π₯2 + 2π₯ β 3; and π¦ < -2π₯ β 1.

d. π¦ β€ 7 β π₯2; π¦ > π₯2.

2)

a.

π₯ > -2
π¦ β€ 6
π₯ + π¦ β€ 5 or π¦ β€ 5 β π₯
$\boldsymbol{y>\frac{1}{2}x+3}$

b.

π₯ + π¦ β€ 2 or π¦ β€ 2 β π₯
π¦ > π₯2

3)

The unshaded region of the graph represents the region included in all three inequalities.

The integer solutions are (0, 0), (0, 1), (-1, 0) and (-1, 1).