# Solving Inequalities and Set Notation – A Level Maths Revision

The basics of solving inequalities at A Level are much the same as at GCSE, with the addition of set notation to represent your solved inequalities.

The blog below explains how to solve inequalities and express your answer using set notation, with worked examples and questions at the end.

The solution of an inequality is the set of real numbers for which the inequality holds true. You will need to be able to represent this solution on a number line, on a graph or using set notation.

You will have seen some set notation at GCSE, for example, the symbols β and β for union and intersect. At A level, the range of set notation you need to know increases.

Set Notation

{π₯ β β: π₯ β₯ π} is often shortened to {π₯: π₯ β₯ π}.

Use a square bracket to show a particular value is included (a closed interval) and a rounded bracket to show it isnβt (an open interval). For example; (2, 3] is the range of values 2 < π₯ β€ 3 while [2, 3) is the range of values 2 β€ π₯ < 3.

When considering inequalities such as π₯ > -4 or π₯ β€ 2, use β or -β as the second bound. This symbol represents the concept of infinity or limitlessness. This isnβt a particular value and you canβt say π₯ = β, so these intervals are always open-ended. The interval π₯ > -4 can be written as (-4, β) and π₯ β€ 2 can be written as (-β, 2].

The final symbol in the list, β, means that there are no real values that satisfy the conditions. For example, if you are looking for π₯ β β such that π₯ > 4 and π₯ < 0, there are no values that are greater than 4 and less than 0, so the set of values for π₯ is denoted by β.

### Example Question 1

Write the set of values, π₯, such that π₯ β€ -2 or (π₯ > 0 and -1 < π₯ β€ 3).

The easiest way to compare two or more inequalities is by drawing them on a number line.

Open circles represent an open interval (> or <), while the closed circles represent a closed interval (β€ or β₯).

The set of solutions will include all values of π₯ β€ -2 along with all values of π₯ where (π₯ > 0 and -1 < π₯ β€ 3). Start with the brackets β what set of values satisfy π₯ > 0 and -1 < π₯ β€ 3?

Considering the diagram above, you can see that values of π₯ greater than 0 and less than or equal to 3 satisfy both these conditions; write this as (0, 3].

You are also interested in values of π₯ β€ -2; this is written as (-β, -2].

The wording of the questions tells us that you are interested in more than one set of numbers – the set which includes (0, 3] and the set which includes (-β, -2]. Write this using the union symbol: the set of values, π₯, such that π₯ β€ -2 or (π₯ > 0 and -1< π₯β€ 3) is:

(-β, -2] β (0, 3]

You could also write {π₯: 0 < π₯ β€ 3} β {π₯: π₯ β€ -2}. The first option is quicker, but you need to be aware of both types of notation.

### Example Question 2

Find the set of values of π₯ for which 7 β 5π₯ β€ 22

Start by solving the inequality as usual:

7 β 5π₯ β€ 22
-5π₯ β€ 15

Remember, when dividing by a negative, you need to reverse the inequality symbol:

π₯ β₯ -3

In set notation, this is {π₯: π₯ β₯ -3} or [-3, β).

### Example Question 3

Find the set of values of π₯ for which 3π₯ β 2 β₯ 7 and 3π₯ β 4 < π₯ + 7

Start by solving each inequality separately:

3π₯ β 2 β₯ 7
3π₯ β₯ 9
π₯ β₯ 3

3π₯ β 4 < π₯ + 7
2π₯ < 11
π₯ < $\frac{11}{2}$

To help see the values which satisfy both inequalities, draw them on a number line:

The values that satisfy both 3π₯ β 2 β₯ 7 and π₯ + 4 < 3π₯ β 7 are:

3 β€ π₯ < $\frac{11}{2}$

In set notation, this can be written in three different ways:

{π₯: π₯ β₯ 3} β {π₯: π₯ < $\frac{11}{2}$}

{π₯: 3 β€ π₯ <$\frac{11}{2}$ }

[3, $\frac{11}{2}$)

### Example Question 4

Find the set of values of π₯ for which 2π₯ β 5 < 2 β π₯ or 2(3π₯ β 5) > 4π₯ + 4

Again, start by solving each inequality:

2π₯ β 5 < 2 β π₯
3π₯ < 7
π₯ < $\frac{7}{3}$

2(3π₯ β 5) > 4π₯ + 4
6π₯ β 10 > 4π₯ + 4
2π₯ > 14
π₯ > 7

In this case, there is no overlap between the two values. However, you are not looking for values that satisfy both inequalities, but for the values that solve 2π₯ β 5 < 2 β π₯ or 2(3π₯ β 5) > 4x + 4. These are values less than $\frac{7}{3}$ or greater than 7.

In set notation, we write {π₯: π₯ < $\frac{7}{3}$} β {π₯: π₯ > 7} or (-β, $\frac{7}{3}$) β (7, β).

### Example Question 5

Find the set of solutions such that 3π₯2 β 2 β₯ 5π₯

In this example, you are looking for the set of solutions that satisfy a quadratic inequality. This particular quadratic can be easily rearranged and factorised:

3π₯2 β 5π₯ β 2 β₯ 0
(3π₯ + 1)(π₯ β 2) β₯ 0

This means that the graph y = 3π₯2 β 5π₯ β 2 crosses the π₯-axis at the points π₯ = $-\frac{1}{3}$ and π₯ = 2. These are called the critical values β you know the inequality will include these two values. Sketch a graph to help work out the solutions:

The graph above is greater than zero in the bold sections. These are the values of π₯ that satisfy
3π₯2 β 2 β₯ 5π₯.
Therefore, the two inequalities are π₯ β₯ 2 and π₯ β€ $-\frac{1}{3}$. Use β₯ and β€, rather than > or <, to stay consistent with the question.

The set of solutions to 3π₯2 β 2 β₯ 5π₯ is (-β, $-\frac{1}{3}$] β [2, β).

### Example Question 6

Find the set of values such that $\frac{4}{x}$ > 1

Your first instinct here may be to multiply each side by π₯. However, multiplying an inequality by a negative number reverses the inequality, and π₯ is an unknown, so could in theory be negative or positive.

To avoid this problem, simply multiply by π₯2.

4π₯ > π₯2
4π₯ β π₯2 > 0
x(4 β π₯) > 0

This means our critical values are π₯ = 0 and π₯ = 4. Consider the graph:

As before, the section in bold represents the values of π₯ that satisfy π₯(4 β π₯) > 0 and hence $\frac{4}{x}$ > 1. This means the inequality is 0 < π₯ < 4 or (0, 4) in set notation.

Note that the sketch above isnβt of the graph of $\frac{4}{x}$ > 1, but it has the same critical values. It is also worth noting that, if the original inequality was $\frac{4}{x}$ β₯ 1, the solution would be 0 < π₯ β€ 4, as $\frac{4}{x}$ is undefined when π₯ = 0.

## Solving Inequalities at A Level

### Practice Questions

1. Find the set of values that satisfy:
a. 6π₯ + 9 β€ 1
b. 2π₯ β 8 β₯ 4π₯ + 2
c. 3 < $\frac{2x}{3}$ β 2 β€ 8

2. Find the set of values that satisfy:
a. 3π₯ > 2π₯ β 4 and 4π₯ + 1 β€ 3π₯ β 2
b. 4 < 8π₯ β 2 or 5π₯ + 2 < 3
c. $\frac{x}{3}$ β 2 β€ 2π₯+ 1 and 4π₯ β 2 > 5

3. Find the set of values that satisfy:
a. 3x2 + 2π₯< 16
b. 2π₯2 + 6π₯ + 4 β₯ 0
c. 8π₯ β π₯2 β€ 12

1. Find the set of values that satisfy:
a. 6π₯ + 9 β€ 1
π₯ β€ $-\frac{4}{3}$
(-β, –$-\frac{4}{3}$]

b. 2π₯ β 8 β₯ 4π₯ + 2
π₯ β€ -5
(-β, -5]

c. 3 < $-\frac{2x}{3}$ β 2 β€ 8
$\frac{15}{2}$ < π₯ β€ 15
($\frac{15}{2}$, 15]

2. Find the set of values that satisfy:
a. 3π₯ > 2π₯ β 4 and 4π₯ + 1 β€ 3π₯ β 2
π₯ > -4 and π₯ β€ -3
(-4, -3]

b. 4 < 8π₯ β 2 or 5π₯ + 2 < 3
π₯ > $\frac{3}{4}$ or π₯ < $\frac{1}{5}$
(-β, $\frac{1}{5}$) β ($\frac{3}{4}$, β)

c. $\frac{x}{3}$ β 2 β€ 2π₯+ 1 and 4π₯ β 2 > 5
π₯ β₯ $-\frac{9}{5}$ and π₯ > $\frac{7}{4}$
($\frac{7}{4}$, β)

3. Find the set of values that satisfy:
a. 3π₯2 + 2π₯< 16
$-\frac{8}{3}$ < π₯ < 2
($-\frac{8}{3}$, 2)

b. 2π₯2 + 6π₯ + 4 β₯ 0
π₯ β€ -2 or π₯ β₯ -1
(-β, -2] β [-1, β)

c. 8π₯ β π₯2 β€ 12
π₯ β€ 2 or π₯ β₯ 6
(-β, 2] β [6, β)

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