Title: Default for 3.5 Solving Inequalities
Course: HELM Workbooks 2023-2024

The inequality symbols

Recall the meaning of the following symbols:

Key Point

The symbols \(>\), \(<\), \(\ge\), \(\le\) are called inequalities

\(>\) means: `is greater than'

\(≥\) means: `is greater than or equal to'

\(<\) means: `is less than'

\(≤\) means: `is less than or equal to'

So, we may state, for example, \[8>7\qquad 9\ge 2\qquad -2<3\qquad 7\le 7\]

A number line is often a helpful way of picturing inequalities. Given two numbers \(a\) and \(b\), if \(b>a\) then \(b\) will be to the right of \(a\) on the number line as shown in the figure below.

[Image]

Note from the figure below that −3 > −5, 4 > −2 and 8 > 5.

[Image]

Inequalities can always be written in two ways. For example in English we can state that 8 is greater than 7, or equivalently, that 7 is less than 8. Mathematically we write \(8>7\) or \(7<8\). Similarly if \(b>a\) then \(a<b\). If \(a<b\) then \(a\) will be to the left of \(b\) on the number line.

Example

Rewrite the inequality \(-{2\over 5}<x\) using only the ‘greater than’ sign, \(>\).

Solution

\(-{2\over 5}<x\) can be written as \(x>-{2\over 5}\)

Example

Rewrite the inequality \(5>x\) using only the ‘less than’ sign, \(<\).

Solution

\(5>x\) can be written as \(x<5\).

Sometimes two inequalities are combined into a single statement. Consider for example the statement \(3~<~x~<~6\). This is a compact way of writing ‘\(3<x\) and \(x<6\)’. Now \(3<x\) is equivalent to \(x>3\) and so \(3<x<6\) means \(x\) is greater than 3 but less than 6.

Inequalities obey simple rules when used in conjunction with arithmetical operations:

Key Point

1. Adding or subtracting the same quantity from both sides of an inequality leaves the inequality sign unchanged.
2. Multiplying or dividing both sides by a positive number leaves the inequality sign unchanged.
3. Multiplying or dividing both sides by a negative number reverses the inequality.

For example, since \(8>5\), by adding \(k\) to both sides we can state \[8 + k > 5 + k\] for any value of \(k\). For example (with \(k=-3\)) \(8-3>5-3\).

Further, by multiplying both sides of \(8>5\) by \(k\) we can state \[8k > 5k\] provided \(k\) is positive.

However, \[8k<5k\] if \(k\) is negative. We emphasise that the inequality sign is reversed when multiplying both sides by a negative number. A common mistake is to forget to reverse the inequality sign when multiplying inequalities by a negative number. For example \(8>5\), but multiplying both sides by \(-1\) gives \(-8<-5\).

Modulus

The modulus or magnitude sign is sometimes used with inequalities. For example \(|x|<1\) represents the set of all numbers whose actual size, irrespective of sign, is less than 1. This means any value between \(-1\) and 1. Thus \[|x|<1 \hbox{ implies } -1<x<1\]

Similarly \(|x|>4\) means all numbers whose size, irrespective of sign, is greater than 4. This means any value greater than 4 or less than \(-4\). Thus \[|x|> 4 \hbox{ implies } x> 4 \hbox{ or } x< -4.\]

In general, if \(k\) is a positive number,

Key Point

\[|x|<k \hbox{ means } -k<x<k\] \[|x|>k \hbox{ means } x>k \hbox{ or } x<-k\]

Solving linear inequalities algebraically

When we are asked to solve an inequality, the inequality will contain an unknown variable, say \(x\). Solving means obtaining all values of \(x\) for which the inequality is true. In a linear inequality the unknown appears only to the first power, that is as \(x\), and not as \(x^2\), \(x^3\), \(x^{1/2}\) and so on. It is possible to solve a linear inequality by making the unknown the subject. Consider the following examples.

Example

Solve the inequality \(4x+3>0\).

Solution

\[\begin{aligned} 4x+3 &> 0&& \\ 4x &> -3 && \hbox{ by subtracting 3 from both sides}\\ x &>-{3\over 4} && \hbox{ by dividing both sides by 4.}\\\end{aligned}\] Hence all values of \(x\) greater than \(\displaystyle{-{3\over 4}}\) satisfy \(4x+3>0\).

Example

Solve the inequality \(-3x-7≤0\).

Solution

\[\begin{aligned} -3x-7 &≤ 0&& \\ -3x &≤ 7 && \hbox{ by adding 7 on both sides}\\ x &≥-{7\over 3} && \hbox{ by dividing both sides by -3 and reversing the inequality.}\\\end{aligned}\] Hence all values of \(x\) greater than \(\displaystyle{-{7\over 3}}\) satisfy \(-3x-7≤0\).

Example

Solve the inequality \(|5x-2|<4\) and depict the solution graphically.

Solution

We treat each part of the inequality separately: \[\begin{aligned} -4 &< 5x-2 \\ -2 &< 5x \qquad \hbox{ by adding 2 to both sides} \\ -{2\over 5} &< x \qquad \hbox{ by dividing both sides by 5}\end{aligned}\] So \(x>-{2\over5}\). Now consider the second part: \(5x-2<4\). \[\begin{aligned} 5x-2 &< 4 \\ 5x &< 6 \qquad \hbox{ by adding 2 to both sides}\\ x &< {6\over 5} \qquad \hbox{ by dividing both sides by 5}\end{aligned}\] Putting both parts of the solution together we see that the inequality is satisfied when
\(\displaystyle{-{2\over 5}<x<{6\over 5}}\). This range of values is shown in the figure below.

[Image]

Solving inequalities using graphs

Graphs can be used to help solve inequalities. This approach is particularly useful if the inequality is not linear as, in these cases solving the inequalities algebraically can often be very tricky. Graphics calculators or software can help save a lot of time and effort here.

Example

Solve graphically the inequality \(5x+2<0\).

Solution

[Image]

We consider the function \(y=5x+2\) whose graph is shown in the figure above. The values of \(x\) which make \(5x+2\) negative are those for which \(y\) is negative. We see directly from the graph that \(y\) is negative when \(x<-{2\over 5}\).

Example

Find the range of values of \(x\) for which \(x^2-x-6<0\).

Solution

[Image: A graph of the curve y=x^2-x-6.]

We consider the graph of \(y=x^2-x-6\) which is shown in the figure above. Note that the graph crosses the \(x\) axis when \(x=-2\) and when \(x=3\). Now \(x^2-x-6\) will be negative when \(y\) is negative. Directly from the graph we see that \(y\) is negative when \(-2<x<3\).

Example

By plotting a graph of \(y= 20x^4-4x^3-143x^2+46x+165\) find the range of values of \(x\) for which \[20x^4-4x^3-143x^2+46x+165<0\]

Solution

A software package has been used to plot the graph which is shown in the figure below. We see that \(y\) is negative when \(-2.5<x<-1\) and when \(1.5<x<2.2\).

[Image]

Exercise

Find the result of multiplying both sides of the inequality \([value] < [value]\) by the number \([value]\).
Solution:
Feedback: When multiplying both sides of an inequality by a negative number, the inequality sign changes. So \([value] < [value]\) multiplied by \([value]\) is written as: \[[value]([value]) > [value]([value])\] which can be simplified to \[[value]\]

Exercise

Solve the inequality \([value][value]+[value]<[value][value]+[value]\).

This is done by making \([value]\) the subject and obtain it on its own on the left-hand side.

Start by subtracting \([value][value]\) from both sides to remove quantities involving \([value]\) from the right:

Now subtract \([value]\) from both sides to remove the \([value]\) on the left:

Finally, the range of values of \([value]\) is:
Feedback: We have the inequality \[[value][value]+[value]<[value][value]+[value]\] We start by subtracting \([value][value]\) from both sides to remove quantities involving \([value]\) from the right. Hence, we get: \[ [value]\] Now subtract \([value]\) from both sides to remove the \([value]\) on the left: \[ [value] \] Finally, the range of values of \([value]\) is: \[ [value] \]

Exercise

Solve the inequality \(|[value]-[value]x|<[value]\).

First of all rewrite the inequality without using the modulus sign:

Write each answer of the form "\(a \le x\le b\)" as a logical sequence of inequalities, such as "\( x\ge a\) and/or \(x < b\)"

\(|[value]-[value]x|<[value]\) is equivalent to

Treat each part separately. Firstly, find the upper bound of \([value]\):

\([value]<\)

Now, find the lower bound of \([value]\):

\([value]>\)

Finally, give the solution as one statement:
Feedback: We are given \(|[value]-[value]x|< [value]\). Then we can interpret this as \([value]\) being a lower bound on \([value]-[value]x\) and [value] being an upper bound on \([value]-[value]x\). In other words we have \[[value]\]. If \([value]<[value]\) then we can rearrange to get an upper bound on \(x\). So \( [value]< [value]\) which implies \(x < \displaystyle [value]\). Similarly, if \([value]<[value]\) then we can rearrange to get a lower bound on \(x\). So \( [value]>[value]\) which implies \(x> \displaystyle [value]\). Putting this altogether in a single statement we obtain \[[value].\]

Exercise

Find the range of values of \([value]\) for which \([value]\gt 0\). We require \(y=[value]\) to be positive. Plot the function and use the graph to solve the problem.

\([value]\lt\)

\([value]\gt\)
Feedback: The graph crosses the \([value]\)-axis when \([value]=[value]\) and \([value]=[value]\). The graph of \(y=[value]\) is negative when \([value]\in ([value],[value])\), so the correct answer is: \([value] \lt [value]\) and \([value] \gt [value]\). [value]

Exercise

In questions (a)-(e) solve each inequality using the graph of the function:

1. \([value]\lt 0\)
\(x<\)
2. \([value]\lt 0\)
\(x<\)
3. \([value]\gt 0\)
\(x>\)
4. \([value]\gt 0\)
\(x>\)
5. \([value]\lt 0\)
\(x<\) , \(x>\)

Exercise

State which of the following statements are true and which are false.

1. \([value]>[value]\)
2. \([value]>[value]\)
3. \([value]\geq[value]\)
4. \([value]<10^[value]\)
5. \(|[value]|<[value]\)
6. \(|[value]|>[value]\)
7. \([value]\geq 10^[value]\)
Feedback: 1. We are given \([value]>[value]\), i.e. [value] is greater than [value] The answer is therefore false. 2. We are given \([value]>[value]\), i.e. [value] is greater than [value]. The answer is therefore false. 3. We are given \([value]\geq[value]\), i.e. [value] is greater than or equal to [value]. Therefore the answer is true. 4. We are given \([value] < 10^[value]\), i.e. [value] is less than \([value]\). Therefore the answer is false. 5. We are given \(|[value]|<[value]\), i.e. \([value]\) is less than [value]. Therefore the answer is true. 6. We are given \(|[value]|>[value]\), i.e. \([value]\) is greater than [value]. Therefore the answer is true. 7. We are given \([value]\geq 10^[value]\), i.e. [value] is greater than or equal to \([value]\). Therefore the answer is true.

Exercise

In the following questions rewrite each of the statements without using a modulus sign:

Write each answer of the form "\(a \le x\le b\)" as a logical sequence of inequalities, such as "\( x\ge a\) and/or \(x < b\)", where by "and" you mean both statements are true, and by "or" you mean at least one statement is true. You can write \(\le\) as "<=" and \(\ge\) as ">=".

1. \([value]\)
2. \([value]\)
3. \([value]\)
4. \([value]\)
5. \([value]\)
Feedback: - The modulus of \(x\) is defined as \[|x|=\begin{cases} x & \text{if } x>0 \\ -x & \text{if } x<0 \end{cases}, \] Therefore, if \(|x|>0\) then \(x>0\) or \(x<0\) - clearly both cases cannot occur at the same time. Similarly, if \(|x|<0\) then \(x>0\) or \(x<0\). - The modulus of \(x\) minus an arbitrary constant \(a\) is defined as \[ |x-a| = \begin{cases} x-a & \text{if } x>a \\ a-x & \text{if } x<a \\ 0 & \text{if } x=a \end{cases}, \] Therefore, if \(|x-a|<b\), then \(x-a<b\) and \(a-x<b\). - The modulus of \(x\) plus an arbitrary constant \(a\) is defined as \[ |x+a| = \begin{cases} x+a & \text{if } x+a>0 \\ -x-a & \text{if } x+a<0 \\ 0 & \text{if } x=-a \end{cases}, \] Therefore, if \(|x+a|<b\), then \(x+a<b\) and \(-x-a<b\).

Exercise

In the following questions solve the given inequality algebraically:

1. \([value]\)
2. \([value]\)
3. \([value]\)
4. \([value]\)
5. \([value]\)
6. \([value]\)
7. \([value]\)
8. \([value]\)
9. \([value]\)
10. \([value]\)
11. \([value]\)
12. \([value]\)
13. \([value]\)
14. \([value]\)
15. \([value]\)
16. \([value]\)

Exercise

In the following questions solve the given inequality algebraically:

1. \([value]\)
2. \([value]\)
3. \([value]\)
4. \([value]\)
5. \([value]\)
6. \([value]\)
7. \([value]\)
8. \([value]\)
9. \([value]\)

Exercise

In the following questions solve the given inequality algebraically:

Write each answer of the form "\(a \le x\le b\)" as a logical sequence of inequalities, such as "\( x\ge a\) and/or \(x ≤ b\)"

1. \([value]\)
2. \([value]\)
3. \([value]\)
4. \([value]\)
5. \([value]\)
6. \([value]\)

Exercise

State which of the following statements are true and which are false.

1. \([value]>[value]\)
2. \([value]>[value]\)
3. \([value]\geq[value]\)
4. \([value]<10^[value]\)
5. \(|[value]|<[value]\)
6. \(|[value]|>[value]\)
7. \([value]\geq 10^[value]\)
Feedback: 1. \([value]\) is less than \([value]\), therefore the statement is false. 2. \([value]\) equals \([value]\), therefore the statement is false. 3. \([value]\) equals \([value]\), therefore the statement is true. 4. \([value]\) is more than \(10^[value]\), therefore the statement is false. 5. Even though the modulus of a negative number is positive, \(|[value]|\) is still less than \([value]\), therefore the statement is true. 6. The modulus of a negative number is positive, which means that \(|[value]|\) is more than \([value]\), therefore the statement is true. 7. \([value]\) is more than \(10^[value]\), therefore the statement is true.

Exercise

In the following questions solve the given inequality algebraically:

1. \([value]\)
2. \([value]\)
3. \(\displaystyle[value]\)
4. \([value]\)
5. \([value]\)
Feedback: 1. \([value] \Rightarrow [value]\) 2. \([value] \Rightarrow [value] \) 3. \(\displaystyle[value] \Rightarrow [value] \) 4. \([value] \Rightarrow [value]\) 5. \([value] \Rightarrow [value]\)

Exercise

In the following questions solve the given inequality algebraically:

1. \([value]\)
2. \([value]\)
3. \([value]\)
Feedback: 1. \([value] \Rightarrow [value] \) 2. \([value] \Rightarrow [value] \) 3. \([value] \Rightarrow [value] \)

Exercise

In the following questions solve the given inequality algebraically and simplify your answers:

Write each answer of the form "\(a \le x\le b\)" as a logical sequence of inequalities, such as "\( x\ge a\) and/or \(x ≤ b\)"

1. \([value]\)
2. \([value]\)
3. \([value]\)
4. \([value]\)
Feedback: If we have an inequality of the form \(|ax + b| > c\), where \(a\), \(b\) and \(c\) are non-zero constants, then we have \(ax+b > c\) or \(ax + b < - c\). Similarly, if we have an inequality of the form \(|ax+b| \geq c\), where \(a\), \(b\) and \(c\) are non-zero constants, then we have \(ax+b \geq c\) or \(ax+b \leq -c\). If we have an inequality of the form \(|ax+b|\leq 0\), where \(a\) and \(b\) are non-zero constants, then we have \(ax+b \geq 0\) and \(ax+b \leq 0\). Since both must be true, we have \(ax + b = 0\).

Exercise

In questions (a)-(c) solve each inequality using the graph of the function:

1. \([value]\lt 0\)
\(x<\)
2. \([value]\gt 0\)
\(x>\)
3. \([value]\lt 0\)
\(x<\)   , \(x>\)
Feedback: 1. The graph crosses the \([value]\)-axis when \([value]=\frac{-([value])}{[value]}\). The graph of \(y=[value]\) is negative for all \(x\) less than this value. [value] 2. The graph crosses the \([value]\)-axis when \([value]=\frac{[value]}{[value]}\). The graph of \(y=[value]\) is positive for all \(x\) greater than this value. [value] 3. The graph crosses the \([value]\)-axis when \([value]=[value]\) and \([value]=[value]\). The graph of \(y=[value]\) is negative when \([value]\in ([value],[value])\), so the correct answer is: \([value] \lt [value]\) and \([value] \gt [value]\). [value]