Inequalities are used to compare two values or expressions. An inequality is used when we don't know exactly what an expression is equal to. For example, we might know that $x$ is greater than $y$ and that $y$ is greater than $z$, but not the actual values of $x, y$ and $z$. Using inequality notation we would write this as: \[z\lt y\lt x\] When an inequality contains variables, *solving* the inequality means finding the values of the variables which make the inequality true (see the worked examples below).

There are 4 inequality symbols:

\[\lt\] |
less than |

\[\gt\] |
greater than |

\[\leq\] |
less than or equal to |

\[\geq\] |
greater than or equal to |

- $\lt$ and $\gt$ are called
**strict inequalities**as the expression on the left of the symbol must be less than ($\lt$) or greater than ($\gt$) that on the right. For example, $a\lt b$ says that $a$ is less than $b$.

- $\leq$ and $\geq$ are called
**weak inequalities**as the expression on the left of the symbol may be equal to or it may be less/greater than the expression on the right. For example, $c\geq d$ says that $c$ may be greater than or equal to $d$.

- $a\leq x\leq b$, $a\lt x\lt b$ and $a\lt x\leq b$ are all examples of
**double inequalities**. $a\lt x\lt b$ says that $x$ is greater than $a$ but less than $b$. Note that $a\lt x\lt b$ and $b\gt x\gt a$ mean the same thing.

**Note**: the phrase “$a$ is greater than $b$” means that $a$ is further to the right of the number line than $b$. “$c$ is less than $d$” means that $c$ is further to the left of the number line than $d$.

Sometimes we are interested only in the **absolute magnitude** (also called the **absolute value** or **modulus**) of a number. This is the value of the number without paying attention to whether the number is positive or negative. For any real number $x$ the absolute value or modulus of $x$ is denoted by $|x|$ and defined as

\[|x|=\left\{ \begin{align} &x\;\;\;\text{when } x\geq 0\\ -&x\;\;\;\text{when } x\leq 0 \end{align} \right.\]

For example, the absolute value of $(-4)$ is denoted by $|4|$ and is equal to $4$. The absolute value of $4$ is also $4$.

The following properties can be thought of as rules we must use when solving inequalities involving unknown variables.

If $x\gt y$ and $y\gt z$, then $x\gt z$.

**Example**: If Bob is older than Mary and Mary is older than Tom, then Bob must also be older than Tom.

Adding the same number to both sides does not change direction of the inequality.

If $x\lt y$, then $x+z\lt y+z$.

**Example**: If Sam is older than Mark then in $5$ years time Sam will still be older than Mark.

Subtracting the same number from both sides does not change direction of the inequality.

If $x\lt y$, then $x-z\lt y-z$.

**Example**: If Jim is older than Tom now then Jim must also have been older than Tom last year.

When we multiply or divide both $x$ and $y$ by a positive number, the direction of the inequality stays the same.

- If $x\lt y$, and $z\gt 0$ then $x\times z\lt y\times z$.
- If $x\lt y$, and $z\gt 0$ then $\Large{\frac{x}{z} \lt \frac{y}{z}}$.

**Example**: Suppose that Tom and Harry are playing a game and Tom has more points than Harry. If both players double their score then Tom will still have more points than Harry.

When we multiply or divide both $x$ and $y$ by a negative number, the inequality changes direction!

- If $x\lt y$, and $z\lt 0$ then $x \times z \gt y\times z$
- If $x\lt y$, and $z\lt 0$ then $\Large{\frac{x}{z}\gt \frac{y}{z}}$

If we take the inverse of both sides, the inequality changes direction.

If $y\gt x$, then $\frac{1}{y} \lt \frac{1}{x}$.

**Example**: Max is older than Hayley, so Hayley is younger than Max.

Find all values of $x$ such that $3x-6 \geq 4-2x$.

This graph shows the lines $f(x)=3x-6$, in red, and $g(x)=4-2x$, in blue. We are looking for the values of $x$ for which the red line is greater than the blue line.

We solve linear inequalities in the same way that we solve linear equations: isolate the variable on one side of the inequality symbol and move all other terms to the other side.

\[3x-6 \geq 4-2x\] Add $6$ to both sides. \[3x \geq 10-2x\] Add $2x$ to both sides. \[5x\geq 10\] Divide through by $5$. As this is a positive number, the inequality symbol remains unchanged. \[x \geq \frac{10}{5} = 2\]

Hence the solutions to this inequality are all $x$ such that $x \geq 2$ i.e. the interval $[2,\infty)$. This agrees with the graph.

Find all $x$ such that

**(a)** $x^2-2x-8\gt0$

**(b)** $x^2-2x-8\lt0$

See Quadratic Equations and Functions if you need a refresher on quadratic equations.

This is the graph of $f(x)=x^2-2x-8$. When the curve is above the $x$ axis, this satisfies part **(a)** of the question. When the curve is below the $x$-axis, this satisfies part **(b)**.

We can solve quadratic inequalities in the same way that we solve quadratic equations. First try to factorise the expression:

\[x^2-2x-8 = (x-4)(x+2)\]

This is zero at $x=4$ and $x=-2$, so these are the points at which the curve crosses the $x$-axis.

**(a)** For $x^2-2x-8\gt0$, we have \[(x-4)(x+2) \gt 0\]

For the product on the left hand side be greater than zero, both brackets must be positive, or both brackets must be negative.

To make both brackets positive, we must have $x\gt 4$ and $x \gt -2$, so $x\gt 4$.

To make both brackets negative, we must have $x \lt 4$ and $x \lt -2$, so $x\lt -2$.

So we must have

\[x\lt-2 \text{ or } x\gt4\]

**(b)** For $x^2-2x-8\lt0$, we have \[(x-4)(x+2) \lt 0\] For this to be less than zero, we must have one negative bracket and one positive bracket.

We cannot make the first bracket positive and the second negative as we would need $x\gt 4$ and $x\lt -2$, which cannot happen at the same time.

To make the first bracket negative and second positive, we must have $x\lt 4$ and $x \gt -2$, so our answer is \[-2 \lt x \lt 4\]

Find all $x$ such that $\dfrac{14}{1+5x} \gt 4$.

This shows the graph of $f(x)=\dfrac{14}{1+5x}$, in red, and the line $g(x)=4$ in blue. We are looking for values of $x$ for which the red line is greater than the blue line.

When looking to solve this inequality, the obvious first step would be to multiply by $1+5x$. However, this could be either positive or negative depending on the value of $x$; recall that when we multiply by a negative number, the inequality sign changes. So we must consider two cases separately:

**Case 1**: $1+5x \lt 0$**Case 2**: $1+5x \gt 0$

First, look at **Case 1**.

Solve for $x$. \begin{align} 1+5x &\lt 0\\ 5x &\lt -1\\ x &\lt -\frac{1}{5} \end{align}

Now look at the whole inequality. \[\frac{14}{1+5x} \gt 4\] Multiply through by $1+5x$, remembering that we are treating it as negative so the inequality symbol changes. \[14 \lt 4(1+5x)\] Expand the brackets and simplify. \[14 \lt 4+20x\] \[10\lt 20x\] Solve for $x$. \[x\gt \frac{10}{20}=\frac{1}{2}\] We cannot have $x\lt -\dfrac{1}{5}$ and $x\gt\dfrac{1}{2}$ at the same time, so we cannot have $1+5x \lt 0$.

Now consider **Case 2**.

Solve for $x$. \begin{align} 1+5x &\gt 0\\ 5x &\gt -1\\ x&\gt -\frac{1}{5} \end{align}

Now look at the whole inequality. \[\frac{14}{1+5x} \gt 4\] Multiply through by $1+5x$. This time we are treating it as positive so the inequality symbol will remain the same. \[14 \gt 4(1+5x)\]

Expand and simplify.

\begin{align} 14 &\gt 4+20x \\ 10 &\gt 20x \end{align}

Solve for $x$.

\[x\lt \frac{10}{20}=\frac{1}{2}\]

Combining the two inequalities we have derived for $x$, the solutions are given by $-\dfrac{1}{5} \lt x \lt \dfrac{1}{2}$.

Prof. Robin Johnson solves the linear inequality $2x-3\geq 2-5x$.

Prof. Robin Johnson solves the quadratic inequalites $t^2-2t-3 \gt 0$ and $\lt 0$.

Prof. Robin Johnson solves the inequality $1 \lt \vert x-1 \vert \lt 3$.

Prof. Robin Johnson finds all the values for $x$ that satisfy $\dfrac{1}{1+2x} \gt 1$.

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.