Inequalities
Things you need to know about Inequalities
 If the question has inequalities, the solution must include inequalities. If your solution has an equals sign then you have gone wrong.
 We solve linear inequalities in very much the same way as linear equations. See Example 1.
 We solve quadratic inequalities by first finding the roots of the quadratic. See more on Quadratics. Use the graph to identify where the quadratic is positive or negative. If the graph has two separate regions then the answer must have two separate regions. See Example 2.
 You can plot inequalities on a number line. So, if you want to see where two hold simultaneously look for where they overlap. See Example 3. Set notation may also be given or required in some questions. See Example 2. For instance, $x< a$ OR $x> b$ is equivalent to $\left\lbrace x:x< a\right\rbrace\cup \left\lbrace x:x>b\right\rbrace$. Similarly, $x< c$ AND $x> d$ is equivalent to $\left\lbrace x:x< c\right\rbrace\cap \left\lbrace x:x>d\right\rbrace$. See Venn Diagrams for more information on union and intersection.
Representing Inequalities Graphically
We can represent Inequalities graphically. Consider $ y> x+1$ and $y\geq 4x^22$ and imagine them on a graph. The blue region represents the former. The red region represents the latter. Notice the use of the dashed line to show inequality while the solid lines shows equality. When the red solid line is above the blue dashed line the following inequality holds: $4x^22> x+1$. Use standard techniques to find where the curves intersect to identify the appropriate regions.
When the Inequality is reversed
The following points are very important facts about inequalities in addition to the above bullet points:
 Multiplying or dividing both sides by a negative number reverses the sign of the inequality. To solve $ax> b$ where $a$ is positive, add $ax$ to both sides to get $0> b+ax$. Then, subtract $b$ from both sides to get $b> ax$ or $\frac{b}{a}> x$ since a is positive. Written as $x< \frac{b}{a}$ we can think of the sign as being reversed.
 Taking reciprocals of both sides also reverses the inequality. In order to solve $\frac{a}{x}< b$ we must multiply both sides by $x^2$. This is because we don’t know the sign of $x$ but we do know $x^2$ is positive. It follows that $ax< bx^2$ – we don’t reverse the sign of the inequality. Finally, we have $axbx^2=x(abx)< 0$ which can be solved as any other quadratic inequality (see above). Be careful to take note of the signs of a and b.
See Example 4.
Questions by Topic
Examples of Indices
Solve the inequality $2x+3\hspace{2pt}\geq \hspace{2pt}5(x2)+1$.
We solve linear inequalities. in very much the same way as linear equations – effectively making x the subject. Expand the brackets and simplify the right hand side: $2x+3\geq 5x9.$ Take $2x$ from both sides of the inequality to get $3\hspace{2pt}\geq\hspace{2pt} 3x9$. Then add 9 to both sides of the inequality giving $12\hspace{2pt}\geq\hspace{2pt} 3x$. Dividing both sides by 3 gives the solution $x\hspace{2pt}\leq\hspace{2pt}4$.
Solve the following: $x^2x\hspace{2pt}>\hspace{2pt} 6$.
We solve quadratic inequalities in very much the same way as quadratic equations. First of all, set one side to 0 and find the roots. It is very important to remember here that quadratic equations have specific solutions, i.e. where the quadratic cuts the xaxis. On the other hand, when solving quadratic inequalities you are not looking for roots, you are looking to see where the graph is positive or negative. Finding the roots is helpful as it tells you where the graph changes from being positive to negative or vice versa. $x^2x6\hspace{2pt}>\hspace{2pt}0$ can be factorised to $(x3)(x+2)\hspace{2pt}>\hspace{2pt}0$. The roots are $x=3$ and $x=2$. It follows that the graph is positive if we choose any $x$ more than 3 or any $x$ less than 2. Hence, the solution is $x\hspace{2pt}>\hspace{2pt}3$ OR $x\hspace{2pt}<\hspace{2pt}2$. Alternatively, the solution can be written as $\left\lbrace x:x<2\right\rbrace\cup \left\lbrace x:x> 3\right\rbrace$
Find the region where $2x+3\hspace{2pt}\geq \hspace{2pt}5x9$ AND $x^2x\hspace{2pt}>\hspace{2pt} 6$.
The first inequality holds when $x \leq 4$, see Example 1. The second inequality holds when $x\hspace{2pt}>\hspace{2pt}3$ or $x\hspace{2pt}<\hspace{2pt}2$, see Example 2. It follows that both inequalities can only be true if $x\hspace{2pt}>\hspace{2pt}3$ AND $x\hspace{2pt}\leq\hspace{2pt}4$. Hence, the solution is $3<\hspace{2pt}x\hspace{2pt}\leq\hspace{2pt}4$.
Solve the following inequalities:
 $x\leq 4$
 $\frac{1}{x}\hspace{2pt}>\hspace{2pt}7$

 Multiply both sides of the inequality by 1 in order to get a solution range for $x$. Note that 1 is a negative number and the inequality must be reversed. See the fourth bullet point above for more details. Hence, we get the solution $x\geq 4$.
 Taking ‘one over’ on both sides of this inequality in order to get a solution range for $x$. Note that taking a reciprocal means that the inequality must be reversed. See the fourth bullet point above for more details. Hence, we obtain the solution $x\hspace{2pt}<\hspace{2pt}\frac{1}{7}$.