### Number lines

Suppose that you are told that x<4. This means that each possible value of x must be less than 4. Therefore, x can be 3, 2, 1, 0, -1, -2…….but that is not all.

3.2 is also less than 4, as is 3.999, 2.43, -3.4, -100……..

If we draw a number line, you can use an arrow to represent the set of numbers. This allows us to show the possible values of x so you can’t write them all down. Notice that the open circle above the four is not filled in. This symbol is used because it’s not possible for x to be equal to four.

Now if we suppose that $\displaystyle x\ge -2$. This tells us that $\displaystyle x$ can be greater than or equal to -2.You can show that $\displaystyle x$ can be equal to -2 by ‘filling in’ the circle above -2 on the number line. The following example show that more than one inequality symbol can appear in a question.

Example 1: Show the set of values that satisfy each of the following inequalities on a number line.

a) $\displaystyle x>3$

The values of $\displaystyle x$ have to be larger than 3. $\displaystyle x$ cannot be equal to 3, so do not fill in the circle. ‘Greater than’ means ‘to the right’ on the number line. b) $\displaystyle 4<y<8$

Notice that $\displaystyle y$ is now being used as the variable and this should be clearly labelled on your number line. Also to inequality symbols have been used. In fact there are two inequalities, and both must be satisfied.

$\displaystyle 4<y$ tells you that $\displaystyle y$ is greater than (but not equal) to 4.

$\displaystyle y<8$ tells you that $\displaystyle y$ is also less than (but not equal) to 8.

So $\displaystyle y$ lies between 4 and 8 (not inclusive) c) $\displaystyle -1.4<x\le 2.8$

This example has to inequalities that must be both satisfied. $\displaystyle x$ is greater than( but not equal to) -1.4, and $\displaystyle x$ is less than or equal to 2.8 d) List of all integers that satisfy the inequality $\displaystyle 4.2<x\le 10.4$

Here $\displaystyle x$ must be greater than, but not equal to 4.2. So the smallest possible value of $\displaystyle x$ is 5. x must also be less than or equal to 10.4. The largest that  $\displaystyle x$ can be is therefore 10.

So the possible values of $\displaystyle x$ are 5, 6, 7, 8, 9 or 10.

### Solving inequalities algebraically

Consider the inequality $\displaystyle 3x>6$.

Now suppose that $\displaystyle x=2$, than $\displaystyle 3x=6$ but this doesn’t quite satisfy the inequality! Any value that is larger than 2 will work however. For example

If $\displaystyle x=2.1$, then $\displaystyle 3x=6.3$, which is greater than 6.

In the same way that you could divide both sides of an equation by 3, both sides of the inequality can be divided by 3 to get the solution.

$\displaystyle 3x>6$

$\displaystyle \frac{{3x}}{3}>\frac{6}{3}$

$\displaystyle x>2$

Notice that this solution is a range of values of $\displaystyle x$ rather than a single value. Any value that is greater than 2 works!

In fact you can solve any linear inequality in much the same way as you would solve a linear equation, thought there are important exceptions. But you should always remember that what you do to one side of the inequality you must do to the other.

Example 2: Find the set of values of x for which each of the following inequalities holds.

a) $\displaystyle 3x-4<14$

$\displaystyle 3x-4<14$   Add 4 to both sides

$\displaystyle 3x<18$

$\displaystyle \frac{{3x}}{3}<\frac{{18}}{3}$Divide both sides by 3

So, $\displaystyle x<6$

b) $\displaystyle 4(x-7)\ge 16$

$\displaystyle 4(x-7)\ge 16$

$\displaystyle 4x-28\ge 16$   Expand the brackets

$\displaystyle 4x\ge 44$   Add 48 to both sides

$\displaystyle \frac{{4x}}{4}\ge \frac{{44}}{4}$  Divide both sides by 4

So, $\displaystyle x\ge 11$

$\displaystyle 4(x-7)\ge 16$   Notice that you can always solve this inequality by dividing both sides by 4 at the beginning:

$\displaystyle x-7\ge 4$    Divide both sides by 4

$\displaystyle x\ge 11$     Add 7 to both sides to get the same answer as before

### c) $\displaystyle 5x-3\le 2x+18$

$\displaystyle 5x-3\le 2x+18$

$\displaystyle 5x-3-2x\le 2x+18-2x$   Subtract 2x from both sides

$\displaystyle 3x-3\le 18$    Simplify

$\displaystyle 3x\le 21$     Add 3 to both sides

$\displaystyle x\le 7$      Divide both sides by 3

d) $\displaystyle 4-7x\le 53$

$\displaystyle 4-7x\le 53$

$\displaystyle 4\le 53+7x$    Add 7x to both sides

$\displaystyle -49\le 7x$    Subtract 53 from both sides

$\displaystyle x\ge -7$        Divide both sides by 7

Notice that the x is on the right-hand side of the inequality in this answer. This is perfectly acceptable. You can reverse the entire inequality to place the x on the left without changing its meaning, but you must remember to reverse the actual inequality symbol.

Be careful! You should be aware that there is one further rule to remember. Consider this inequality:

$\displaystyle 3-5x>18$

$\displaystyle -5x>15$

If you divide both sides of this by -5 it appears that the solution will be,

$\displaystyle x>-3$

This is satisfied by any value of $\displaystyle x$ that is greater than -3, for example -2,-1, 2, 4, 3.5, 10……

If you calculate the value of $\displaystyle 3-5x$ for each of this values you get 13, 8, -9, -14.5, -47……and not one of this works in the original inequality as they are smaller than 18.

But here is an alternate solution:

$\displaystyle 3-5x>18$

$\displaystyle 3>18+5x$

$\displaystyle -15>5x$

$\displaystyle -3>x$ or $\displaystyle x<-3$

This is a correct solution and the final answer is very similar to the ‘wrong’ one above. The only difference is that the inequality symbol has been reversed.

Don’t forget! If you multiply or divide both sides of an inequality by a negative number then you must reverse the direction of the inequality.

Example 3: Solve each of the following inequalities.

a) $\displaystyle \frac{{y+6}}{4}>9$

$\displaystyle \frac{{y+6}}{4}>9$

$\displaystyle y+6>36$Multiplyboth sides by 4

$\displaystyle y>30$   Subtract 6 from both sides

b) $\displaystyle \frac{1}{2}(x+5)\le 2$

$\displaystyle \frac{1}{2}(x+5)\le 2$

$\displaystyle x+5\le 4$   Multiply both sides by 2

$\displaystyle x\le -1$      Subtract 5 from both sides

c) $\displaystyle 6(n-4)-2(n+1)<3(n+7)+1$

$\displaystyle 6(n-4)-2(n+1)<3(n+7)+1$

$\displaystyle 6n-24-2n-2<3n+21+1$       Expand the brackets

$\displaystyle 4n-26<3n+22$          Simplify

$\displaystyle 4n-3n<22+26$   Group the terms by subtracting 3n and adding 26 to both sides

$\displaystyle n<48$

d) $\displaystyle \frac{r}{2}+\frac{1}{3}<2$

$\displaystyle \frac{r}{2}+\frac{1}{3}<2$

$\displaystyle \frac{{3r+2}}{6}<2$Add the fractions by finding the common  denominator

$\displaystyle 3r+2<12$Multiply both sides by 6

$\displaystyle 3r<10$    Subtract 2 from both sides

$\displaystyle r<\frac{{10}}{3}$  Divide both sides by 3

Note! If you can avoid negatives, by adding or subtracting terms, then try to do so.