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Mr Barton’s Maths Notes
Algebra
9. Inequalities
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9. Inequalities
What are Inequalities?
Inequalities are just another time-saving device invented by lazy mathematicians.
They are a way of representing massive groups of numbers with just a couple of numbers and a
fancy looking symbol.
Good News: So long as you can solve equations and draw graphs, you already have all the skills
you need to become an expert on inequalities!
1. What those funny looking symbols mean
< means “is less than”
 means “is less than or equal to”
> means “is greater than”
 means ”is greater than or equal to”
For Example:
1. x < 5
Means x is less than 5
So x could be 4, 0.6, -23… but NOT 5!
2. p  100
Means p is greater than or
equal to 100
So p could be 104, 10000, 201.5… AND 100!
3. m > -2
Means x is greater than -2
So x could be -1.9, 0, 4.3… but NOT -2!
2. Representing Inequalities on a Number line
These are very common questions, and pretty easy ones too.
Method:
Draw a line over all the numbers for which the inequality is true (the ones you can see, anyway)
At the end of these lines, draw a circle, and colour it in if the inequality can equal the number,
and leave it blank if it cannot.
•
•
x  -2
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-4
-3
-2
-1
0
1
2
3
4
5
6
0
1
2
3
4
5
6
x<3
-6
-5
x > -4
-6
-5
x≤0
and
-4
-3
-2
-1
3. Solving Linear Inequalities
Good News: The rule for solving linear inequalities is exactly the same as that for solving linear
equations – whatever you do to one side of the inequality, do exactly the same to the other
Just one thing: if you multiply or divide by a NEGATIVE, the inequality sign swaps around!
Why on earth does the sign swap around?
Imagine you have the inequality that says: 8 is greater than 5:
Let’s multiply both sides by 4…
32 > 20
x4
Now, let’s divide both sides by -2…
÷ -2
-16 > -10
8 > 5
which is still true!
which is NOT true!
And the only way to make the inequality true is to switch the sign around!...
Example 1
6x  3  27
1. Okay, so just like when solving equations, we unwrap our
unknown letter, by thinking about what was the last thing
done to it, and doing the inverse to both sides!
2. Just need to divide both sides by 6, and we have our
answer… and because 6 is positive, no need to swap any signs
around!
-16 < -10
6x  3  27
-3
÷6
6x  24
x  4
Example 2
5x  6  2 x  9
1. Again, we do exactly the same as we would if this was an
equation. Start by collecting all your x’s on the side which
starts off with the most x’s!
5x  6  2 x  9
- 2x
3x  6  9
2. Now we have a nice easy inequality to unwrap!
+6
3x  15
3. Which gives us our answer
÷3
Example 3
2(5 x  4)  98
2(5 x  4)  98
10x  8  98
1. Let’s get those brackets expanded, being extremely
careful with our negative numbers!
2. Now we begin to unwrap!
3. Notice here that we are dividing by a negative number,
and so we must make sure we remember to switch our
inequality sign around!
x  5
-8
÷ -10
10x > 90
x  9
Example 4
4  3x  5  8
4  3x  5  8
1. This looks complicated, but all you are trying to do is
unwrap the unknown letter in the middle, and whatever you
do the middle, you must also do to both ends!
-5
9  3x  3
2. Careful unwrapping gives us our answer:
÷3
3  x  1
3. But what does that mean?... Well, it may become clearer
when written like this:
x  3
and
x  1
So, x must be greater than -3 and less than or equal to 1, so
x must be between -3 and 1!
4. Solving Linear Inequalities Graphically
The examiners love asking these ones, and my pupils hate doing them!
Basically, you are given one or more inequalities and you are asked to show the region on a graph
which satisfies them all (i.e. every inequality works for every single point in your region)
Now, before we go on, I am going to assume you are an expert on drawing straight line graphs.
If this is not the case, read Graphs 1. Straight Line Graphs before carrying on…
Method
1. Pretend the inequality sign is an equals sign and just draw your line
2. Look at the inequality sign and decide whether your line is dashed or solid
3. Pick a co-ordinate on either side of the line to help decide which region you want
e.g. 1
x  2
1. Draw the line x = 2
2. Notice it is a solid line
as x CAN be 2
e.g. 2
y  2x 1
1. Draw the line y = 2x - 1
2. Notice it is a dashed
line as y CANNOT be
2x - 1
3. Choose a co-ordinate
on one side of the line:
e.g. (-3, 1).
x  3
x  2
y 1
3  2 √
So our point is on the
side of the line we want!
3. Choose a co-ordinate
on one side of the line:
e.g. (4, 2).
x4
y2
y  2x 1
4  8 1
4  7 x
So we want the other
side of the line!
e.g. 3
x  1
y  2
5 x  8 y  40
For questions like this, just deal with each inequality in turn, shading as you go!
Note: When you have got more than one inequality like this, it’s normally best to shade the region
you DON’T WANT, so you can leave the region you do want blank!
x  1
You should be able to do
this one all in one.
The points where x is
greater than 2 are to the
right, so shade the left!
5 x  8 y  40
1. Draw the line 5x+8y = 40
2. Notice it is a solid line as
5x+8y CAN be equal to 40
y  2
The big y values are all
above the line, so let’s
shade the ones we don’t
want below the line!
3. Choose a co-ordinate
on one side of the line:
e.g. (2, 1).
x2
y 1
5 x  8 y  40
10  8  40
18  7 √
So we shade the other
side
Putting it all together leaves us
the blank region in the middle
that satisfies all the inequalities!
5. Solving Quadratic Inequalities – warning, these are hard!
Now, I have a way of doing these which may be different to how you have been taught, so feel
free to completely ignore my method… but I must admit I think it’s pretty good!
How Mr Barton Solves Quadratic Inequalities
1. Do the same to both sides to make the quadratic inequality as simple as possible
2. Sketch the simplified quadratic inequality
3. Use the sketch to find the values which satisfy the inequality
Example
2 x 2  3  53
1. Okay, so let’s use our algebra skills to get this quadratic
inequality as simple as possible:
-3
÷2
2 x 2  3  53
2 x 2  50
x 2  25
2. Now, let’s think about what this inequality is saying: “we want all
the values of x where x2 is greater than our equal to 25”
Let’s sketch that!
y  x2
y  25
Now all we need to ask ourselves is: “for
what values of x is x2 bigger than 25?”...
Well, from our graph it looks like the answer
is when x is either bigger than 5 or smaller
than -5, which gives our answer:
x  5
or
x  5
Good luck with
your revision!