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Work through this MyMaths activity to remember

how to find all the outcomes of combined events

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test your probability knowledge here.

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Test your understanding of probability with this self marking quiz.

This is level 1: calculating the probability of a single event occuring. Type in your answers as fractions in their lowest terms. You can earn a trophy if you get at least 7 questions correct.

There are 7 balls on a snooker table. Three of the balls are red, one is blue, one pink, one yellow and one white.

Description of Levels

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Level 1 - Calculating the probability of a single event occuring

Level 2 - Calculating the probability of a single event occuring from a two-way table

Level 3 - Calculating the probability of a single event occuring then deciding the expected number of times it will occur in a given number of trials

Level 4 - Finding the probability of a combined event occuring by drawing a table to represent the possibility space

Level 5 - Calculating probability of an event occuring by listing all the possible outcomes

Exam Style questions are in the style of GCSE or IB/A-level exam paper questions and worked solutions are available for Transum subscribers.

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Example

Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly.

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If you are not familiar with standard playing cards here is a guide.

There are 52 cards in a pack (usually called a deck). The cards are grouped into four suits of 13 cards each. these suits are called clubs, diamonds, hearts and spades. Click on a tab below to see the cards in each of these suits.

Hearts ♥Diamonds ♦Clubs ♣Spades ♠

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Test your understanding of probability with this self marking quiz.

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Instructions and a chance to practice

and check your answers.

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This covers ALL of trigonometry. VERY THOROUGH, but maybe watch in 5 minute chunks to break it down

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Maths Tutorial: Trigonometry SOH CAH TOA (trigonometric ratios)

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Some hints on solving inequalities - what to do BEFORE you show it on a number line.

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Sometimes we need to solve Inequalities like these:

Solving

Our aim is to have x (or whatever the variable is) on its own on the left of the inequality sign:

We call that "solved".

Example: x + 2 > 12

Subtract 2 from both sides:

x + 2 − 2 > 12 − 2

Simplify:

x > 10
Solved!
How to Solve

Solving inequalities is very like solving equations ... we do most of the same things ...

... but we must also pay attention to the direction of the inequality.

Direction: Which way the arrow "points"

Some things can change the direction!

Safe Things To Do

These things do not affect the direction of the inequality:

Add (or subtract) a number from both sides

Multiply (or divide) both sides by a positive number

Simplify a side

Example: 3x < 7+3

We can simplify 7+3 without affecting the inequality:

3x < 10

But these things do change the direction of the inequality ("<" becomes ">" for example):

Here are the details:

Adding or Subtracting a Value

We can often solve inequalities by adding (or subtracting) a number from both sides (just as in Introduction to Algebra), like this:

Example: x + 3 < 7

If we subtract 3 from both sides, we get:

x < 4

And that is our solution: x < 4

In other words, x can be any value less than 4.

What did we do?

We went from this:

To this:

x+3 < 7

x < 4

And that works well for adding and subtracting, because if we add (or subtract) the same amount from both sides, it does not affect the inequality

Example: Alex has more coins than Billy. If both Alex and Billy get three more coins each, Alex will still have more coins than Billy.

What If I Solve It, But "x" Is On The Right?

No matter, just swap sides, but reverse the sign so it still "points at" the correct value!

Example: 12 < x + 5

If we subtract 5 from both sides, we get:

7 < x

That is a solution!

But it is normal to put "x" on the left hand side ...

... so let us flip sides (and the inequality sign!):

x > 7

Do you see how the inequality sign still "points at" the smaller value (7) ?

And that is our solution: x > 7

Note: "x" can be on the right, but people usually like to see it on the left hand side.

Multiplying or Dividing by a Value

Another thing we do is multiply or divide both sides by a value (just as in Algebra - Multiplying).

But we need to be a bit more careful (as you will see).

Positive Values

Everything is fine if we want to multiply or divide by a positive number:

Example: 3y < 15

If we divide both sides by 3 we get:

y < 5

And that is our solution: y < 5

Negative Values

When we multiply or divide by a negative number
we must reverse the inequality.

Why?

Well, just look at the number line!

For example, from 3 to 7 is an increase,
but from −3 to −7 is a decrease.

−7 < −3

7 > 3

See how the inequality sign reverses (from < to >) ?

Let us try an example:

Example: −2y < −8

Let us divide both sides by −2 ... and reverse the inequality!

−2y< −8

y > 4

And that is the correct solution: y > 4

(Note that I reversed the inequality on the same line I divided by the negative number.)

So, just remember:

When multiplying or dividing by a negative number, reverse the inequality

Multiplying or Dividing by Variables

Here is another (tricky!) example:

Example: bx < 3b

It seems easy just to divide both sides by b, which gives us:
x < 3
... but wait ... if b is negative we need to reverse the inequality like this:
x > 3
But we don't know if b is positive or negative, so we can't answer this one!

To help you understand, imagine replacing b with 1 or −1 in the example of bx < 3b:

The answer could be x < 3 or x > 3 and we can't choose because we don't know b.

So:

Do not try dividing by a variable to solve an inequality (unless you know the variable is always positive, or always negative).

A Bigger Example

First, let us clear out the "/2" by multiplying both sides by 2.

Because we are multiplying by a positive number, the inequalities will not change.

x−3 < −10

Now add 3 to both sides:

x < −7

And that is our solution: x < −7

Two Inequalities At Once!

How do we solve something with two inequalities at once?

Example:

First, let us clear out the "/3" by multiplying each part by 3.

Because we are multiplying by a positive number, the inequalities don't change:

−6 < 6−2x < 12

Now subtract 6 from each part:

−12 < −2x < 6

Now divide each part by 2 (a positive number, so again the inequalities don't change):

−6 < −x < 3

Now multiply each part by −1. Because we are multiplying by a negative number, the inequalities change direction.

And that is the solution!

But to be neat it is better to have the smaller number on the left, larger on the right. So let us swap them over (and make sure the inequalities point correctly):

Summary

Many simple inequalities can be solved by adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own.

But these things will change direction of the inequality:

Multiplying or dividing both sides by a negative number

Swapping left and right hand sides

Don't multiply or divide by a variable (unless you know it is always positive or always negative)

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Sometimes we need to solve Inequalities like these: Our aim is to have x (or whatever the variable is) on its own on the left of the inequality sign: We call that "solved". Subtract 2 from both sides: x + 2 − 2 > 12 − 2 Simplify: x > 10 Solved!

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An example of showing inequalities on a number line. Do you know why one has an open circle and one has a closed circle at the start? Can you tell which way the arrow points by looking at the direction of the < symbol?

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How to find the biggest and smallest possible values of a number that has been rounded.

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Upper and lower bounds - Approximation - Edexcel - GCSE Maths Revision - Edexcel - BBC Bitesize

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Learn about and revise approximation using a range of rounding and estimation techniques with this BBC Bitesize GCSE Maths Edexcel study guide.

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GAME

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QUIZ

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A self marking exercise on the application of Pythagoras' Theorem.

Here are some questions which can be answered using Pythagoras' Theorem. You can earn a trophy if you get at least 9 questions correct. Each time you finish a question click the 'Check' button lower down the page to see if you got it right!

Description of Levels

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Level 1 - Finding the hypotenuse

Level 2 - Finding a shorter side

Level 3 - Mixed questions

Level 4 - Pythagoras coordinates

Level 5 - Pythagoras' Theorem exercise

Level 6 - Pythagoras' Theorem harder exercise

Exam Style questions requiring an application of Pythagoras' Theorem and trigonometric ratios to find angles and lengths in right-angled triangles.

Three Dimensions - Three dimensional Pythagoras and trigonometry questions

More on this topic including lesson Starters, visual aids, investigations and self-marking exercises.

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Curriculum Reference

Pythagoras' Theorem

The area of the square on the hypotenuse of a right angled triangle is equal to the sum of the areas of the squares on the two shorter sides.

You may have learned the theorem using letters to stand for the lengths of the sides. The corners (vertices) of the right-angled triangle is labelled with capital (upper case) letters. The lengths of the sides opposite them are labelled with the corresponding small (lower case) letters.

Alternatively the sides of the right-angled triangle may me named using the capital letters of the two points they span.

As triangle can be labelled in many different ways it is probably best to remember the theorem by momorising the first diagram above.

To find the longest side (hypotenuse) of a right-angled triangle you square the two shorter sides, add together the results and then find the square root of this total.

To find a shorter side of a right-angled triangle you subtract the square of the other shorter side from the square of the hypotenuse and then find the square root of the answer.

Example

AB ² = AC ² - BC ²
AB ² = 4.7 ² - 4.1 ²
AB ² = 22.09 - 16.81
AB ² = 5.28
AB = √5.28
AB = 2.3m (to one decimal place)

The diagrams aren't always the same way round. They could be rotated by any angle.

The right-angled triangles could be long and thin or short and not so thin.

Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly.

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A self marking exercise on the application of Pythagoras' Theorem. Here are some questions which can be answered using Pythagoras' Theorem. You can earn a trophy if you get at least 9 questions correct. Each time you finish a question click the 'Check' button lower down the page to see if you got it right!

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Very detailed - but done well!

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How to write numbers in standard form.wmv

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Shows the ELIMINATION METHOD on Page 1 and the GRAPHICAL METHOD on Page 2

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Equations that have more than one unknown can have an infinite number of solutions. For example, \(2x + y = 10\) could be solved by:

To be able to solve an equation like this, another equation needs to be used alongside it. That way it is possible to find the only pair of values that solve both equations at the same time. These are known as simultaneous equations .

An example of this is:

The unknowns of \(x\) and \(y\) have the same value in both equations. This fact can be used to help solve the two simultaneous equations at the same time and find the values of \(x\) and \(y\) .

Solving simultaneous equations by elimination

The most common method for solving simultaneous equations is the elimination method which means one of the unknowns will be removed from each equation. The remaining unknown can then be calculated. This can be done if the of one of the letters is the same, regardless of sign.

Solve the following simultaneous equations:

First, identify which unknown has the same coefficient. In this example this is the letter \(y\) , which has a coefficient of 1 in each equation.

Either add or subtract the two equations from each other to eliminate the letter \(y\) . In this example the equations will need to be subtracted from each other as \(y - y = 0\) .

If the equations were added together, then \(y + y = 2y\) , and so the letter \(y\) would not be eliminated.

\[\begin{array}{ccccc} 3x & + & y & = & 11 \\ - && - && - \\ 2x & + & y & = & 8 \\ = && = && = \\ x &&& = & 3 \end{array}\]

\[3~\mathbf{\times~3} + y = 11\]

Check the answers by substituting both values into the other original equation. If the equation balances, then the answers are correct:

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Simultaneous equations - Solving simultaneous equations - Edexcel - GCSE Maths Revision - Edexcel - BBC Bitesize

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Equations that have more than one unknown can have an infinite number of solutions. For example, \(2x + y = 10\) could be solved by: To be able to solve an equation like this, another equation needs to be used alongside it.