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Here we will learn about Venn diagrams, including what they are, what they represent, their uses and key set notation.

There are also Venn diagram worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you're still stuck.

What is a Venn diagram?

A Venn diagram is a diagrammatic representation of two or more sets.

For example,

Above are examples of a two set and a three set Venn diagram with the following features:

The universal set is a rectangle outlining the space in which all values within the smaller sets are found. The universal set is denoted using the symbol \xi.

The set A, shown using a circle and labelled A.

The set B, shown using a circle and labelled B.

The set C, shown using a circle and labelled C (Three set Venn diagram).

Set A and set B (and set C) overlap, showing the items which are in set A and in set B. This is called the intersection.

To analyse data using Venn diagrams, all of the values within each set must be correctly allocated into the correct part of the Venn diagram. The number of sets is usually outlined or logically deduced from the information provided.

● Constructing a Venn diagram

To construct a Venn diagram, we draw a region containing two or more overlapping circles (or ellipses), each representing a set, and fill in the relevant information that is either given, or can be logically deduced.

For example, this Venn diagram shows the set of numbers \xi=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} which have been sorted into factors of 10 (F) and even numbers (E).

All numbers 1-10 are represented on the Venn diagram.

The factors of 10 appear within F.

The even numbers appear within E.

The numbers that are both factors of 10 and even numbers appear in the intersection of F and E.

The numbers that are not factors of 10 or even numbers are outside of the circles.

Step-by-step guide: Constructing Venn diagrams

● Set notation

To describe a subset, we need to understand key symbols and set notation for different sets including the intersection of sets, the union of sets and the absolute complement of sets.

Here are some common examples of set notation and their meaning,

We can use combinations of these, for example

A^{\prime} \cup B^{\prime} \text { or }(A \cap B)^{\prime} .

Step-by-step guide: Venn diagram symbols

● Subsets

To calculate the number of items in a subset of a Venn diagram, we need to add together the frequencies of the required subset.

For example, this Venn diagram shows the number of people who own a cat (C) or a dog (D).

To find the total number of people who own a dog, we need to include everyone in set D.

The number of people who own a dog is 12+6=18 .

To find the total number of people
who own a dog or a cat, we need
to include everyone in set C or set D.

The number of people who own
a cat or a dog is 12+6+9=27 .

Step-by-step guide: Set notation

● Calculating probabilities

To calculate a Venn diagram probability, we need to know each frequency within a subset and the frequency of the larger set which the probability is conditioned on.

For example, this Venn diagram shows the number of people who like Chinese food (C) and Indian food (I).

We can find the probability that a person likes Chinese and Indian food.

In general, \text{probability}=\frac{\text{number of desired outcomes}}{\text{total number of outcomes}} .

Here, the probability that a person chosen at random likes Chinese and Indian food, written \text{P}(\text{C} \cap \text{I}), is \frac{34}{80} since 34 people like Chinese and Indian food out of 80 people in total.

Step-by-step guide: Venn diagram probability

See also: Venn diagram union and intersection

What do Venn diagrams include?

Venn diagram worksheet

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Venn diagram worksheet

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Example 1: constructing a two set Venn diagram

The set \xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \}. Draw a Venn diagram that shows the distribution of the values in the set \xi into the two following sets:

Prime Numbers

Factors of 12

Draw the rectangular universe.

2 Draw the overlapping circles within the universe and label.

The first circle should be labelled with the name of the set Prime Numbers. The second circle should be labelled Factors of 12.

3 Fill in the information for each subset within the Venn diagram.

Place each item in the universal set into the correct part of the Venn diagram.

Top tip: Go through each item in order so you don't miss any values.

As 1 is a factor of 12 and not a prime number, the number 1 would be placed in the right crescent of the set Factors of 12.

2 is a factor of 12 and it is a prime number. This would go in the overlapping part of the two circles (the intersection).

3 is a factor of 12 and it is a prime number so this would also go into the intersection of the two sets.

4 is a factor of 12 and not a prime number so that will go into the right crescent.

5 is a prime number but not a factor of 12 and so this would go into the left crescent.

So far we have the following Venn diagram,

Continuing this process, we get the final solution,

Example 2: constructing a three set Venn diagram

A research group divided 40 construction workers into three groups based on the following criteria,

A = {Aged over 30 years old}

B = {Builder}

C = {Owns a car}

10\% of the participants satisfy all three criteria.

\frac{1}{2} of the construction workers satisfy at least 2 of the three criteria.

Complete the following three circle Venn diagram to represent the distribution of the participants in the research study.

Draw the rectangular universe.

The universal set is already given and so we can move on to the next step.

Draw the overlapping circles within the universe and label.

The three sets have already been given and so we can move on to the next step.

Fill in the information for each subset within the Venn diagram.

As 10\% of participants are within all three sets, the very centre of the Venn diagram where all three sets overlap would contain the value,

(40 \div 100) \times 10=4 .

As \frac{1}{2} of the participants are in at least 2 of the three sets, 40\times\frac{1}{2}=20.

The missing value in the intersection between A and C is therefore 20-(2+4+5)=9.

The final missing value for set B is the number of remaining participants. This is equal to

40-(8+3+2+9+4+5+2)=7.

This gives us the completed Venn diagram.

Example 3: describing a subset of a two set Venn diagram

State the name of the subset that is the shaded region in the Venn diagram below,

Every value in set A and every value in set B are included in the shaded region.

Every value not in the set A or B is not included in the shaded region.

As every region within set A or set B is shaded, this is the union of set A and set B, written as A \cup B.

Solution: A \cup B

Example 4: set notation

A meteorologist recorded the daily weather patterns over the month of November.

The two factors that he wants to look at are

S={number of days with at least 4 hours of sunshine}

T={Temperature dropped below 0^{o}C between 8pm and 8am }

Below is a Venn diagram of his results.

Determine the frequency of the set S \cup T'.

The subset S \cup T' is the union of all of the values in the set S or all of the values not in the set T. This is the following shaded region:

Do not count any duplicated values twice. We only add each value within the two sets once, to get 8+3+2=13.

Solution: S \cup T' =13.

Example 5: probability

40 musicians were asked if they could play the Violin (V) or the Piano (P). The results are shown in the Venn diagram below.

Calculate the probability of selecting a musician at random that can only play the violin.

The subset of musicians that can only play the violin is the set,

Example 6: conditional probability

70 students were asked about whether they have B={biscuits} or C={crisps} for their break time snack. The results are shown in the Venn diagram below.

Calculate the probability of a student not having crisps given that they have biscuits.

We need to identify the subset which is the part of B that is not part of C,

In this case, we know the student has biscuits, so the population we are looking at is those in set B and we can ignore those not in set B.

The frequency of values in set B is 18+15 = 33.

Common misconceptions

Placing every item in a set in the intersection

The intersection is the overlap between two sets. If the item is only in Set A and not Set B, it must go in the subset that is just A, and not the intersection.

(This is the intersection of sets \mathrm{A} and \mathrm{B}^{\prime}. We write this as \mathrm{A} \cap \mathrm{B}^{\prime} and say A and not B, or just A)

Make sure that every value within all of the sets is written into the Venn diagram. Count the items or add the frequencies to check your answer.

The universal set contains all of the items within every set given. If an item does not go in either circle, the item is written in the outer region, still within the rectangular universe.

It is possible that a set contains no items, or is the empty set. The frequency should therefore be written as 0, and not left blank.

Sets incorrectly labelled

Take for example the following two sets,

☐ Odd numbers less than 10
☐ Prime numbers less than 10

The Venn diagram is drawn so that the two sets are given as,

whereas the Venn diagram should look like this,

Using probability terminology incorrectly

The set of A is written as A which consists of a list of items or a frequency.

The probability of A is written as P(A) and is a fraction. These two should not be confused.

For conditional probability, the denominator of the fraction will be a subset of the population, rather than the universal set.

The items that are placed in the intersection are sometimes incorrectly added twice, as it is assumed that the values are doubled as they belong to two sets. They only need to be counted once.

1. Shade the region in the Venn diagram that represents the set (A \cup B) \cap (A \cup C).
(1 mark)

(1)

2. (a) A marine life park is opening a new underwater world consisting of 16 different tanks.

\frac{1}{4} of the tanks have coral and fish in them, \frac{1}{2} of the tanks contain fish only, and 1 tank does not contain fish or coral.

Draw a Venn diagram to show the distribution of Fish and Coral in the new underwater world tanks.

(b) How many tanks contain Coral?
(5 marks)

(a)

4 written in the intersection of Fish and Coral
(1)
8 written in Fish only
(1)
3 written in Coral only
(1)
1 written in the universal set
(1) Solution:
(b)

4+3=7
(1)

3. (a) Members of a mountain rescue team were asking local walking groups whether they had any of the following emergency items: F = {First aid kit}, B = {Bivvy bag}, W = {Whistle}.

The results of their research were written into a Venn diagram:

How many walking groups had at least 2 of the three emergency items with them?

(b) Calculate P(F \cap B \cap W). Give your answer as a simplified fraction.
(4 marks)

(a)

9+12+11+2=34 groups
(1)
(b)

F \cap B \cap W =12
(1)
P(F \cap B \cap W) =\frac{12}{64}
(1)
\frac{3}{16}
(1)

Learning checklist

You have now learned how to:

A brief history of diagrams

John Venn (1834-1923) was a mathematician, logician and philosopher who introduced the Venn diagram which we use widely in modern day statistics, probability, set theory and computer science. Clarence Irving wrote about the Venn diagram in his book ' A survey of symbolic logic ' in 1918 after which Venn diagrams became more popular.

The Venn diagram below shows the relationship between the data set of Factors of 18, and the data set of Prime numbers less than 18.

Leonhard Euler (1707-1783) was a pioneering Swiss mathematician who has greatly influenced today's fundamental understanding of mathematics in areas such as fluid mechanics, complex analysis and calculus. An Euler diagram (sometimes referred to as Eulerian circles) is different to a Venn diagram because they only show relevant relationships, whereas a Venn diagram shows all possible relationships.

Take for example the set of 2D shapes and 3D shapes. These two sets cannot overlap as a 2D shape cannot be a 3D shape, however Quadrilaterals are a subset of 2D shapes and so they are contained entirely within the set of 2D shapes.

A Carroll diagram (named after the author and mathematician Lewis Carroll) is similar to a two-way table which categorises items within a set into smaller subsets.

For example, below is a Carroll diagram showing the Humans and Animals that did/not attend the tea party in the book 'Alice in Wonderland':

Note: A Venn diagram is named after John Venn and so 'Venn' should be capitalised (this is the same for Euler diagrams and Carroll diagrams as they are named after Leonhard Euler, and Lewis Carroll respectively).

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Free Venn diagram GCSE maths revision guide, including step by step examples, exam questions and free worksheet.

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\text{P}(G|H)=\dfrac{12}{14}

a) The first thing we need to do is draw a rectangle labelled \xi to represent all the year 9 students. Then we need to draw two circles which overlap, one labelled 'football' and the other labelled 'rugby'.

There is some simple information which we can input immediately. We are told that 14 students do not play rugby or football, so we can write '14' outside the circles, but still inside the rectangle. We are also told that 42 students play both sports, so we can write '42' in the intersection of the two circles.

The only information which we need to complete is the number of students who play football only and the number of students who play rugby only. Since there are 126 students in total, of which 14 play neither sport and of which 42 play both sports, we can work out how many students play football or rugby only:

126-42-14 = 70\text{ students}

Now that we know that a total of 70 students play only football or only rugby, we are now in a better position to work out exactly how many students fall into each category. We are told that the ratio of the number of students who play football only to the number of students who play rugby only is 7 : 3. This means that \frac{7}{10} of the 70 students play football only and \frac{3}{10} of the 70 students play rugby only. (We are dealing in tenths here because the sum of the ratio is 10.).

The number of students who play football only can be calculated as follows:

\dfrac{7}{10}\times70 = 49\text{ students}

The number of students who play rugby only can be calculated as follows:

\dfrac{3}{10}\times70 = 21\text{ students}

Therefore we need to insert '49' inside the football circle (but not in the intersection) and '21' inside the rugby circle (but not in the intersection).

Our completed Venn diagram should be similar to the below:

b) We know that 21 students play rugby only out of the total 126 students. We can write this as the following fraction:

\dfrac{21}{126}

This fraction can be simplified to:

\dfrac{1}{6}

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a) The number of students in year 10 is the total of:

All the students that study geography only (40)

All the students that study history only (52)

All the students that study both geography and history (14)

All the students that study neither history nor geography (36)

40+52+14+36=142\text{ students}

b) We know that there are 142 students in total and, of these, 14 study both history and geography. We can express this as the following fraction:

\dfrac{14}{142}

This fraction can be simplified to:

\dfrac{7}{71}

(You could give the probability as a decimal or as a percentage (the question doesn't state how the answer should be given), but since both the decimal (0.0985....) and the percentage (9.85...\%) have several decimal places, leaving the probability as a fraction would probably be the most sensible option here, but the decimal and percentage answers are not incorrect.)

c) The phrase 'given that the student selected does not study geography' tells us that we need to work out how many students do not study geography. The students that don't study geography are those who study history only (52 students) or those who study neither subject (36 students). The total that do not study geography is therefore 52+36 = 88 students.

Of these 88 students, we need to find how many do not study history. The 36 that study neither subject are the ones that do not study history, so 36 out of the 88 who do not study geography do not study history. We can write this as the following fraction:

\dfrac{36}{88}

This fraction can be simplified to:

\dfrac{18}{44}

This fraction can be simplified again to:

\dfrac{9}{22}

(The decimal answer of 0.409..... or the percentage answer of 40.909....\% are both acceptable ways to express the probability also.)

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a) What need to input the data. Some of the data we are going to input will be easy, other parts may be more confusing. Let's start with one fact at a time:

All \bf{80} students like at least \bf{1} of the animals. There is not much we can do with this information. However, this is a key piece of information since it tells us that once we have filled in the Venn diagram, the numbers should add up to 80.

\bf{15} students like all \bf{3} animals. If 15 students like all 3 animals, then we can insert '15' in the intersection of all 3 circles.

\bf{14} students like sharks and crocodiles but do not like hippos. With this information we can insert '14' in the section where sharks and crocodiles intersect (which should be outside the hippos circle).

\bf{23} students like crocodiles and hippos. This is the piece of information which can cause confusion. Although that statement says that 23 students like crocodiles and hippos, this doesn't mean that 23 students like crocodiles and hippos only. This figure of 23 also includes students who like sharks as well as crocodiles and hippos. This is the slightly confusing part of the question! 15 students like sharks, crocodiles and hippos, so that means that these 15 students definitely like crocodiles and hippos. So, if there are 23 students in total who like crocodiles and hippos, that means that there are 8 students who like crocodiles and hippos only. Therefore we can now insert '8' in the intersection of crocodiles and hippos (but not in the intersection of sharks, crocodiles and hippos). You may need to read this paragraph a few times: it will sink in and make sense eventually!

\bf{21} students like sharks and hippos. This is a very similar piece of information to the above piece of information about crocodiles and hippos. 15 students like sharks, crocodiles and hippos, therefore that means that there are at least 15 students who like sharks and hippos. Since we have been told that the total number of students who like sharks and hippos is 21, then this means that 6 students like sharks and hippos, but not crocodiles. Therefore we can insert '6' in the intersection of sharks and hippos.

\bf{44} students like crocodiles. We know that 14 students like crocodiles and sharks, that 15 students like sharks, crocodiles and hippos, and that a further 8 that like crocodiles and hippos, so if we add these numbers up we know that 37 students like crocodiles. Since we are told that 44 students like crocodiles in total, this means that there are 7 students who like crocodiles only. Therefore we can insert '7' in the part of the crocodile circle that does not overlap with any other section.

\bf{12} students only like sharks. All we need to do is insert '12' in the sharks only section.

All we need to do is work out how many students only like hippos. We know that there are 80 students in total, and we have filled in every section of the Venn diagram except this final hippos only section. If we subtract all of the numbers from our total of 80, we will be able to work out how many students only like hippos:

80-12-14-7-6-15-8=18

Therefore our completed Venn diagram will look similar to the below:

b) There are 80 students in total, of which 18 only like hippos. This means that 18 out of the 80 only like hippos which can be written as the following fraction:

\dfrac{18}{80}

This fraction can be simplified to:

\dfrac{9}{40}

So, the probability of a person being selected at random and only liking hippos is \frac{9}{40} (or 9\div40=0.225 if you prefer to give your answer as a decimal. The answer of 22.5\% is also acceptable.)

c) There are 44 students in total that like crocodiles. There are 14 students that like crocodiles and sharks and 8 students that like crocodiles and hippos, so there are 22 students that like crocodiles and one other animal.

This means that 22 out of the 80 only like crocodiles and one other type of animal, which can be written as the following fraction:

\dfrac{22}{80}

This fraction can be simplified to:

\dfrac{11}{40}

So, the probability of a person being selected at random and only liking hippos is \frac{11}{40} or 11\div40=0.275 if you prefer to give your answer as a decimal. The answer of 27.5\% is also acceptable.

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Venn Diagrams Worksheets, Questions and Revision

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Venn Diagram questions, revision notes and worksheets. Maths Made Easy, making your Venn Diagram revision more effective.

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Let's do a little bit of probability with playing cards. And for the sake of this video, we're going to assume that our deck has no jokers in it. You could do the same problems with the joker, you'll just get slightly different numbers. So with that out of the way, let's first just think about how many cards we have in a standard playing deck. So you have four suits, and the suits are the spades, the diamonds, the clubs, and the hearts. You have four suits and then in each of those suits you have 13 different types of cards-- and sometimes it's called the rank. You have the ace, then you have the two, the three, the four, the five, the six, seven, eight, nine, ten, and then you have the Jack, the King, and the Queen. And that is 13 cards. So for each suit you can have any of these-- you can have any of the suits. So you could have a Jack of diamonds, a Jack of clubs, a Jack of spades, or a Jack of hearts. So if you just multiply these two things-- you could take a deck of playing cards, take out the jokers and count them-- but if you just multiply this you have four suits, each of those suits have 13 types. So you're going to have 4 times 13 cards, or you're going to have 52 cards in a standard playing deck. Another way you could have said, look, there's 13 of these ranks, or types, and each of those come in four different suits-- 13 times 4. Once again, you would have gotten 52 cards. Now, with that of the way, let's think about the probabilities of different events. So let's say I shuffle that deck. I shuffle it really, really well and then I randomly pick a card from that deck. And I want to think about what is the probability that I pick a Jack. Well, how many equally likely events are there? Well, I could pick any one of those 52 cards. So there's 52 possibilities for when I pick that card. And how many of those 52 possibilities are Jacks? Well you have the Jack of spades, the Jack of diamonds, the Jack of clubs, and the Jack of hearts. There's four Jacks in that deck. So it is 4 over 52-- these are both divisible by 4-- 4 divided by 4 is 1, 52 divided by 4 is 13. Now, let's think about the probability. So I'll start over. I'm going to put that Jack back and I'm going to reshuffle the deck. So once again, I still have 52 cards. So what's the probability that I get a hearts? What's the probability that I just randomly pick a card from a shuffled deck and it is a heart? Well, once again, there's 52 possible cards I could pick from. 52 possible, equally likely events that we're dealing with. And how many of those have our hearts? Well, essentially 13 of them are hearts. For each of those suits you have 13 types. So there are 13 hearts in that deck. There are 13 diamonds in that deck. There are 13 spades in that deck. There are 13 clubs in that deck. So 13 of the 52 would result in hearts, and both of these are divisible by 13. This is the same thing as 1/4. One in four times I will pick it out, or I have a one in four probability of getting a hearts when I randomly pick a card from that shuffled deck. Now, let's do something that's a little bit more interesting, or maybe it's a little obvious. What's the probability that I pick something that is a Jack-- I'll just write J-- and it is a hearts? Well, if you are reasonably familiar with cards you'll know that there's actually only one card that is both a Jack and a heart. It is literally the Jack of hearts. So we're saying, what is the probability that we pick the exact card, the Jack of hearts? Well, there's only one event, one card, that meets this criteria right over here, and there's 52 possible cards. So there's a one in 52 chance that I pick the Jack of hearts-- something that is both a Jack and it's a heart. Now, let's do something a little bit more interesting. What is the probability-- you might want to pause this and think about this a little bit before I give you the answer. What is the probability of-- so I once again, I have a deck of 52 cards, I shuffled it, randomly pick a card from that deck-- what is the probability that that card that I pick from that deck is a Jack or a heart? So it could be the Jack of hearts, or it could be the Jack of diamonds, or it could be the Jack of spades, or it could be the Queen of hearts, or it could be the two of hearts. So what is the probability of this? And this is a little bit more of an interesting thing, because we know, first of all, that there are 52 possibilities. But how many of those possibilities meet these conditions that it is a Jack or a heart. And to understand that, I'll draw a Venn diagram. Sounds kind of fancy, but nothing fancy here. So imagine that this rectangle I'm drawing here represents all of the outcomes. So if you want, you could imagine it has an area of 52. So this is 52 possible outcomes. Now, how many of those outcomes result in a Jack? So we already learned, one out of 13 of those outcomes result in a Jack. So I could draw a little circle here, where that area-- and I'm approximating-- represents the probability of a Jack. So it should be roughly 1/13, or 4/52, of this area right over here. So I'll just draw it like this. So this right over here is the probability of a Jack. There's four possible cards out of the 52. So that is 4/52, or one out of 13. Now, what's the probability of getting a hearts? Well, I'll draw another little circle here that represents that. 13 out of 52 cards represent a heart. And actually, one of those represents both a heart and a Jack. So I'm actually going to overlap them, and hopefully this will make sense in a second. So there's actually 13 cards that are a heart. So this is the number of hearts. And actually, let me write this top thing that way as well. It makes it a little bit clearer that we're actually looking at the number of Jacks. And of course, this overlap right here is the number of Jacks and hearts-- the number of items out of this 52 that are both a Jack and a heart-- it is in both sets here. It is in this green circle and it is in this orange circle. So this right over here-- let me do that in yellow since I did that problem in yellow-- this right over here is a number of Jacks and hearts. So let me draw a little arrow there. It's getting a little cluttered, maybe I should draw a little bit bigger number. And that's an overlap over there. So what is the probability of getting a Jack or a heart? So if you think about it, the probability is going to be the number of events that meet these conditions, over the total number events. We already know the total number of events are 52. But how many meet these conditions? So it's going to be the number-- you could say, well, look at the green circle right there says the number that gives us a Jack, and the orange circle tells us the number that gives us a heart. So you might want to say, well, why don't we add up the green and the orange, but if you did that, you would be double counting, Because if you add it up-- if you just did four plus 13-- what are we saying? We're saying that there are four Jacks and we're saying that there are 13 hearts. But in both of these, when we do it this way, in both cases we are counting the Jack of hearts. We're putting the Jack of hearts here and we're putting the Jack of hearts here. So we're counting the Jack of hearts twice, even though there's only one card there. So you would have to subtract out where they're common. You would have to subtract out the item that is both a Jack and a heart. So you would subtract out a 1. Another way to think about it is, you really want to figure out the total area here. And let me zoom in-- and I'll generalize it a little bit. So if you have one circle like that, and then you have another overlapping circle like that, and you wanted to figure out the total area of both of these circles combined, you would look at the area of this circle. And then you could add it to the area of this circle. But when you do that, you'll see that when you add the two areas, you're counting this area twice. So in order to only count that area once, you have to subtract that area from the sum. So if this area has A, this area is B, and the intersection where they overlap is C, the combined area is going to be A plus B-- -- minus where they overlap-- minus C. So that's the same thing over here, we're counting all the Jacks, and that includes the Jack of hearts. We're counting all the hearts, and that includes the Jack of hearts. So we counted the Jack of hearts twice, so we have to subtract 1 out of that. This is going to be 4 plus 13 minus 1, or this is going to be 16/52. And both of these things are divisible by 4. So this is going to be the same thing as, divide 16 by 4, you get 4. 52 divided by 4 is 13. So there's a 4/13 chance that you'd get a Jack or a hearts.

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Probability with Venn diagrams (video) | Khan Academy

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Let's do a little bit of probability with playing cards. And for the sake of this video, we're going to assume that our deck has no jokers in it. You could do the same problems with the joker, you'll just get slightly different numbers.

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Find the missing side length - SOHCAHTOA

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Find the missing side length - SOHCAHTOA

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Here we will learn about y = mx + c including how to recognise the gradient and y-intercept from the equation y = mx + c and rearrange an equation into the standard form of y = mx + c.

There are also y = mx + c worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you're still stuck.

What is y=mx+c?

The equation y = mx + c is the general equation of any straight line where m is the gradient of the line (how steep the line is) and c is the y-intercept (the point in which the line crosses the y-axis).

y = mx + c is a linear equation and the variables x and y relate to coordinates on the line.
When we input a value for x into the equation y = mx + c, we get a result for y.
This means that x is an independent variable, and y is a dependent variable as it is determined by the value of x.

E.g.

Let's look at the line y = 2x + 1.

This graph has a gradient of 2 and a y-intercept of 1, the coordinate (0,1).

The term linear is used to describe a straight line where the variables are raised to a power no higher than 1

If the variables are raised to a power no higher than 2 we refer to it as a quadratic

If the variables are raised to a power no higher than 3 we refer to it as a cubic, and so on.

The gradient 'm'

The gradient of the line tells us how steep the line is.

We use the letter m to denote the gradient.

Imagine climbing a ladder. If the ladder is really close to the wall, the gradient of the ladder is really steep (you would almost be climbing vertically). Taking the base of the ladder away from the wall means that the gradient of the slope decreases reaching a lower point on the wall.
The bigger the gradient the steeper the line.

E.g.
A gradient of 2 will generate a line that is much steeper than the gradient of {\frac{1}{3}} .

See the diagram below.

The gradient of a line can be many different types of number, i.e. fractions, decimals, negatives etc.

The y-intercept 'c'

The y-intercept is the point of intersection between the straight line y = mx + c, and the y-axis. The y-intercept is the value of y when x = 0.

E.g.
Let's look at the equation y = 5x + 7.

To find the y intercept we substitute x = 0 into the equation.

\begin{aligned} &y=(5 \times 0)+7\\\\ &y=7 \end{aligned}

So when x = 0, y = 7, giving us the coordinate (0,7).

This is the y-intercept.

Here is a quick summary of some equations in the form y = mx + c with the gradient and y intercept highlighted.

y=mx+c

y=2x+4

y=6x-3

y=-4x

y=6-x

gradient m

2

6

-4

-1

y -intercept c

4

-3

0 (the origin)

6

How to use y=mx+c

In order to state the gradient and y-intercept of a line:

1. Rearrange the equation to make y the subject

2. Substitute x = 0 into the equation to find the y-intercept

3. State the coefficient of x (the gradient)

Explain how to use y=mx+c

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Related lessons on straight line graphs

y=mx+c is part of our series of lessons to support revision on straight line graphs. You may find it helpful to start with the main straight line graphs lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

Example 1: y=mx+c form

State the gradient and y-intercept of the line y = −3x + 8.

Rearrange the equation to make y the subject

The equation y = −3x + 8 is already in the general form of y=mx+c so we can progress to step 2 straight away.

2 Substitute x = 0 into the equation to find the y-intercept

When x = 0,

\begin{aligned} &y=(-3\times{0})+8\\\\ &y=0+8\\\\ &y=8 \end{aligned}

The y-intercept of the line y = −3x + 8 is 8, or c = 8.

3 State the coefficient of x (the gradient)

The coefficient of x is −3.

The gradient of the line y = −3x + 8 is −3, or m = −3.

Solution: m = −3, c = 8

Example 6: changing the subject including decimals

State the gradient and y-intercept of the line

\[x=\frac{y+0.85}{0.2}.\]

State the coefficient of x (the gradient)

The coefficient of x is 0.2.

The gradient of the line

is 0.2, or m = 0.2.

Common misconceptions

Incorrect inverse operation

When rearranging equations, instead of applying the inverse operation to the value being moved, the value is simply moved to the other side of the equals sign.

E.g.
y + 5 = 2x is rearranged to make y = 2x + 5 ✘

Stating the value of m and c

A common error is to incorrectly state the values of m and c as a result of not rearranging the equation so that it is in form of y=mx+c.
Take example 3 above when x = y + 10. The gradient would be stated as 1 as this is the coefficient of x however the value of the y-intercept would be incorrectly stated as 10.

The correct answer for the y intercept is −10 as we can rearrange the equation to make y the subject. See below.

Mixing up the gradient and the y-intercept

Take, for example, the equation y = 10 + 3x. Here, the coefficient of x is 3 and the value of y when x = 0 is 10. This means that the gradient is 3 and the y-intercept is 10, not the other way around.

Practice y=mx+c questions

y=mx+c GCSE questions

Substitute x=3 and y=4 into y=3x+c to get

4=(3\times3)+c (1mark)
(1)
4=9+c so

c=-5 (1mark)
(1)

2. (a) The coordinate A=(0,2) lies on a straight line. The gradient of the line is 5 . Using this information, state the equation of the straight line.

(b) Write the equation of a line that is parallel to a) in the form y=mx+c .
(4 Marks)

(a)

A is the y -intercept so c=2 or when x=0, y=2 so c=2
(1)
y=5x+c (1mark)
(1)
y=5x+2 (1mark)
(1)
(b)

y=5x+c where
c ≠ 2 (1mark)
(1)

3. Show m=2 for the straight line 8x-4y=12.
(3 Marks)

(3)

Learning checklist

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Free y=mx+c GCSE maths revision including step by step guide and examples, plus a free y=mx+c worksheet and exam questions.