I am a medical student at the university of St Andrews. I have always loved maths and sciences, and hope to share that passion with you too! I am patient and very friendly, so hopefully you will enjoy our sessions and get the most out of them. Of course, **you will guide which topics we cover **in our sessions.

With maths and science the most important thing is that you completely understand the ideas and concepts, so I will use as many different ways to try and help you learn (diagram, graphs, etc.) until you feel comfortable enough with the material that you can explain it to me. Once you've grasped one idea, we can build on it with the next one! I hope that the **sessions will be very fun**!

We can cover a lot of material, especially if it's exciting. **Maths is fun**, and I hope to show you that! If you have any quedstions at all, please feel free to book a 'Meet the Tutor Session'. **Make sure that you tell me your exam board and what topics you'd like to go through.**

**I look forward to meeting you!**

I am a medical student at the university of St Andrews. I have always loved maths and sciences, and hope to share that passion with you too! I am patient and very friendly, so hopefully you will enjoy our sessions and get the most out of them. Of course, **you will guide which topics we cover **in our sessions.

With maths and science the most important thing is that you completely understand the ideas and concepts, so I will use as many different ways to try and help you learn (diagram, graphs, etc.) until you feel comfortable enough with the material that you can explain it to me. Once you've grasped one idea, we can build on it with the next one! I hope that the **sessions will be very fun**!

We can cover a lot of material, especially if it's exciting. **Maths is fun**, and I hope to show you that! If you have any quedstions at all, please feel free to book a 'Meet the Tutor Session'. **Make sure that you tell me your exam board and what topics you'd like to go through.**

**I look forward to meeting you!**

Everybody learns differently. I completely understand that, so make sure that every tutoring session that we do is tailored more towards which way of learning best suits you. Whether it's lots of questions, or diagrams, or silly drawings and pnemonics (I find these work best!), my aim is to make the session as best suited as possible for you, so we can have fun and learn together.

I've always found that you are best learning the basics and then moving on to more intricate topics, but that you should always recap over the basics to make sure you have completely grasped it. So that would be the approach I'd take, where with every session we have, we would briefly recap over everything else that has come before, to build your confidence, and show you how much you really know!

Most importantly, it doesn't need to be dull, **learning can be fun!**

Everybody learns differently. I completely understand that, so make sure that every tutoring session that we do is tailored more towards which way of learning best suits you. Whether it's lots of questions, or diagrams, or silly drawings and pnemonics (I find these work best!), my aim is to make the session as best suited as possible for you, so we can have fun and learn together.

I've always found that you are best learning the basics and then moving on to more intricate topics, but that you should always recap over the basics to make sure you have completely grasped it. So that would be the approach I'd take, where with every session we have, we would briefly recap over everything else that has come before, to build your confidence, and show you how much you really know!

Most importantly, it doesn't need to be dull, **learning can be fun!**

Enhanced DBS Check

28/10/20155from 16 customer reviews

Sheila (Parent)

January 14 2018

Great!

James (Parent)

August 28 2017

Excellent tutorial William, thank you.

Rory (Student)

January 14 2018

Helen (Parent)

January 14 2018

Right, so the **Young's modulus** of a material is a measure of it's rigidity. The higher the value of the Young's modulus, the more rigid the material is.

Ok, so the Young's modulus, E, is equal to the **stress **that the material experiences divided by the **strain **experienced by the material.

The **stress, σ**, of a material is equal to the force, F, applied to this material, divided by the area, A, upon which this force is applied. The equation for this is:

**σ = F/A**

On the other hand, the **strain, ε**, of the material is equal to the **extension **of the material, **e** (how much the material has extended by having this force applied to it), divided by the original **length, l**, of the material. The equation for strain is

**ε = e/l**

The Young's modulus is equal to the stress divided by the strain, so:

**E= σ/ε = (F/A)/(e/l) = (F*L)/(A*e)**

**So, E = (F*l)/(A*e)**

For example, If a metal wire of original length 2cm

Right, so the **Young's modulus** of a material is a measure of it's rigidity. The higher the value of the Young's modulus, the more rigid the material is.

Ok, so the Young's modulus, E, is equal to the **stress **that the material experiences divided by the **strain **experienced by the material.

The **stress, σ**, of a material is equal to the force, F, applied to this material, divided by the area, A, upon which this force is applied. The equation for this is:

**σ = F/A**

On the other hand, the **strain, ε**, of the material is equal to the **extension **of the material, **e** (how much the material has extended by having this force applied to it), divided by the original **length, l**, of the material. The equation for strain is

**ε = e/l**

The Young's modulus is equal to the stress divided by the strain, so:

**E= σ/ε = (F/A)/(e/l) = (F*L)/(A*e)**

**So, E = (F*l)/(A*e)**

For example, If a metal wire of original length 2cm

Ok, so we would call this a quadratic equation because it is written in the form of ax^2+bx+c=0 (in our case, a=1, b=6, c=8).

Luckily, this type of quadratic equation can be factorised, so we can solve it easily!

Ok, so we're trying to factorise x^{2}+6x+8=0 so that it is in the form of (x+p)(x+q)=0.

Let's try expanding (x+p)(x+q)=0.

If you multiply the brackets together, you're left with:

x^{2}+px+qx+pq=0

We can tidy this up a little bit to give us:

x^{2}+(p+q)x+pq=0

This looks very similar to x^{2}+6x+8=0, doesn't it?

Yes, it does! If we compare these two equations, we find out two things:

p+q=6

pq=8

So we're looking for two numbers which when they are multiplied by each other will give 8, and when they are added together will give 6.

What we're left with is that p=4 and q=2 (or the other way round, it doesn't really matter).

Let's plug this back into our original equation: (x+p)(x+q)=0

Of course, now we have:

(x+4)(x+2)=0

This is much easier to solve than what we started with!

So in this case, either the first bracket is equal to 0 or the second bracket is equal to zero - this gives us two solutions for x.

Either x+4=0, meaning that x=-4

Or x+2=0, meaning that x=-2

So now we have the answers x=-2 or x=-4

Of course a quicker way to do this would be to look at x^{2}+6x+8=0 and to find two numbers that are factors of 8 (they multiply together to make 8) and also add together to make 6.

Ok, so we would call this a quadratic equation because it is written in the form of ax^2+bx+c=0 (in our case, a=1, b=6, c=8).

Luckily, this type of quadratic equation can be factorised, so we can solve it easily!

Ok, so we're trying to factorise x^{2}+6x+8=0 so that it is in the form of (x+p)(x+q)=0.

Let's try expanding (x+p)(x+q)=0.

If you multiply the brackets together, you're left with:

x^{2}+px+qx+pq=0

We can tidy this up a little bit to give us:

x^{2}+(p+q)x+pq=0

This looks very similar to x^{2}+6x+8=0, doesn't it?

Yes, it does! If we compare these two equations, we find out two things:

p+q=6

pq=8

So we're looking for two numbers which when they are multiplied by each other will give 8, and when they are added together will give 6.

What we're left with is that p=4 and q=2 (or the other way round, it doesn't really matter).

Let's plug this back into our original equation: (x+p)(x+q)=0

Of course, now we have:

(x+4)(x+2)=0

This is much easier to solve than what we started with!

So in this case, either the first bracket is equal to 0 or the second bracket is equal to zero - this gives us two solutions for x.

Either x+4=0, meaning that x=-4

Or x+2=0, meaning that x=-2

So now we have the answers x=-2 or x=-4

Of course a quicker way to do this would be to look at x^{2}+6x+8=0 and to find two numbers that are factors of 8 (they multiply together to make 8) and also add together to make 6.

Ok, so here are the equations

5x+3y=24

3x-4y=26

So let's multiply the first equation by 3, which gives us:

15x+9y=72

Now let's multiply the second equation by 5, which gives us:

15x-20y=130

So we're now left with:

15x+9y=72

15x-20y=130

Let's rearrange both of these equations to make 15x the subject. So now we're left with:

15x = 72-9y

15x= 130+20y

Now we can compare these two equations, to give us:

72-9y=130+20y (=15x)

If we rearrange this new equation, we find that:

20y + 9y = 72 - 130

29y = -58

y = -2

Since we now have a value for y, we can substitute this back into 5x + 3y = 24

5x + 3(-2) = 24

5x -6 = 24

5x = 30

x = 6

So, are final answer is x = 6 and y = -2

Ok, so here are the equations

5x+3y=24

3x-4y=26

So let's multiply the first equation by 3, which gives us:

15x+9y=72

Now let's multiply the second equation by 5, which gives us:

15x-20y=130

So we're now left with:

15x+9y=72

15x-20y=130

Let's rearrange both of these equations to make 15x the subject. So now we're left with:

15x = 72-9y

15x= 130+20y

Now we can compare these two equations, to give us:

72-9y=130+20y (=15x)

If we rearrange this new equation, we find that:

20y + 9y = 72 - 130

29y = -58

y = -2

Since we now have a value for y, we can substitute this back into 5x + 3y = 24

5x + 3(-2) = 24

5x -6 = 24

5x = 30

x = 6

So, are final answer is x = 6 and y = -2