I am very passionate about the study of mathematics and enjoy breaking down complex problems into simple solutions whilst explaining the process of doing so. I adapt my teaching methods to the each student to try and maximise their grades and understanding. I am currently a student studying at Imperial College London who has a deep understanding in mathematics and I think that as a young student myself, I am able to relate to my students and offer help and advice through my tutoring.

Having been through exams like GCSE's and A-Levels recently, i feel like i will be able to relate to my students and provide them with the best technique necessary in order to maximise their grades.

My Qualifications:

GCSE's: 2A*s and 10As. Obtaining an A* in Mathematics.

A-Levels: 3A*s and an A. Obtaining an A* in Mathematics and another A* in Further Mathematics.

I am very passionate about the study of mathematics and enjoy breaking down complex problems into simple solutions whilst explaining the process of doing so. I adapt my teaching methods to the each student to try and maximise their grades and understanding. I am currently a student studying at Imperial College London who has a deep understanding in mathematics and I think that as a young student myself, I am able to relate to my students and offer help and advice through my tutoring.

Having been through exams like GCSE's and A-Levels recently, i feel like i will be able to relate to my students and provide them with the best technique necessary in order to maximise their grades.

My Qualifications:

GCSE's: 2A*s and 10As. Obtaining an A* in Mathematics.

A-Levels: 3A*s and an A. Obtaining an A* in Mathematics and another A* in Further Mathematics.

No DBS Check

There are two ways to approach this problem. In face there are two ways to approach every single set of simultanious equation. Firstly we can try eliminating terms. Or we can also try substituting terms. Lets take a look at both now.

Starting off with the elimination method. We seek to get the same number of a single unknown. May it be x or y. So in this example, lets try to get 2y in both equations as its the easier than obtaining the same number of x's. So we would obtain the following;

10x + 2y = 8 (we multiplied every term in the first equation by 2)

3x + 2y = 5

We can now subtract (3x + 2y = 5) from (10x + 2y = 8) to result in (7x + 0y = 3). This is done on a term by term basis. 10x - 3x, 2y - 2y and 8 - 5 would be the right calculations to do. Now we can see that 7x = 3. So x = 3/7. We can now use our algebra skills to substitute the known x into one of the equations earlier.

As we know that x = 3/7, we can now say 3(3/7) + 2y = 5. By using algebra, we can rearrange to obtain the following;

9/7 + 2y = 5 (Multiply 3 by 3/7)

2y = 16/5 (Subtract 9/7 from 5)

y = 8/5

Now lets take a look at the substitution method. For this method, we need to isolate one of the unknowns so that there is only of it. So for example rewrite 5x + y = 4 in such a way that we show x = something. Lets find it.

5x + y = 4

x + y/5 = 4/5 (divide every term by 5)

x = 4/5 - y/5 (take y/5 to the other side of the = sign)

Now we would need to substitute this expression into the other simultanious equation like so.

3x + 2y = 5 becomes

3(4/5 - y/5) + 2y = 5

12/5 - 3y/5 + 2y = 5

By futher simplifying and rearranging, we obtain that y = 8/5. Therefore we can now substitute the known y into one of the equatoins above to obtain x. Just like in the previous method. So substituting y = 8/5 into 3x + 2y = 5 would lead us to finding out that x = 3/7

That's how we find the solutions to simultanious equations.

There are two ways to approach this problem. In face there are two ways to approach every single set of simultanious equation. Firstly we can try eliminating terms. Or we can also try substituting terms. Lets take a look at both now.

Starting off with the elimination method. We seek to get the same number of a single unknown. May it be x or y. So in this example, lets try to get 2y in both equations as its the easier than obtaining the same number of x's. So we would obtain the following;

10x + 2y = 8 (we multiplied every term in the first equation by 2)

3x + 2y = 5

We can now subtract (3x + 2y = 5) from (10x + 2y = 8) to result in (7x + 0y = 3). This is done on a term by term basis. 10x - 3x, 2y - 2y and 8 - 5 would be the right calculations to do. Now we can see that 7x = 3. So x = 3/7. We can now use our algebra skills to substitute the known x into one of the equations earlier.

As we know that x = 3/7, we can now say 3(3/7) + 2y = 5. By using algebra, we can rearrange to obtain the following;

9/7 + 2y = 5 (Multiply 3 by 3/7)

2y = 16/5 (Subtract 9/7 from 5)

y = 8/5

Now lets take a look at the substitution method. For this method, we need to isolate one of the unknowns so that there is only of it. So for example rewrite 5x + y = 4 in such a way that we show x = something. Lets find it.

5x + y = 4

x + y/5 = 4/5 (divide every term by 5)

x = 4/5 - y/5 (take y/5 to the other side of the = sign)

Now we would need to substitute this expression into the other simultanious equation like so.

3x + 2y = 5 becomes

3(4/5 - y/5) + 2y = 5

12/5 - 3y/5 + 2y = 5

By futher simplifying and rearranging, we obtain that y = 8/5. Therefore we can now substitute the known y into one of the equatoins above to obtain x. Just like in the previous method. So substituting y = 8/5 into 3x + 2y = 5 would lead us to finding out that x = 3/7

That's how we find the solutions to simultanious equations.

So lets have a quick recap of differenciation to start off. In informal terms, differenciation is carried out by dropping the power to be in front of the term we are differenciating to as well as decrementing the power. So in practice, x^3 would differenciate to 3x^2 when differenciating in terms of x.

We drop the power (being 3) so that it is in front of the term (3x). Then decrement the power (so x^3 becomes x^2). Putting it all together, we get the result 3x^2. Similarly, if we were to differenciate 2x^3. We would obtain (2)(3)x^2 being 6x^2. Don't forget to multiply your result by the number that was in front previously! Which in this case was 2 being in front of x.

Now lets look at the question in hand, lets differenciate each term separately and add them up.

1. x^2 would become 2x^1 if we were to drop the power and decrement the power. 2x^1 is easily rewritten as 2x

2. 6x. This is actually 6x^1. So if we again drop the power and decrement the power, we obtain 6x^0. Which is 6

3. Lastly, 2 which is actually 2x^0. So if we do the same again, we get 0x^-2. 0 multiplied by anything is 0. Therefore 2 differenciated with respect to x is 0.

Adding up add the differenciated terms results in 2x + 6 + 0. So dy/dx = 2x + 6.

So lets have a quick recap of differenciation to start off. In informal terms, differenciation is carried out by dropping the power to be in front of the term we are differenciating to as well as decrementing the power. So in practice, x^3 would differenciate to 3x^2 when differenciating in terms of x.

We drop the power (being 3) so that it is in front of the term (3x). Then decrement the power (so x^3 becomes x^2). Putting it all together, we get the result 3x^2. Similarly, if we were to differenciate 2x^3. We would obtain (2)(3)x^2 being 6x^2. Don't forget to multiply your result by the number that was in front previously! Which in this case was 2 being in front of x.

Now lets look at the question in hand, lets differenciate each term separately and add them up.

1. x^2 would become 2x^1 if we were to drop the power and decrement the power. 2x^1 is easily rewritten as 2x

2. 6x. This is actually 6x^1. So if we again drop the power and decrement the power, we obtain 6x^0. Which is 6

3. Lastly, 2 which is actually 2x^0. So if we do the same again, we get 0x^-2. 0 multiplied by anything is 0. Therefore 2 differenciated with respect to x is 0.

Adding up add the differenciated terms results in 2x + 6 + 0. So dy/dx = 2x + 6.

Firstly, we can take a look at the degree x's in each term. It seems like they share a common degree of 0. In other words, we will not be able to factor out the x's in the expression.

Secondly, we need to find out if we can factor out a number out of the expression. Therefore we can identify that 10 is a common factor of both terms. So we will be able to factor out 10 from the expression to get 10(2 + x).

Firstly, we can take a look at the degree x's in each term. It seems like they share a common degree of 0. In other words, we will not be able to factor out the x's in the expression.

Secondly, we need to find out if we can factor out a number out of the expression. Therefore we can identify that 10 is a common factor of both terms. So we will be able to factor out 10 from the expression to get 10(2 + x).