I'm currently a first year student at Loughborough University studying systems engineering. I love systems engineering because there are so many different aspects to it! Everything is a system whether its your mobile phone, a car, plane, boat, road network, house, tiny microprocessors, everything is a system! And its a systems engineers job to look at the bigger picture to make sure all of the sub-systems and components fit together to ensure that the final product fulfills its purpose and the stakeholders needs. Systems engineering involves lots of maths and physics and critical thinking and analysis.

I've taken exams throughout my life and i tend to do better under high pressure situations which means usually i have done well in all my exams. I have plenty of experience and a wealth of knowledge dealing with the pressure of exams and how to do well in them. I used to compete at a very high level in swimming so i'm used to high-pressure situations and i believe i can draw on this to help you out during your exams. I also went to a very good high school and college whom provided me with lots of exams tips that i can give to you as well.

As i mentioned previously, i used to swim competitively, now i have taken up karting and squash, and i still like to go to the gym regularly to stay in shape. I also enjoy going out and playing ps4 in my spare time.

I'm currently a first year student at Loughborough University studying systems engineering. I love systems engineering because there are so many different aspects to it! Everything is a system whether its your mobile phone, a car, plane, boat, road network, house, tiny microprocessors, everything is a system! And its a systems engineers job to look at the bigger picture to make sure all of the sub-systems and components fit together to ensure that the final product fulfills its purpose and the stakeholders needs. Systems engineering involves lots of maths and physics and critical thinking and analysis.

I've taken exams throughout my life and i tend to do better under high pressure situations which means usually i have done well in all my exams. I have plenty of experience and a wealth of knowledge dealing with the pressure of exams and how to do well in them. I used to compete at a very high level in swimming so i'm used to high-pressure situations and i believe i can draw on this to help you out during your exams. I also went to a very good high school and college whom provided me with lots of exams tips that i can give to you as well.

As i mentioned previously, i used to swim competitively, now i have taken up karting and squash, and i still like to go to the gym regularly to stay in shape. I also enjoy going out and playing ps4 in my spare time.

My lessons will usually be based around a specific topic each session. I shall start by explaining the problem at hand and how to go about solving it. I'll be spending the most time on this as from my own experience i believe this is by far the most important stage in learning. Teachers often explain things once, and only in one way, which can be very confusing as i have experienced in the past. I aim to ensure you have a solid understanding of the concept first, going through example questions step by step, and if you don't understand it at first, i can try different ways of explaining until you do understand the problem. After this i shall set different kinds of problems, gradually encouraging you to do more of the work until you are confident to solve a number of different kinds of questions by yourself. I have a deep understanding of lots of different areas, and have access to all of my old notes, questions, and examples which will be of great help. We can set sessions up based around your needs, whether it's just a one off to help clarify a particular aspect of a problem, or more regular sessions to help you with a whole concept.

My lessons will usually be based around a specific topic each session. I shall start by explaining the problem at hand and how to go about solving it. I'll be spending the most time on this as from my own experience i believe this is by far the most important stage in learning. Teachers often explain things once, and only in one way, which can be very confusing as i have experienced in the past. I aim to ensure you have a solid understanding of the concept first, going through example questions step by step, and if you don't understand it at first, i can try different ways of explaining until you do understand the problem. After this i shall set different kinds of problems, gradually encouraging you to do more of the work until you are confident to solve a number of different kinds of questions by yourself. I have a deep understanding of lots of different areas, and have access to all of my old notes, questions, and examples which will be of great help. We can set sessions up based around your needs, whether it's just a one off to help clarify a particular aspect of a problem, or more regular sessions to help you with a whole concept.

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The question is asking you to find an expression for the gradient of the curve. This means you should be looking to differentiate the equation. As it has y values, you'll need to differentiate implicitly. Implicit differentiation is when you differentiate a term with respect to another different term. In this case we are differentiating y with respect to x where as usually you just differentiate x with respect to x. The general rule for implicit differentiation is to differentiate the term normally but at the end you multiply it by the differential of the term, (ie times it by dy/dx). You leave this dy/dx attached until you have finished differentiating and then you can simplify the expression to get what dy/dx is equal to. The question: Differentiate x^{2}+y^{2}-3x+4y=6. dy/dx x^{2} = 2x. dy/dx y^{2} = 2y(dy/dx). dy/dx -3x = -3. dy/dx 4y = 4(dy/dx). dy/dx 6 = 0. Now we have all of the differentiated terms we can write it out like this: dy/dx = 2x+2y(dy/dx)-3+4(dy/dx)=0. Now we need to get dy/dx on its own on one side so we have a clear expression for the gradient of the curve. Let's move all the dy/dx 's onto the left and all the other terms to the right. Now we have: 2y(dy/dx)+4(dy/dx)=3-2x. Now we can simplify by taking dy/dx out as a factor. This leaves us with dy/dx(2y+4)=3-2x. To get dy/dx on its own we simply divide (3-2x) by (2y+4). This gives us the final answer that the gradient of the curve (dy/dx) is: dy/dx = (3-2x)/(2y+4).The question is asking you to find an expression for the gradient of the curve. This means you should be looking to differentiate the equation. As it has y values, you'll need to differentiate implicitly. Implicit differentiation is when you differentiate a term with respect to another different term. In this case we are differentiating y with respect to x where as usually you just differentiate x with respect to x. The general rule for implicit differentiation is to differentiate the term normally but at the end you multiply it by the differential of the term, (ie times it by dy/dx). You leave this dy/dx attached until you have finished differentiating and then you can simplify the expression to get what dy/dx is equal to. The question: Differentiate x^{2}+y^{2}-3x+4y=6. dy/dx x^{2} = 2x. dy/dx y^{2} = 2y(dy/dx). dy/dx -3x = -3. dy/dx 4y = 4(dy/dx). dy/dx 6 = 0. Now we have all of the differentiated terms we can write it out like this: dy/dx = 2x+2y(dy/dx)-3+4(dy/dx)=0. Now we need to get dy/dx on its own on one side so we have a clear expression for the gradient of the curve. Let's move all the dy/dx 's onto the left and all the other terms to the right. Now we have: 2y(dy/dx)+4(dy/dx)=3-2x. Now we can simplify by taking dy/dx out as a factor. This leaves us with dy/dx(2y+4)=3-2x. To get dy/dx on its own we simply divide (3-2x) by (2y+4). This gives us the final answer that the gradient of the curve (dy/dx) is: dy/dx = (3-2x)/(2y+4).

When solving simultaneous equations the first step is to take one of the equations and rearrange it to make one of the terms the subject. In this case lets take the equation 1+y=2x. Take the 1 over to the other side to make y the subject. y=2x-1. The next step is to substitute the expression for the term you have just worked out into the other equation that you didn't use at the beginning. In our case this gives us 2(2x-1)+x=8. Expand the brackets: 4x-2+x=8. Collect like terms: 5x=10. Simplify: x=2. Now we have one of the unknown values, we can substitute it back into one of the original equations to work out the final unknown. In our case: 1+y=2(2). Expand the brackets: 1+y=4. Collect like terms: y=3. Now we have both the answers x=2, y=3. You can check you answers by substituting the values back into the original equations and see if the two sides equal each other. 2(3)+(2)=8. 8=8. The answer is correct.When solving simultaneous equations the first step is to take one of the equations and rearrange it to make one of the terms the subject. In this case lets take the equation 1+y=2x. Take the 1 over to the other side to make y the subject. y=2x-1. The next step is to substitute the expression for the term you have just worked out into the other equation that you didn't use at the beginning. In our case this gives us 2(2x-1)+x=8. Expand the brackets: 4x-2+x=8. Collect like terms: 5x=10. Simplify: x=2. Now we have one of the unknown values, we can substitute it back into one of the original equations to work out the final unknown. In our case: 1+y=2(2). Expand the brackets: 1+y=4. Collect like terms: y=3. Now we have both the answers x=2, y=3. You can check you answers by substituting the values back into the original equations and see if the two sides equal each other. 2(3)+(2)=8. 8=8. The answer is correct.