Hello everyone,thanks for looking on my profile ! I'm on my second year at University of Bristol, studying mathematics. From an early age I started to participate in lots of contests and maths olympiads, and the experience I achieved along the way enriched both my knowledge and my confidence. But, what recommends me most for the job are the following beliefs I got: 1.Anyone can learn anything. At the beginning I struggled, I didn't know how to approach a particular problem and I failed, many times. But also, in time, I achieved things I could had only dream about years ago. You can do maths,and you can be the best at it. 2.Maths can be fun. Trust me, you'll think so, when you'll manage to solve problem after problem with ease. Your passion for maths will thrive with my help and then you'll understand the beauty of it. 3.Formulas might be useful, but real maths is about thinking. I can't count the number of times I had no idea about an useful formula, but I still solved the problem I had to. Thinking is the best formula. 4.Every correct solution must be apreciated. I've seen more solutions of the same problem,and why every such solution had its own importance. You can't say a problem should be solved only by a fixed method. Maths is flexible. If you think I could help you, please, let me know !Hello everyone,thanks for looking on my profile ! I'm on my second year at University of Bristol, studying mathematics. From an early age I started to participate in lots of contests and maths olympiads, and the experience I achieved along the way enriched both my knowledge and my confidence. But, what recommends me most for the job are the following beliefs I got: 1.Anyone can learn anything. At the beginning I struggled, I didn't know how to approach a particular problem and I failed, many times. But also, in time, I achieved things I could had only dream about years ago. You can do maths,and you can be the best at it. 2.Maths can be fun. Trust me, you'll think so, when you'll manage to solve problem after problem with ease. Your passion for maths will thrive with my help and then you'll understand the beauty of it. 3.Formulas might be useful, but real maths is about thinking. I can't count the number of times I had no idea about an useful formula, but I still solved the problem I had to. Thinking is the best formula. 4.Every correct solution must be apreciated. I've seen more solutions of the same problem,and why every such solution had its own importance. You can't say a problem should be solved only by a fixed method. Maths is flexible. If you think I could help you, please, let me know !

No DBS Check

Many students ask how they can they find 2 unknowns,given 2 simple equations. I'll try to focus in this answer on giving students a tool which they can use to solve ANY 2 simple equations with one solution.Let's take an example:

5*x+y=22 and 3*x+4*y=20

In short,the plan is to:

1.) Write y in terms of x from the first equation

2.) Substitute it in the second one, so that we will get only an equation in function of x

3.)Then, find x from it.

4.)Now, we can substitute x's value in the first equation and find y.

Concrete:

1.) Write y in terms of x from the first equation

From 5*x+y=22 we get that y=22-5*x.

2.) Substitute it in the second one, so that we will get only an equation in function of x

Substituing y in the second equation gives :

3*x+4*(22-5*x)=20.

3.)Then, find x from it.

We now rearrange it ,so:

3*x+88-20*x=20(we opened the parenthesis)

Therefore:

88-17*x=20(we gave x as common factor and had x*(3-20) which is -17*x)

Therefore by adding 17*x and subtracting 20 we get :

68=17*x.

By dividing the equation with 17 we have x=4.

4.)Now, we can substitute x's value in the first equation and find y.

So we substitute it in the first equation,so 5*x+y=22 gives 5*4+y=22,so y=2.

The beauty of this method stays in the fact that it can be used to solve any problem like that.

Now,with some practice,you should be able to find the solution of a similar problem. Here are some exercises which you could use to practice some more :

1.) 3*x+7*y=10 and x+5*y=6

2.) x+y=9 and 3*x+4*x=32

3.) x+y=6 and x+5*y= 26

4.) 4*y=28 and 2*x+y=9

5.) 6*x-2*y=72 and x+2*y=12

I would finally recommend not to memorise the steps of this method,but to understand them. Good luck !

Solutions:

1.) x=1 and y=1

2.) x=4 and y=5

3.) x=1 and y=5

4.) x=1 and y=7

5.) x=12 and y=0

Many students ask how they can they find 2 unknowns,given 2 simple equations. I'll try to focus in this answer on giving students a tool which they can use to solve ANY 2 simple equations with one solution.Let's take an example:

5*x+y=22 and 3*x+4*y=20

In short,the plan is to:

1.) Write y in terms of x from the first equation

2.) Substitute it in the second one, so that we will get only an equation in function of x

3.)Then, find x from it.

4.)Now, we can substitute x's value in the first equation and find y.

Concrete:

1.) Write y in terms of x from the first equation

From 5*x+y=22 we get that y=22-5*x.

2.) Substitute it in the second one, so that we will get only an equation in function of x

Substituing y in the second equation gives :

3*x+4*(22-5*x)=20.

3.)Then, find x from it.

We now rearrange it ,so:

3*x+88-20*x=20(we opened the parenthesis)

Therefore:

88-17*x=20(we gave x as common factor and had x*(3-20) which is -17*x)

Therefore by adding 17*x and subtracting 20 we get :

68=17*x.

By dividing the equation with 17 we have x=4.

4.)Now, we can substitute x's value in the first equation and find y.

So we substitute it in the first equation,so 5*x+y=22 gives 5*4+y=22,so y=2.

The beauty of this method stays in the fact that it can be used to solve any problem like that.

Now,with some practice,you should be able to find the solution of a similar problem. Here are some exercises which you could use to practice some more :

1.) 3*x+7*y=10 and x+5*y=6

2.) x+y=9 and 3*x+4*x=32

3.) x+y=6 and x+5*y= 26

4.) 4*y=28 and 2*x+y=9

5.) 6*x-2*y=72 and x+2*y=12

I would finally recommend not to memorise the steps of this method,but to understand them. Good luck !

Solutions:

1.) x=1 and y=1

2.) x=4 and y=5

3.) x=1 and y=5

4.) x=1 and y=7

5.) x=12 and y=0

There is a simple way of making calculations with percentages.

Say you want to subtract from a numerical value x,20%. Then multiply it with 80/100.

Say you want to add 20%.Then multiply with 120/100.

Say you want to add 70%. Then multiply with 170/100.

In short,if you want to add x%,then multiply with (100+x)/100. If you want to subtract y% then multiply with (100-y)/100.Note that you can't subtract more than 100%.

Easy,right? :P

Now we will look on a particular tricky question. We have a TV that costs 1000 pounds.Then,the price of TV is reduced with 20%.After some time,the TV is increased with 20%. What is the final price of a TV?

Don't fall into the trap of saying that the final price is the same with the initial one,therefore 1000 punds.

1.) We reduce with 20% : 1000*80/100=800 pounds

2.) We increase the current price with 20%,so: 800*120/100=960 pounds.

So the final price is 960 pounds.

There is a simple way of making calculations with percentages.

Say you want to subtract from a numerical value x,20%. Then multiply it with 80/100.

Say you want to add 20%.Then multiply with 120/100.

Say you want to add 70%. Then multiply with 170/100.

In short,if you want to add x%,then multiply with (100+x)/100. If you want to subtract y% then multiply with (100-y)/100.Note that you can't subtract more than 100%.

Easy,right? :P

Now we will look on a particular tricky question. We have a TV that costs 1000 pounds.Then,the price of TV is reduced with 20%.After some time,the TV is increased with 20%. What is the final price of a TV?

Don't fall into the trap of saying that the final price is the same with the initial one,therefore 1000 punds.

1.) We reduce with 20% : 1000*80/100=800 pounds

2.) We increase the current price with 20%,so: 800*120/100=960 pounds.

So the final price is 960 pounds.

To solve a quadratic equation we need to find the value of x for which a*x^{2}+b*x+c=0,where a,b,c are constants.There are some well known formulas which you should get accustomed with.However,in this answer,I'll try to make you understand where from this formulas actually come from,and make you understand how to solve a quadratic equation.

If we take the common factor a,we get a*(x^{2}+(b/a)*x+c/a)=0.(Note that a must be different from 0,otherwise the equation becomes b*x+c)

We try to write this equation of order 2 as a product of two equations of order 1.

Therefore,a*(x^{2}+(b/a)*x+c/a)=a*(x-t)(x-s)=0. If we can find s and t for which the first equality holds true, we solved the initial equation,and we would have solutions s and t.

So (x-s)(x-t)=x^{2}-x(s+t)+s*t,and must be equal with our equation(we divide by a),so:

x^{2}-x(s+t)+s*t=x^{2}+(b/a)*x+c/a. If we are able to find s and t such that s+t=-b/a and s*t=c/a then the equality mult be true and we solved our initial equation.(This 2 relations are also called Vieté relations.)

So we square the first relation and get s^{2}+t^{2}+2*s*t=b^{2}/a^{2}.We subtract 4*s*t=4*c/a and get:

(s-t)^{2}=s^{2}+t^{2}-2*s*t=b^{2}/a^{2}-4*c/a.=b^{2}/a^{2}-4*a*c/a^{2}(we amplified a/c with a). So :

(s-t)^{2}=(b^{2}-4*a*c)/a^{2},so if we take the square root of this we get:

|s-t|=(b^{2}-4*a*c)^{1/2}/a. The expression b^{2}-4*a*c got the name of determinant and is noted with a triangle. However,I'll note it with p.So, s-t = p^{1/2}/a or s-t = -p^{1/2}/a.

For the first case :

If we add s+t=-b/a,we obtain:

s=(-b+p^{1/2})/(2*a). and t=(-b-p^{1/2})/(2*a). Make yourself the calculations and find that

s=(-b-p^{1/2})/2*a and t+(-b+p^{1/2})/2*a for the seond case as well. This is in fact the same case.(As both tells us the same thing, x^{2}+(b/a)*x+c/a=( x- (-b-p^{1/2}) / (2*a) ) ( x- ( b - p^{1/2}) / ( 2*a) ) ) with the first one.

Now we can verify it by calculations,and get that the equality is indeed true.(a necessary step as we made a lot of assumptions along the way.However,as long as we can verify this by direct calculation,nothing else matters anymore.)

Therefore,we found out the values of s and t in terms of a,b and c,so therefore we found the solutions we were looking for.

REMEMBER:

The solutions are: s=(-b+p^{1/2}) / (2*a). and

t=(-b-p^{1/2}) / (2*a) , where p=b^{2}-4*a*c.(Only in the case where this solutions exist and are real)

Challenge: Use this to solve the following:

1.) x^{2}+8*x+7=0

2.)x^{2}+11*x+28=0

Good luck!

Solutions:

1.) The solutions are -1 and -7.

2.) The solutions are -4 and -7.

To solve a quadratic equation we need to find the value of x for which a*x^{2}+b*x+c=0,where a,b,c are constants.There are some well known formulas which you should get accustomed with.However,in this answer,I'll try to make you understand where from this formulas actually come from,and make you understand how to solve a quadratic equation.

If we take the common factor a,we get a*(x^{2}+(b/a)*x+c/a)=0.(Note that a must be different from 0,otherwise the equation becomes b*x+c)

We try to write this equation of order 2 as a product of two equations of order 1.

Therefore,a*(x^{2}+(b/a)*x+c/a)=a*(x-t)(x-s)=0. If we can find s and t for which the first equality holds true, we solved the initial equation,and we would have solutions s and t.

So (x-s)(x-t)=x^{2}-x(s+t)+s*t,and must be equal with our equation(we divide by a),so:

x^{2}-x(s+t)+s*t=x^{2}+(b/a)*x+c/a. If we are able to find s and t such that s+t=-b/a and s*t=c/a then the equality mult be true and we solved our initial equation.(This 2 relations are also called Vieté relations.)

So we square the first relation and get s^{2}+t^{2}+2*s*t=b^{2}/a^{2}.We subtract 4*s*t=4*c/a and get:

(s-t)^{2}=s^{2}+t^{2}-2*s*t=b^{2}/a^{2}-4*c/a.=b^{2}/a^{2}-4*a*c/a^{2}(we amplified a/c with a). So :

(s-t)^{2}=(b^{2}-4*a*c)/a^{2},so if we take the square root of this we get:

|s-t|=(b^{2}-4*a*c)^{1/2}/a. The expression b^{2}-4*a*c got the name of determinant and is noted with a triangle. However,I'll note it with p.So, s-t = p^{1/2}/a or s-t = -p^{1/2}/a.

For the first case :

If we add s+t=-b/a,we obtain:

s=(-b+p^{1/2})/(2*a). and t=(-b-p^{1/2})/(2*a). Make yourself the calculations and find that

s=(-b-p^{1/2})/2*a and t+(-b+p^{1/2})/2*a for the seond case as well. This is in fact the same case.(As both tells us the same thing, x^{2}+(b/a)*x+c/a=( x- (-b-p^{1/2}) / (2*a) ) ( x- ( b - p^{1/2}) / ( 2*a) ) ) with the first one.

Now we can verify it by calculations,and get that the equality is indeed true.(a necessary step as we made a lot of assumptions along the way.However,as long as we can verify this by direct calculation,nothing else matters anymore.)

Therefore,we found out the values of s and t in terms of a,b and c,so therefore we found the solutions we were looking for.

REMEMBER:

The solutions are: s=(-b+p^{1/2}) / (2*a). and

t=(-b-p^{1/2}) / (2*a) , where p=b^{2}-4*a*c.(Only in the case where this solutions exist and are real)

Challenge: Use this to solve the following:

1.) x^{2}+8*x+7=0

2.)x^{2}+11*x+28=0

Good luck!

Solutions:

1.) The solutions are -1 and -7.

2.) The solutions are -4 and -7.