Finding Mind Boggling Real Pairs (x,y) || A Hard System of Non-Linear Algebraic Equations || AMS

Hello everyone. In this video, we are given these two equations and we have to find all real pairs (x,y). So let we move toward the solution. The given equations are these one. Let we call these equation as equation number one and equation number two. Here we have to note that x and y cannot be zero because if we take x and y both zero then these term in both of the equations will be indeterminate form and so (0,0) is not a solution. So to solve these equation let we multiply equation number one with y and 2 with x. So when we multiply this equation number one with y we will get this one and when we multiply equation number two with x we will get this one. Adding these two equations, we will get if we add these two terms then we will get 2xy and then sum of these two terms and that is equal to 0 + 3y that is equal to 3y. Taking LCM of these two terms that is x² + y² and in the numerator we will get this, that is equal to 3y, and from here we can see that 3xy will be cancel out and we will have 2xy - (y² + x²)/(x²+y² ) = 3y. Again this will cancel out. x² + y² in numerator and denominator will be cancel out and we will get a very simple equation 2xy - 1 = 3y. Taking this one towards right we will get 2xy = 3 y + 1 and dividing both sides by 2y we will get x= (3 y + 1)/2y. So this is what the value of x is. We can substitute this value either in equation number one or two. So I will submit it in equation number two because equation number two is bit easier because of having zero on the right hand side. So, I have to calculate this term (x+3y)/(x² + y²). So first I calculate x + 3 y so x + 3 y will be equal to this is value of x and then + 3y and after simplification we will get this one. Now the value of x² + y² will be this one + y². This is value of x and after simplification x² + y² will be equal to this one. So here we can see that in denominator we have 4y². So in x + 3y we make it 4y². So multiplying and dividing by 2y so x + 3 y will become as, in the denominator 4y² and if we multiply this by 2y we will get 12y^3+6y²+2y. So dividing this and this we will get (x + 3 y)/(x² + y²). That is equal to this relation. In this relation 4y² will be cancel out and we will get this value as this one. So substituting this value in equation number two which is this one. We will get this equation. Taking this term towards right we will get y is equal to this one. Now here we can see that, we can take y common from the numerator on the right hand side and so we take it common and y will be equal to this relation. On the left and right we can see that y will be cancel out and we have one is equal to this whole relation. By cross multiplying we will get this one here. 6y on both side will be cancel out and from here we can see that 4y^4, 9y² and here 12y². This will become minus so this will become minus 3y². This is +1 and when 2 will shift it to the left it will become -2 so it will become -1 that is equal to 0. This equation is a quadratic equation in y² and we can factorize it by breaking its midterm. So 4y^4 - 3y² will be decomposed into -4y² + y² and - 1 is equal to 0. Taking 4y² common we will get y²-1 and from here just + one and it will give y²-1. So taking y² - 1 common we will get other Factor as 4y² + 1 is equal to 0 since Y is a real number and we can see that this Factor 4y² + 1 cannot be equal to 0. Because y² is positive so 4y² is also positive and a positive number + one is always a nonzero number so this Factor cannot be equal to zero. And the product is zero so it means the only factor which is z that is y s - 1 so y s - 1 is equal to 0 from here we can find out y² is equal to 1 and taking square root on both sides we will get y=+ -1. So there are only two values of y that are +1 and -1. Now we have 6 = 3 y + 1 / 2 y we will submit these value here for case one when y= 1. We have x = 2 and the pair (x,y) will be equal to (2,1) and for the case 2 when y = 2. After submitting this value is equal to 1 and the pair (x,y) will be equal to 1 and minus one. So there are only two pairs which are mentioned here are the solution of given set of equation. That is the answer.