Very good... It can also be solved in this way : xy=12 i.e., 2xy=24...(i) x²+y²=25... (ii) Firstly adding equation(i) and(ii), we get : (x+y)²= 49 i.e., x+y=±(7)... (iii) Secondly, equation(ii)- equation(i), we get : (x-y)²=1 i.e., x-y=±(1)...(iv) Solving equation (iii) and equation(iv), we now get: {x, y} ={4, 3}, {4, -3}, {-4, 3}, {-4, -3}
5:49 xy=12 x^2+y^2=25, Shorter solution: 2*xy = 24. (x+y)^2=49;(x-y)^2= 1 then system equations x+y=±7, x-y=±1. From symmetry only 4 real solution (±3,±4) and (±4,±3). But there are complex solution to.
What complex solutions ??? You did not restricted yourself to real solutions and the relations: xy = 12, x^2 + y^2 = 25 result from the given equatios, implying (as you have shown) the only 4 solutions: {(x,y)} = { ±(4,3), ±(3,4) }. There are no aditional solutions, not real and not complex.
You dont get 4 pairs of solutions, just 2. Product is 12 so they are both negative or both positive. Because of this it dors matter which of the two equations of the system you use after solving the square equation :)
At 5:30 you could have sped it up by: x^2 + 2xy + y^2 = 25 + 2*12 = 49 = (x + y)^2. So x + y = ±7 x^2 - 2xy + y^2 = 25 - 2*12 = 1 = (x - y)^2. So x - y = ±1 etc
Trigonometric Substitution: x = r cos theta, y = r sin theta Dividing the equations we get the value of sin 2 theta. Using sin 2 theta = 2 tan theta/ 1 + tan^2 theta, we can find tan theta (y/x). r can be found by substituting in sin 2 theta
Divide both equations, as both of them don't contain any zero, you get x = ±3y/4 and x = ±4y/3. Given the symmetry of the problem, we only have to solve for x = ±3y/4 and pick all permutations of the answer. Make the substitution on the second equation and you get |y^3| = 27 which is the same as y = ±3, hence x = ±4. So the solutions are (-4, -3), (-3, -4), (3, 4), (4, 3).
Really liked the problem. But I couldn't follow it after about 4:41 mark, probably due to short cuts that everyone else would get. Anyway, I have to educate myself more to follow your vids. But I really like your channel.
When you see xx+yy=25 You could.guess a 3,4,5 triangle and when xy=12 that confirms it. You just have to remember to take the negative pair -3,-4 and note the symmetry in x&y so that They can be swapped .
xy = 12, x^2 + y^2 = 25
corresponds to the Pythagorean triple (3,4,5)
thus (3,4) (4,3) (-3,-4) (-4,-3)
are solutions
Nice!
great
Very good... It can also be solved in this way :
xy=12 i.e., 2xy=24...(i)
x²+y²=25... (ii)
Firstly adding equation(i) and(ii), we get :
(x+y)²= 49 i.e., x+y=±(7)... (iii)
Secondly, equation(ii)- equation(i), we get :
(x-y)²=1 i.e., x-y=±(1)...(iv)
Solving equation (iii) and equation(iv), we now get:
{x, y} ={4, 3}, {4, -3}, {-4, 3}, {-4, -3}
5:49 xy=12 x^2+y^2=25, Shorter solution: 2*xy = 24. (x+y)^2=49;(x-y)^2= 1 then system equations x+y=±7, x-y=±1.
From symmetry only 4 real solution (±3,±4) and (±4,±3). But there are complex solution to.
Nice!
What complex solutions ???
You did not restricted yourself to real solutions and the relations: xy = 12, x^2 + y^2 = 25 result from the given equatios, implying (as you have shown) the only 4 solutions: {(x,y)} = { ±(4,3), ±(3,4) }. There are no aditional solutions, not real and not complex.
You dont get 4 pairs of solutions, just 2. Product is 12 so they are both negative or both positive. Because of this it dors matter which of the two equations of the system you use after solving the square equation :)
Right 👍
There will be two more solutions interchanging x and y. As there is a symmetry between x and y. Hence, altogether there are grout solutions.
At 5:30 you could have sped it up by:
x^2 + 2xy + y^2 = 25 + 2*12 = 49 = (x + y)^2. So x + y = ±7
x^2 - 2xy + y^2 = 25 - 2*12 = 1 = (x - y)^2. So x - y = ±1 etc
Trigonometric Substitution: x = r cos theta, y = r sin theta
Dividing the equations we get the value of sin 2 theta. Using sin 2 theta = 2 tan theta/ 1 + tan^2 theta, we can find tan theta (y/x).
r can be found by substituting in sin 2 theta
Fantastic problem and solution.
Thanks!
Commendable and praise worthy work
Thank you!
Divide both equations, as both of them don't contain any zero, you get x = ±3y/4 and x = ±4y/3. Given the symmetry of the problem, we only have to solve for x = ±3y/4 and pick all permutations of the answer. Make the substitution on the second equation and you get |y^3| = 27 which is the same as y = ±3, hence x = ±4. So the solutions are (-4, -3), (-3, -4), (3, 4), (4, 3).
Very nice!
Really liked the problem. But I couldn't follow it after about 4:41 mark, probably due to short cuts that everyone else would get. Anyway, I have to educate myself more to follow your vids. But I really like your channel.
Hi
If you divide the first equation by the second you have :
x=(3/4)y OR x=(4/3)y
Good thinking!
I also solve in the same way.
VERY VERY NICELY DONE!
😊
I liked your way to solving equations:) thanks
Glad to hear that!
When you see xx+yy=25
You could.guess a 3,4,5 triangle and when xy=12
that confirms it.
You just have to remember to take the negative pair -3,-4 and note the symmetry in x&y so that
They can be swapped .
This is cool!
Is {-4,-3} false?
shouldn't be
Aren’t (3,4) and (4,3) both solutions? Also their opposites?
Yes! That's correct! I forgot to switch x and y.
Yep they are symmetric in x and y
@@LOL-gn7kv cool. I like the video
Great content. Thank You 😊
My pleasure!
Why you dont use t = 9
because that case is already covered. 9+16=25
Good afternoon sir
Actually since the equations are symmetric wrt x and y, we get 4 solutions by writing (x, y) as (y, x) too....
Good
Author didn't find two more solutions (3;4) and (-3;-4). Тhis follows from the fact that the original equations are symmetric with respect to x and y
Why can't (3,4) , (-3,-4) be solutions ???
They are solutions!
I guess you're missing 2 other solutions which are (3;4) and (-3;-4)
Man where do you live?
Good question! Perhaps I will publicize that one day, maybe Monday! 😊
There are 4 solutions to this equation : (3, 4), (-3, -4), (4, 3), (-4. -3)
hypobolic sub let r ::= the radical
(±3,±4) and (±4,±3)
Good! (3,-4) doesn't work, does it?
@@SyberMath (3,4).(4,3).(-3,-4).(-4,-3)
@@karunakarreddyg3395 That's right!