Solve this System of rational Equations | Quick & Easy Explanation

[huongtoituonglai]

Mashallah amazing video & very beautiful place 😍 💕 🌷

[j.r.]

After factoring the cubics as in the official explanation, and getting xy = 6, I then did y = 6/x. Plugging this back into the x^2 + y^2 equation yielded x^2 + (6/x)^2 = 13, then eventually x^4 - 13x^2 + 36 = 0. This is easily factorable to give x^2 - 4 = 0 and x^2 - 9 = 0. Then take the four values for x back to y = 6/x to calculate the y values.

[sngmn]

x + y = ± 5
xy = 6
From Vieta's formula we can construct a quadratic eqn which roots are our "x" and "y":
t² - (x + y)t + xy = 0
t² ± 5t + 6 = 0
(t ± 5/2)² - 25/4 + 24/4 = 0
(t ± 5/2)² = 1/4
t = ± 5/2 ± 1/2
t = {2, 3, -2, -3} => (x₁, y₁) = (2, 3); (x₂, y₂) = (-2, -3).
And from symmetry (x < - > y) we will get more such solutions:
t = {3, 2, -3, -2} => (x₃, y₃) = (3, 2); (x₄, y₄) = (-3, -2).
Answer: (x, y) = {(2, 3), (-2, -3), (3, 2), (-3, -2)}.

[govindashit]

Simple but more exciting problem.
I enjoy it.
Thanks sir .

[HassanLakiss]

Thank you for a nice algebraic challenge and a well executed solution. This may be an ideal opportunity for those teaching higher Mathematics to draw the graphs of the circle x² + y² = 13 (centre origin, radius √13)
and x+y=±5 as well as y=6/x
and show that the points of intersection are the 4 points in your solution. Thank you

[mikezilberbrand]

Instead of picking up 2 and 3 use Viet. Solve x^2-5x+6=0, then x^2+5x+6=0.

[divyanshuraghav]

Nice question sir. I solve this question in 1minute☺️.

[Gargaroolala]

I was so close! I managed to obtain equations 3, 4, 5, 6. And then I got stuck. Didn’t know the last step was just to do an observation. Good qns

[math]

very well explained, thanks for sharing this system of rational equations, happy holidays

[bentels]

Solved it almost the same way. Did a few things in a different order, but essentially the same.

The thought occurs that there's an elegant hint in there about there being four solutions: the first equation simplifies to x^ +y^2 -xy = 7, the second to x^2+y^2+xy = 19. Both those functions describe ovals, so the solution space is the intersection of two ovals.

[georgesadler]

Thank you for another fantastic video.

[nezamasarie]

First problem solved
I did a similiar solution
So i factored the 2 equations
then got that x^2+y^2=13
plugging it in the first equation gives xy=6
so i added another xy to get 2xy
so x^2+2xy+y^2=25
this factors into (x+y)^2
so x+y=+/-5
and since xy=6
the solutions are 3, 2 and -3, -2

[pranavamali]

Fantastic question as well as well explained thnx a lot

[sameerqureshi-khcc]

Prenath journey continues to Postmath 😊👍

[lazaremoanang]

Simple, S={(3;2)}: with those two equations you can see easily that xy=6 so it's easy to choose (3, 2) because x has to be bigger.

[philipkudrna]

I don‘t know how to solve it the proper way, but simply by testing some obvious alternatives out, I arrived at x=2 and y=3, as 2+3=5 and 35/5=7 and 8+27=35. If you check you arrive at -19/-1=19.
Now I watch the „long way“ of the proper solution (without having to have luck upon trying…)
Maybe there are further solution sets, but (2; 3) is definitely one of them…
After watching: I definitely need to remember the a^3+b^3 formula…! 🤔😀

[youspin]

You can of course solve this easily in your head by a little thought and simple testing of suitable integers. It is immediately obvious that x and y are integers, and that they must be small integers like 1, 2, 3, 4. Then in seconds you see it is 3 and 2.

[badalmondal]

Sir I solved this in 1 minute .
Answers -- 2, 3
3, 2
-2, -3
-3, -2
Sir your method and my method is same.
Please bring more fantastic questions like this.

[chandiprasadmishra]

Solution can also be found by evaluating ( x-y)=plus/minus 1.

[abazcanolli]

Sorry, can i ask something? Can you help me to solve some limit exercises, please?