Dominating the channel diversification game!

@PegasusTenma1

2 жыл бұрын

Fancy seeing you here

We Vietnamese students call this a basic "type 1 symmetric system of equations", which is when the equations remain the same if the two variables are interchanged. The standard way to solve these systems is to substitute S = x + y and P = xy, solve for S and P, and use Vieta's formula to solve for x and y.

@thaitrieu791

2 жыл бұрын

Yeah I recalled doing that in grade 9 studying for highschool entrance exam lol

@kubogi

2 жыл бұрын

oh yeah, those system of equations on the "toán chuyên" (specialized math?) side is way more insane than this one lol, especially the ones on the HUS exam (aka chuyên KHTN)

@phamnguyenductin

2 жыл бұрын

@@kubogi Not this one though. Even for ninth graders, this would only be considered as a tiny bit of challenge. Students don't even need to know the "formula" to solve it; a simple factorisation is all they need. Knowing the steps only makes it a bit faster to arrive at the solution.

@kubogi

2 жыл бұрын

@@phamnguyenductin i agree, i'm a 9th grader and this looks so relaxing compared to other things i have to do

@cynthia7927

2 жыл бұрын

Are there other types of symmetric systems? That's nifty. I've never heard of them here

GOODNESS! So, I went the REALLY long way around by substituting from the start. I ended up with a fourth degree polynomial that I had to divide synthetically. PHEW! I had fun doing it, but you always amaze me with your methods. Part of me is mad that you made it so easy. haha I love it

When you were doing the systems of equations with a and b, and x any y (not the Q itself) you could multiply the multiplication (ab or xy) by 2 and then add or subtract the equations to get a perfect square

@SteelM4chine

2 жыл бұрын

PL is illegal(™)

@Professor_Sargeant_JAMS

Жыл бұрын

You have to go a little further. Square Equation 1, then substitute 2 times. Equation 2. Combine some like terms (xy)^2 + (x + y)^2 = 61 61 is a prime equivalent to 1 mod 4. There's a unique way to write primes of such form as a sum of squares. In this case, 5^2 + 6^2.

you are amazing Mr I appreciate for your clarification.

I'm feeling good because all I needed to solve the question is the question itself, my brain, and a single finger that drew random stuff in the air

Before watching the video: Given: x + y + xy = 11 x²y + xy² = 30 xy(x + y) = 30 xy = 30/(x + y) x + y + xy = 11 x + y + 30/(x + y) = 11 (x + y)² + 30 = 11(x + y) (x + y)² -11(x + y) + 30 = 0 (x + y) = (11 ± √(121 - 120))/2 (x + y) = (11 ± 1)/2 (x + y) = 5 or 6. correspondingly, from xy = 30/(x + y), xy = 6 or 5. ((x + y),(xy)) = (5,6) or (6,5) x + y = 6 xy = 5 x(6 - x) = 5 x² - 6x + 5 = 0 x = (6 ± √(36 - 20))/2 x = (6 ± 4)/2 x = 1 or 5 Correspondingly, y = 5 or 1. x + y = 5 xy = 6 x(5 - x) = 6 x² -5x + 6 = 0 x = (5 ± √(25 - 24))/2 x = (5 ± 1)/2 x = 2 or 3 Correspondingly, y = 3 or 2 (x,y) = (1,5), (5,1), (2,3), or (3,2)

@collegemathematics6698

2 жыл бұрын

Reading your comment is much faster than watching the actual video 😁

@roar2789

2 жыл бұрын

@@collegemathematics6698 well done👍 How old are u

Didn't know you had a second channel ! Amazing as usual !

Modify the second equation to be xy(x+y), then change xy=a and x+y=b. Sorry for bad English, I just forgot the lexicon

After the substitution a=x+y, b=xy you get a symmetric system which can have at most 2 roots (rearranging the first eq for a and substituting will give you a quadratic), thus you can guess just 1 solution. It's clearly 5,6 and 6,5. Now you have 2 systems in x and y of the same sort and for a,b= 5,6 you guess x,y=1,5, or 5,1, for a,b=6,5 you guess 2,3 and 3,2. So actually this can be solved in 30 seconds without even pen and paper!

@bharat9525

Жыл бұрын

Yes why not 🙂👍👍

I haven't decided yet what you are more talented at. Solving difficult math problems, or changing colors while writing.😅👍❤️

Bro I love this guy

There's an easier way that I used, let (x+y) and xy be roots of quadratic equation z^2 - 11z + 30 = 0 We get that z = 6 or 5 Make cases for x+y and xy Thus, a few more quadratic equations can be written Solving them we get (x, y) € {(1,5) , (5,1) , (3,2) , (2,3) } Done. Easy!

guess and check underrated

I was actually able to solve this pretty quickly: Let s=x+y, and p=xy s+p=11 sp=30 So s or p is equal to 5, and the other is 6. If p=5 and s=6, then x and y are 1 and 5. If s=5 and p=6, then x and y are 2 and 3. Therefore the solutions are (1,5), (2,3), (3,2) and (5,1).

I rewrote it as y=-2x+11 x^2(-2x+11)+x(-2x+11)=30 I then found three solutions, and assumed x=5 , the only rational solution, to be the correct one. f1:5+(5)(1)+1=11 f2:(25)(1)+(5)(1)=30 Therefore, (5,1) satisfies the nonlinear system of equations.

second equation factor out the xy so you get xy(x+y)=30=2*3*5 so 3 numbers multiply gives 30 which can be broken down to 3 numbers and here you have it 2,3 or 3.2

Nice observation: this system is symmetric one. So, if we have a solution then all permutation of this one will be solutions too.

Ignore the second equation and focus only on the first x+xy+ y =11 since (x+1)(y+1)= x+xy+y+1; hence add 1 to both sides of the equation x+xy+y+1 = 12 (x+1)(y+1) = 3*4 or 6*2 or 12*1 for 3*4 x+1 =3 , hence x=2 y+1= 4 hence y=3 (2, 3) answer for 6*2 x+1 = 6 ; hence x =5 y+1 = 2 ; hence y =1 (5, 1) answer for 12*1 x+1 = 12 hence x=11 y+1 = 1 ;hence y =0 (11,0) cant use this because of the 0 which would turn the second equation in to 0 hence the answer is 2,3 and 5,1

Multiplying the first line by x*y, we have: x^2 y + x^2 y^2+xy^2= 11xy Hence, by the second line, the first line becomes: 30 + x^2 y^2 = 11 xy. By the quadratic formula applyerd for z=xy xy = (11 +/- sqrt( 11^2-4*30 ))/2 Now we have x in terms of y. I would try the first equation.

You could just solve by inspection. Once you got to ab = 30 and a + b = 11 you can easily check for which pair of factors of 30 sum up to 11 and then rinse and repeat for x and y. As long as you don't forget that you're dealing with two symmetrical system of equations, so you should have 4 solutions in general, it's much quicker this way

just peeking at the thumbnail i was able solve it and kinda afraid if it was just a clickbait but it wasn't so I'm gladful

Squaring both sides of Equation 1, x^2 y^2 + 2 x^2 y + x^2 + 2 x y^2 + 2 x y + y^2 = 121. Multiplying both sides of Equation 2, 2x^y + 2xy^2 = 60. Substituting, x^2 y^2 + x^2 + 2 x y + y^2 + 60 = 121. So, (xy)^2 + (x + y)^2 = 61. 61 is a prime number that is equivalent to 1 mod 4. Fermat tells us there can only be a unique expression if 61 as a sum of positive numbers being squared< up to arrangement. 61 = 25 + 36 = 5^2 + 6^2. The first part is a mess, but once we apply the information from Fermat, we are at xy = 5 and x + y = 6 or xy = 6 and x + y = 5. Sorry for replying 1 to 2 years late. Next question is how many in this family of questions could be dealt with this way.

Thank you sir from India

can anyone please answer why didnt we go for finding a-b firstand then adding (a-b) and (a+b) to find a and b?

There is a formula for this kind of systems Solve x *2 -Sx+P=0 S : sum P: product Do it twice

@valvaraad

2 жыл бұрын

It is Viete theorem

@user-ry6cz3vs6g

2 жыл бұрын

I didn't know its name

@valvaraad

2 жыл бұрын

@@user-ry6cz3vs6g consider it a new piece of knowledge

@user-ry6cz3vs6g

2 жыл бұрын

@@valvaraad I knew the method but I didn't know its name

@noedeverchere2833

2 жыл бұрын

It's something we know well in France 😌

Nice!

Factor LHS of second equation to get xy(x+y)=30 From first equation calculate xy 11- (x+y)=xy (11-(x+y))(x+y)=30 Let u=x+y (11-u)u=30 -u^2+11u=30 u^2-11u+30=0 (u-5)(u-6)=0 We have two systems of equations x+y=5 xy = 6 x+y=6 xy=5 These both system of equations are Vieta formulas for quadratics t^2-5t+6=0 // For first system of equations t^2-6t+5=0 // For the second system of equation (x,y) = {(2,3),(3,2),(1,5),(5,1)}

Nothing scares me more than another system of equations

It took me a few secs but here is how I did it: x+xy+y=11 xy+(x+y)=11 x^2y+xy^2=30 xy(x+y)=30 Sub a=xy, b=x+y So, a+b=11, ab=30 Guess and check- a=6, b=5 So, xy=6, x+y=5 Guess and check- x=3, y=2 Non-rigorous but works for me so 🤷‍♂️

xy & x+y add up to 11 and multiply to 30, so xy = 5, and x+y = 6, or xy = 6, x+y = 5 so x = 5, y = 1 for the first one (or vice versa), and x = 3, y = 2 for the second one (or vice versa) thus solutions: {(5;1), (1;5), (3;2), (2;3)}

This is easy question. Set u=x+y and v=xy. Then the original equations are rewritten into addition and multiplication of u and v. Thus we can solve u and v by the roots of another quadratic equation z. Therefore we can solve x and y.

I just chose the first two small integers that came to my head-2 and 3-and substituted them into the equation.

Before watching: I claim that the following pairs: (5,1), (1,5), (2,3), (3,2) are all of the solutions. I was correct. My method basically boiled down to the same as his, though I arrived there in a more roundabout way and had to go through more calculations.

Cool Thank you teacher

while solving i discarded the possibility it was bounding problem and i said to myself "just algebra" LMAO

my method without looking at video. let a=xy, b=x+y given eqns become a+b=11 ab=30 this is a stsndard factorisation problem from which {a,b}={5,6} if a=6 then {x,y}={2,3} if a=xy=5 b=x+y=6 then y=5/x , b=6=x+5/x, mult x, 6x=x^2+5, so x^2-6x+5=0 (x-1)(x-5) x=1,5 so y=5,1 ie {x,y}={1,5} or{2,3}

I have a question: why didn't you turn the signs when you moved them to the other side for ex 30 or 6?

@perqehripzee2881

2 жыл бұрын

Because he did this: 11b-b^2=30 0=30-11b+b^2 Then he just rearranged 0=b^2-11b+30 And then just switches sides because it is the same b^2-11b+30=0

When x+y=a and xy=b, then (x, y) are solutions of t^2-at+b=0.

Ahh nice, my solution also involve substitution of a = xy and b = x+y, but I did something different that involves completing the square a + b = 11 ab = 30 we can square both sides of the equation a + b = 11 a^2 + 2ab + b^2 = 121; substitute ab = 30 a^2 + 60 + b^2 = 121 a^2 + b^2 = 61 Note: ab = 30 2ab = 60 2ab + 1 = 61 (we can use this to substitute 61) a^2 + b^2 = 2ab + 1 a^2 - 2ab + b^2 = 1 (a - b)^2 = 1 a - b = ±1 now we can use a + b = 11 and a - b = ±1 to get the values: a = 5 and b = 6 or a = 6 and b = 5 --------------------- Substitute back a = xy and b = x + y xy = 5 and x + y = 6 or xy = 6 and x + y = 5 Look at that! Both systems of equations are almost the same problem as before but with different values, so we can apply the same technique. On the first equation: --------------------- xy = 5 x + y = 6 (x + y)^2 = 36 x^2 + 2xy + y^2 = 36 x^2 + 10 + y^2 = 36 x^2 + y^2 = 26 Note: xy = 5 2xy = 10 2xy + 16 = 26 (this is the point where I realized that I was actually completing the square) x^2 + y^2 = 2xy + 16 x^2 - 2xy + y^2 = 16 (x - y)^2 = 16 x - y = 4 we can finally use x + y = 6 and x - y = ±4 to get the values: (5, 1) or (1, 5) now for the other equation: --------------------- xy = 5 x + y = 6 (x + y)^2 = 25 x^2 + 2xy + y^2 = 25 x^2 + 12 + y^2 = 25 x^2 + y^2 = 13 Note: xy = 6 2xy = 12 2xy + 1 = 13 x^2 + y^2 = 2xy + 1 x^2 - 2xy + y^2 = 1 (x - y)^2 = 1 x - y = ±1 use x + y = 5 and x - y = ±1 to get the values: (3, 2) or (2, 3) *Final Answer: (5, 1), (1, 5), (3, 2), or (2, 3)*

@someone-wu3vp

2 жыл бұрын

tf you typed up this much!?!?

@santoriomaker69

2 жыл бұрын

@@someone-wu3vp most of that comment is just copy and pasting and changing the variables lmao

What is your FB? I want to connect with you.

when I saw whole numbers on the right I immediately decided to substitute 2 and 3 and voilà!

I started by writing y = 11 - x / (x + 1) and substituted in y in the second equation. I ended up with a fourth degree equation, divided by (x+1)^2, all of it equal to 0. I tested: could x be -1? Well, no because if we substitute x by -1 in the first equation of the system, we find that -1 - y + y = 11. (should have done that earlier, when I first wrote y = thing with x's) So I was allowed to say that my fourth degree equation was equal to 0. I had: x^4 - 11x^3 + 41x^2 - 61x + 30 = 0 I noticed that 1 worked (negative coefficients added to the same sum as the positive ones) so I got: (x-1)(x^3 - 10x^2 + 31x - 30) = 0 What do I do now? Well, 1 doesn't work anymore, but what about 2? Oh it seems 2 is fine: 8 - 40 + 62 - 30 = 0. I then got: (x-1)(x-2)(x^2 - 8x + 15) = 0 This one is pretty straightforward: notice that 8 = 3 + 5 and 15 = 3 * 5. We then got: (x-1)(x-2)(x-3)(x-5) = 0. (we also could find that more easily by testing the value of y for x = 1 because we know that it's symmetrical: if (x;y) is a solution then (y;x) is too) At that point, we're almost done, and I'm not gonna detail how you get (1;5) (2;3) (3;2) and (5;1).

before watching the video solution x+xy+y=11 x²y+xy²=30 (x+y)+(xy)=11 (x+y)(xy)=30 let a=x+y and b=xy a+b=11 ab=30 a+(30/a)=11 a²-11a+30=0 using the quadratic formula a=5 and b=6 or vice versa if a=5 then x+y=5 and xy=6 clearly true when x=5 and y=1 and vice versa if a=6 x+y=6 and xy=5 so x=3 and y=2 and vice versa so the solutions are (x,y); (1,5),(5,1),(3,2)(2,3)

Solving for the value of y, you already found the value of x. No matter if x=2 and y=3 or vice-versa

please give solution for this (x-1)/sqrt(x+1) = 4 + (sqrt(x)-1)/2 then find value of x

Random Guess but X = 1 Y=5... and X=5 Y=1 Both fit the equation... Seems also 2,3 and 3,2 also Solve the equation

(x,y)=(1,5)(2,3)(3,2)(5,1)

Nice; I found the canonical solution.

அருமை.

So simple. S={(2,3),(3,2),(1,5),(5,1)}. After solving X²-11X+30=0, because of the sum of the sum and the product and the product of the sum and the product, it becomes so simple lol. Ok, I don't have the time to watch the video for now but I think that the time you use for that will allow to explain that fluently.

I figured it out within the first few seconds.

I felt like this video was only 1 minute long, went by fast

I still don’t get how 3,2 is a solution because if we multiply then square in the second equation it gives 36 and unless the first solution is negative which it isn’t then our answer would be 47. Maybe I’m just stupid

@deltalima6703

2 жыл бұрын

9x2=18. 3x4=12 18+12=30

Bruh I ended guessing the thumbnail ecuation while the video was still buffering 😂😂

You're a brilliant man. If you ever read this, could you do this as a teen?

I was going to see the video before guessing the answer on my own, but then you said to stop and try to solve and I did recognize a solution that works in this way: On equation 2 i can write x^2y + xy^2 as : xy(x+y) = 30. Then I will transform xy in the form A and the (x+y) term in B i will have the following system : A + B = 11 AB = 30. This system has a sum of two terms as first equation and the product of the same terms as second equation, I can find the solution trough the equation : z^2 - 11z + 30 = 0, the two solutions of my equations are: z1= 5 and z2 = 6. I return on my original unknown : xy = 6 and (x + y) = 5, solving in the same way I obtain : [x =3 ; y=2] and, redoing the same thing considering the opposite case ( xy = 5 ; (x+y) = 6) I obtain a specular result : [x =2 ; y=3] who are the two solutions. EDIT: Sad, I missed the 1,5 solution 😭

I have a question what is 1+1?

i saw the equation, solved the equaton with x=2 y=3, posted this comment and left. Too easy.

@seroujghazarian6343

2 жыл бұрын

1,5 works too

Solved easily

First one tried with 2 and 3 lol then seek with (1 ; ?) And found 1 5

Easy 2 and 3

b^2-11b +30 :)

Imagine you don't know anything about math. The only words you can read on this board are "let" and "answer" lol

Lol I just tried 3 and 2 and it worked

换元法，这题逻辑上比较清晰，不算难

Don’t know how but I just saw the solloution x=1 y=5

It is not hard,it it's easy

I remember I gave it to one of my friend who is not that smart. But he destroyed this question in an instant😳

Oh that’s was easy I tried it for 10mins

I can't believe I substituted eq 1 into eq 2 and solved the 5-term polynomial... 🤦🏽‍♂️

I both hate and love you at the same time. I factored out y in the first equation so I could rearrange into y = (11-x)/((x+1) and then substituted that back into the other equation. After a bit of algebra we get a quadratic which I was able to use synthetic division on to get x = 1,2,3,5 and then just plugging it back into the original equation for y = 5,3,2,1 respectively. Now I was a little skeptical about 4 solutions so I ended up plugging in x and y and did a left side, right side check which ended up being correct. But seeing your method is so musher faster compared to what I did.

We have the same solution : )

I've solve it. It was easy

_6

Iam Vietnamese this is so eassy

😁

Solved it in 10 seconds by the thumbnail just assumed one was 1 and found the other could be 5

@psychotix

2 жыл бұрын

SAME LMAO

@hydrarl3869

2 жыл бұрын

Same but I got 2, 3 and 3, 2

Isn't xy + x + y = 11 just (x + 1)(y + 1) = 12? I did it this way and figured out which pairs of (x, y) out of six fit the second equation...

@phamnguyenductin

2 жыл бұрын

That only works if the solution set consists of integers only.

@Yubin_Lee_Doramelin

2 жыл бұрын

@@phamnguyenductin That's right, though.

I got this answer in half min. 😀

@marktube5732

Жыл бұрын

Unbelievable , nice work

as long as the value of the equation is maintained (so both sides are still equal) then it does not matter how you go about solving the equation x²y=x(x*y/x) Now, the x is being distributed to x and y/x, this is unconventional but it doesn't matter since the equality is maintained. we simplify inside the parentheses x²y=xy x²=x divide both sides by x x=1 once again, though I may have used unstandard methods and ignored a math rule, I still got the correct answer so it doesn't matter too much ^^

no

Multiply the first equation by xy and subtract the second equation from that. You get a quadratic in xy that factors. Its two solutions substituted in the first equation give you two quadratics that factor. You can have all the solutions in about 90 seconds, if you take your time.

@aashsyed1277

2 жыл бұрын

Also use the quadration formula

Cristo viene pronto por su pueblo amigo amiga

2 and 3😂 That was not Asian level. Don't get offended guys I'm just kidding💀

nah this isn't hard, got it in 20 seconds

Sooooooo many unnecessary steps! I loved rest of your videos but here you complicated things unnecessarily