Olympiad Mathematics | Learn how to solve for x and y quickly | Math Olympiad Training

[@herodecesaire8805]

Interesting calculation. Great Job. Be blessed.

[@PreMath]

So nice of you, my friend Herode Glad you enjoyed it! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@monroeclewis1973]

Three powerful problem solving lessons to absorb: pay attention to the givens,I.e., x and y are positive integers; set equation equal to zero; don’t get lost in the algebra and forget to reason.

[@PreMath]

Great points! Thank you for your feedback! Cheers! You are awesome, Monroe. Keep it up 😀

[@schrodingerscat1620]

plz, bring questions from geometry too .

[@PreMath]

Sure, Kumari dear. We are working on it. Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@ramanma9915]

The sum of a fraction and the reciprocal thereof is a whole number since x,y are positive whole numbers. This is possible only if both numerator and denominator equals 1. If either or both of them is not equal to one, then the resultant will not be a whole number.

[@39rama]

This would be a nice general problem to prove

[@renesperb]

I would suggest to note first that the equations are symmetric in x,y. As a consequence we have with a solution x=a also y= a. Besides it takes a few seconds to see (without calculation) that x=1 and hence y=1 are solutins.

[@keinKlarname]

x * y = 4 is also symmetric in x and y. But x = 1, y = 4 is a solution, and 1 not= 4.

[@khoitruongbao5118]

This is my first comment on your channel:)))

[@PreMath]

Welcome aboard Khoi! Thanks for the visit Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@zahidbanatwalla]

I've been teaching my daughter maths for class 10 mostly by watching your videos.

[@PreMath]

That is awesome! Good to know that. Thank you for your feedback! Cheers! You are awesome, Zahid. Keep it up 😀 Love and prayers from the USA!

[@gatemanway]

I have another: (x/y) + (y/x) = x + y = x (y/y) + y (x/x) = (x/y) y + (y/x) x Equalizing coefficients: x = 1 and y = 1

[@PreMath]

Great Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@obadowski1974]

2:12 x²y+xy²=x²+y², you can do a polynomial identity and find that x=1 and y=1, that's a fastest way to solve this problem

[@arthurschwieger82]

Like the last one, I used the bruit force method and started with 1 as it looked like that would work and it did. I also looked to see if -1 would work and it doesn't. It is good to know there is a way to solve this vs guessing.

[@PreMath]

Great tip! Glad you enjoyed it! Thank you for your feedback! Cheers! You are awesome, Arthur. Keep it up 😀 Love and prayers from the USA!

[@KAvi_YA666]

Thanks for video. Good luck sir!!!

[@PreMath]

You're welcome! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@akuntumbal1485]

Understandable, have a great day

[@PreMath]

Thank you for your feedback! Cheers! You are awesome, Akun. Keep it up 😀 Love and prayers from the USA!

[@pdean6140]

Thank you very much for a very good representation

[@luigipirandello5919]

Nice solution. Thank you.

[@PreMath]

Welcome 👍 Glad you enjoyed it! Thank you for your feedback! Cheers! You are awesome, Luigi. Keep it up 😀

[@242math]

got it, thanks so much for the challenge bro, love algebra

[@PreMath]

Thanks for the visit Glad you enjoyed it! Cheers! You are awesome. Keep it up 😀

[@anatolygrudinin892]

It is symmetrical equation in respect to x and y. Therefore x must equal y that immediately leads to 2x= 2 and x=1 and y=1

[@croissant77]

I Believe this is wrong. Symetrical could be 2 solution (a,b) and (b,a) Not necessary x=y

[@jarekferenc1149]

@@croissant77 Normally, you could be right. But in this equation, there are no other numbers that would cause "asymmetry" of solutions. Hence, Anatoly is right, it is a perfectly symmetric equation, so x=y. And both are equal to 1, by the way.

[@phi1111ip]

I went with: without a loss of generality let y = x + a, where a is an integer. substitute in and solve for x. only valid solutions when a is an element of {0}. For a=0, x could be 0 or 1. as x was a positive integer x is an element of {1}. only solution is therefor x=1, y=1.

[@pedroloures3310]

Nice problem!!!!

[@Eduardomeatball]

Assume x

[@everything...k]

Isnt there a possibility that the terms (y-1) and (x-1) are additive inverses of each other?

[@eliasmazhukin2009]

No, because both x and y are positive

[@SuperYoonHo]

you are the greatest

[@PreMath]

Wow So nice of you, Yoonho Cheers! You are awesome. Keep it up 😀

[@SuperYoonHo]

@@PreMath yay

[@hamadwani3006]

Sir you are a brilliant tutor. I am preparing for IIT jee and you have helped me a lot by practicing the best questions of mathematics.

[@PreMath]

So nice of you. Glad you enjoyed it! Thank you for your feedback! Cheers! You are awesome, Hamad. Keep it up 😀 Love and prayers from the USA! Wishing you all the best in your exam.

[@hamadwani3006]

Thank you sir

[@RohitWason]

Nice. I changed it a bit after cross multiplying: ((x+y)²-2xy)/xy = x+y (x+y)²-2 = x+y Letting z=(x+y), gives you a quadratic in z: z²-z-1=0 I.e., z=2 or z= -1. Since the latter isn't possible, the only possibility for z=2 is x=y=1.

[@RohitWason]

NVM...I see a problem with my solution

[@HappyFamilyOnline]

Another interesting video👍 Thanks for sharing💖

[@PreMath]

Thank you! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@guyhoghton399]

The left hand side has the same value if you replace x by 1/x and y by 1/y, so the same must be true on the right hand side since it is symmetric in x and y. ∴ (1/x) + (1/y) = x + y ∴ (x + y)/xy = (x + y) ∴ xy = 1 ∴ x = 1, y = 1 since x, y ∈ ℤ⁺. --- UPDATE --- Having thought about it, I realise that the above argument is wrong! The left hand side will remain the same function of x,y if you replace x by 1/x and y by 1/y. However the right hand side does not have that symmetry. If you make the same change there then (x + y) will become (1/x + 1/y) which is a different function of x, y. There is no reason to equate those two functions and expect to get solutions to the original equation. It happened to yield the wanted values for x and y only because of the coincidence that the correct solutions are x = y = 1 and so x = 1/x and y = 1/y. It was a case of using the result to get the result! Apologies if I misled anyone.

[@PreMath]

Super, Guy Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@keinKlarname]

Guy, I first thougt: very nice idea. And I actually learned something from your clarification. Thanks for this!

[@harshavardhanpechetti]

Sir this is very easy. First do the LCM of LHS then after simplification 2 will reamain so we get x+y=2 then posiible values are x=1,2,0 and y=1,0,2

[@nicogehren6566]

veru nice job

[@PreMath]

So nice of you, Nico Glad you enjoyed it! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@govindashit6524]

Thanks dear.

[@PreMath]

You're welcome! Cheers! You are awesome, Govinda. Keep it up 😀 Love and prayers from the USA!

[@alexgyebi6291]

Good job

[@largestcamil4854]

can be solved for complex group

[@pranavamali05]

Thnku😊

[@PreMath]

You're welcome! Thank you for your feedback! Cheers! You are awesome, Pranav. Keep it up 😀

[@nibeditabaral4639]

Keep it up

[@mahalakshmiganapathy6455]

Nice quite interesting

[@PreMath]

Glad you think so! Thank you for your feedback! Cheers! You are awesome, Mahalakshmi. Keep it up 😀 Love and prayers from the USA!

[@mathsdone2265]

Master key of this sum is. x , y € +ve Z Like your explanation , sir. 👍👍👍👍👍

[@PreMath]

Glad you enjoyed it! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@zplusacademy5718]

Legendary presentation ❤️🙏🙏🙏🙏❤️🙏🙏🙏 Dear sir unsurpassable and praiseworthy job ❤️🙏

[@PreMath]

So nice of you Glad you enjoyed it! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@帝释天-z1w]

两个数的比较大小是否有一个统一的公式可用

[@sigmamaleslogokijalegi6683]

Very easy one

[@PreMath]

Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@jd-gw4gr]

nice problem

[@flexeos]

if you dont limit to integers 2,-2+sqrt(2) is also a solution

[@hgmoon68]

Good!! Surely (x,y)=(1,1) is a solution. By the way Is there other possibility ?? For example, (-19,20) is not a solution but are there any solution that forms (-.+) or (+,-) integer pairs??

[@quanhoang1671]

Why don’t we compare x/y

[@srividhyamoorthy761]

Wow never though to this i did till factoring

[@srividhyamoorthy761]

Nice

[@PreMath]

So nice of you, Moorthy Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@yegnanarayana]

Let x/y=a then y/x=1/a So a+1/a=x+y Then x=a,y=1/a Since x and y are positive integers a cannot be >1 If that is the case y

[@fhffhff]

90,32°,64,9°,24,48°

[@mariamanzanero2337]

interesting

[@wesleyhan3013]

Just by looking, it's no difficult to guess x=1; y=1. Assuming term-term equal, x/y=x and y/x=y. It suggests x/x=y and y=1. Likewise x=1

[@yegnanarayana]

Let x/y=a,then y/x=1/a So a+1/a =x+y then x=a,y=1/a Since x,y are positive integers a cannot be >1 If a>1,y becomes

[@PonLZG]

its maybe so easy question : x/y + y/x = x+y , x=? ,y=? if x=1 , y=1 1/1 + 1/1 = 1+1 1 + 1 = 2 2 = 2 but we will try -1 too if x=-1 , y=-1 -1/-1 + -1/-1 =-1+-1 1 + 1 = -2 2 not equal to -2 so x=1 , y=1 but x,y is not -1

[@imabstrong3726]

My solution before starting video. I struggled a lot with this problem at first since I thought we could use any real number 😅. Soln: Multiply both sides by xy and rearrange: x² + y² = x²y + y²x x²(1 - y) = y²(x - 1). If x > 1 is a solution, then both sides should be positive so 1 > y, which cannot happen. By symmetry, we also can never find a solution with y > 1. The only possibility left is (x, y) = (1, 1), which we can easily see as a solution and by the above is the only solution.

[@davidfromstow]

Without working through it like you did, it seemed to me that the only answer could be for each letter to equal 1; simple substitution of any other number didn't give the correct answer. Great question, though, with your customary elegant solution

[@PreMath]

Glad it helped! Thank you for your nice feedback! Cheers! You are awesome, David. Keep it up 😀

[@provip1276]

that a simple to solve that question =)) and so ez br

[@questrequested9171]

okay, but how do you know x=1 ; y=1 is the only solution? are there really no others? what if y=2 for example?

[@jarekferenc1149]

In such case, if x/y is integer, then y/x is not an integer. If you don't believe, read further. Note that RHS is the sum of integers, which, naturally, must be integer. So, LHS should also be integer. If y=2 (as you propose), x/2 must be integer, which leads you to the fact that x=2 , 4 , 6 , 8 etc. Fine! But y/x must also be integer. Then, 2/4 < 1, 2/6 < 1, 2/8 < 1 etc. If y=2, the only x that satisfies the requirement of integer LHS is x=2. So, 2/2 + 2/2 =? 2+2. No, NOT true. 1+1 does NOT equal to 4. Take any y greater than 2 and the result will be even more obvious that the equation is NOT true. The only solution here is that x=y=1.

[@tombuladam6370]

x²+y² ____ = x+y x ve y pozitif tamsayı ise xy (x+y)²-2xy __________ = x+y xy (x+y)²/xy =x+y+2 (x+y)²/xy tamsayı olmak zorunda ve kareli ifade (x+y)² 4xy,9xy,16xy,25xy..... Olmalıdır 4xy/xy. 4 =x+y+2. x=1 y=1 Pozitif x ve y

[@tankakhati993]

Why not for 9xy or 16xy..? 9=x+y+2 16=x+y+2..

[@PreMath]

Great, Adam Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@nirupamasingh2948]

V nice

[@PreMath]

Thanks for the visit. Thank you for your feedback! Cheers! You are the best Niru. Keep it up 😀 Love and prayers from the USA!

[@nirupamasingh2948]

@@PreMath 🙏🙏

[@emmaass8370]

i just simply remember that any fraction with denominator equals to 1 will be itself so i kind think both will be 1 because of x/y + y/x i never think it can be this complicated 💀

[@КатяРыбакова-ш2д]

Домножить на xy и сгруппировать: x=1, y=1.

[@Creativemathlearning]

Sử dụng tính chất nguyên dương để giải x và y.

[@Copernicusfreud]

With x and y in the equation, I thought I needed two equations to solve the problem with the two unknowns. I followed the steps of the video all the way to (x^2 + y^2) / xy = (x + y) / 1. Instead of doing a cross multiply, I set the numerators as equal and the denominators as equal to give me 2 equations with 2 unknowns. x^2 + y^2 = x + y and xy = 1. I went with y = 1/x and plugged into the other equation. Working through the equation, I ended up with x^3 -1 = 0 or x^3 - 1^3 = 0, so (x-1) * (x^2 + x + 1) = 0. 1st solution is x -1 = 0, so x =1. If xy = 1, then y is also = 1. x =1, y = 1. 2nd and 3rd solutions using the quadratic equations for (x^2 + x + 1) = 0, and solving for x gives [-1 + i * sq-rt (3)] / 2 and [-1 - i * sq-rt (3)] / 2, both of which make no sense, and I rejected both of those solutions as false. I went with x =1 and y = 1.

[@PreMath]

Fabulous job, Mark Thank you for your detailed feedback! Cheers! You are awesome. Keep rocking 😀

[@GodbornNoven]

(x²+y²)/xy = x + y We got no leads, so we'll have to start off with assumptions, assuming x = y 2x²/x²=2x 2=2x x=1 y=1 let's plug these back into original equation (1²+1²)/1=1+1 x=y=1 is a solution.

[@Uncle_Bob27]

X not equal 0 and Y not equal 0 we can see in the beginning, because we have x/y, y/x and we can't devide by 0.

[@manualrepair]

🤘👀👍

[@PreMath]

So nice of you Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@SMARTpeople1135]

why y-1 should be 0????

[@Sanguinium_Light]

You could solve this question by considering the fact that (X/Y) + (Y/X) equals X + Y only if the demominator is 1.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome, Lucas. Keep it up 😀

[@brinzanalexandru2150]

Let x+y=I and XY=v then (u²-2v)/v=u now divide by parts to get u²/v -2=u now move the u to the RHS and the 2 to the LHS and factor u => u(u/v -2)=2 now that's a easy Diophantine equation we just equalise to the integer factors of 2 and get u=2 and v=1 from where x=1 and y=1

[@PreMath]

Very well done! Thank you for your wonderful feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@brinzanalexandru2150]

@@XJWill1 that's my auto corrector😅 but thank u

[@srinivassingarapu7825]

Instant guess x=1 y=1 works

[@xyz9250]

Or can just move everything to the left and factor out like this. x(1/y -1) + y(1/x -1) = 0 and the result becomes obvious

[@PreMath]

Nice Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@mda99das]

you can just know at a glance that they are both 1!!!!

[@tofro4204]

It can also done as If x, y are positive integers and RHS is always integer Then LHS must be an integer Since x, y are positive integers so x/y + y/x is an Integer when x, y are equal to 1

[@mrunalmodak]

Maths is roaming everywhere #mathmarvel

[@edgaracosta6261]

Seriously, it was necessary, so much calculation was necessary, at first glance it was seen that x=1 and y=1

[@राजनगोंगल]

👌👌👌👌👌👌👍👍👍👍👍👍👏👏👏👏👏👏

[@PreMath]

यात्रा के लिए धन्यवाद, राजन। आपका स्वागत है! कितने अच्छे हो। प्रोत्साहित करना! आप कमाल हैं। लगे रहो यूएसए से प्यार और प्रार्थना!

[@alexbaronov4736]

Hmm, not sure this is quite complete. Try x=2+2×sqrt(1÷3), y=2-2×sqrt(1÷3). This should be one more of infinitely many answers.

[@kiyanwest149]

i been out of school for so long i gotta start from scratch because what? tf is you even talking about ? and i graduated with a regents diploma and over 85 in math so i def gotta touch up on the math

[@anatoliy3323]

👍🍷Cheers!

[@PreMath]

Thanks for the visit Cheers! You are awesome, Anatoliy. Keep it up 😀 Love and prayers from the USA!

[@NurHadi-qf9kl]

.x^2+y^2=(x+y)(xy)

[@goranblomberg1173]

Immediately I see that x = y = 1 is one solution.

[@parsashahrestani6590]

In my opinion we could have done it simpler if we look more closer to the equation we wilI understand that what ever has been happended for X has been happended for Y ، I mean because of symmetry X would be equal to Y so they would be 1 Although we can just guess it and it is lot more easier

[@PreMath]

Great job, Parsa Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@CTJ2619]

1,1 is an obvious soluion

[@jarekferenc1149]

Maybe obvious, but try to prove with necessary formalism that it is the only one. No other way than solving like the mathematicians do.

[@devondevon4366]

Answer x=1 , x=0 y=1 y=0

[@devondevon4366]

can't divide by 0 , x=1 and y=1

[@PreMath]

Super good, Devon! Thank you for your nice feedback! Cheers! You are awesome. Keep it up 😀

[@jarekferenc1149]

@@PreMath Devon proposed DIVISION BY ZERO, and you compliment his/her solution? Please, don't be ridiculous. Respect yourself and don't ruin your reputation.

[@PreMath]

@@jarekferenc1149 Dear Jarek, we are all human. Our apologies if any mistake was made unwittingly. Peace! 😀

[@jarekferenc1149]

@@PreMath Errare humanum est, we have known this since the Romans (Seneca's?). But showing them publicly is not the best policy :-) . Your yt channel is too good to make so silly errors. Carry on with fertilising viewers' minds, including mine.

[@ourv9603]

If letters are involved it is not math. It is algebra. !

[@mokhmahdan7190]

X = 1 dan Y =1

[@РомочкаГефнер]

Lol, why did you calculate anything? If x and y are positive integers, they cant be less than one. Lets say one of them, x for example, is greater than one. In this case x/y

[@Germankacyhay]

Интересно, а если бы (x ,y) € Z -

[@PreMath]

Great point! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀

[@jarekferenc1149]

If x0, so left hand side > 0. Regretfully, right hand side is the sum of two negative numbers, which is < 0. So, your proposal leads to false equation, both sides will never be equal. They could be equal if x=y=0, but this is not possible: division by zero is forbidden.

[@yakupbuyankara5903]

X=Y=1.

[@PreMath]

Great job, Yakup Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!

[@gibbogle]

Obvious by inspection.

[@adgf1x]

x=y=1

[@chingyuenyik9436]

Obviously, 1+1=1+1 No need to calculate

[@jarekferenc1149]

Yes, this one is obvious. But try to prove that this is the only solution. Then, you will see that it's a bit more tricky and you will need a little bit of calculations.

[@ИванВоронин-и2м]

x=y.

[@sturilin]

Решение выполнено неаккуратно. Всех благ

[@patpat5135]

It was obvious that 1/1 + 1/1 = 1 + 1 No need to go thru all this waste of time.

[@quangkhai5306]

xàm

[@jalma9643]

What does x, y ∈ ℤ⁺ mean?

[@PreMath]

x and y are positive integers! Thanks for asking.

[@patpat5135]

It was obvious that 1/1 + 1/1 = 1 + 1 No need to go thru all this waste of time.