Math Olympiad | A Nice Algebra Problem | How to solve for X and Y in this problem ?

[@ivansudakov6877]

Как всё же хорошо знать формулы бинома Ньютона для (x + y)^3 и (x + y)^5, теорему Виета и метод подстановки! Сколько бумаги можно сэкономить!

[@odinrossi3353]

Ага, видео - та ещё наркомания)

[@poonkodisrikumar]

😊​@@odinrossi3353

[@justme2012lena]

Да, точно!

[@ВладимирГолованов-й5я]

А сколько ещё ошибок можно допустить во всей этой писанине!

[@muxlisashomirzaeva9963]

Индус все усложнял

[@AliBarisa]

I always feel like you complicate matters when it is actually unnecessary. Eg. X=4-Y ...(1) and XY=14...(2), just substitute (1) into (2) and solve the resultant quadratic equation. For the other case, XY=2...(3), So, substitute (1) into (3) and solve the quadratic equation. period.

[@Ankara_pharao]

Sure, i was expected the end of video at this point but suddenly he was in the middle of explanation.

[@tyronekim3506]

I agree. 😅

[@SalmanAhmed-zo4pi]

Same thought! 😊

[@justme2012lena]

That what I thought as well.

[@alessandracampelli6090]

Exactly 🙁

[@himanshupaliwal2026]

a very simple solution can be to use binomial therorem (x+y)^5 and then create the expansion in terms of x+y and xy, u will directly get the quadratic equation to calculate xy.

[@assat10]

Agreed

[@krwada]

binomial theorem is the fastest way to solve this thing.

[@knotwilg3596]

The elementary way to solve this is reducing the higher powers to lower powers by substitution. Since x^5+y^5 = (x+y) (x^4-x^3y+x^2y^2-xy^3+y^4) => x^4-x^3y+x^2y^2-xy^3+y^4 = 116 At the same time (x+y)^4 = x^4+4x^3y+6x^2y^2+4xy^3+y^4 = 256 Subtracting gives 5x^3y+5x^2y^2+5xy^3 = 140 => xy (x²+xy+y²) = 28 We further reduce x²+xy+y² = (x+y)² - xy = 16-xy This gives xy (16-xy) = 28 and this is a quadratic equation in xy with solutions 2 and 14 This gives 2 sets of equations (x+y=4, xy = 2) and (x+y=4, xy=14) which we can transform into quadratic equations by substitution Those give solutions (2+v2, 2-v2) and (2+2i v10, 2-2i v10) where the latter exists in the complex field. Even if you show this step by step, this can be done in 5 minutes. You are math literate but you take unnecessary long roads to the solution.

[@jiqin4298]

this is the real method that the question setter want the solvers to find out, not the way in video

[@THESURAFELB]

Thanks🔔

[@BP-gn2cl]

Correct

[@sparkgamer6625]

This is no different from what the poster did. Same process of reducing powers.

[@damirdukic]

Another nice way to solve this equation, and I think a quicker one, is to substitute "x" with "2 + u" and "y" with "2 - u". This way, when we expand "(2 + u)^5 + (2 - u)^5 = 464" we get a biquadratic equation, which is easy to solve.

[@IvanPetrov-rs1kq]

The equation is solved in 10 minutes not in 25 in the clip.

[@stantackett107]

Yeah, I hate how these equations are made way more complicated than they need to be. Some I can solve in my head in 10 seconds. This one, needs paper​@@IvanPetrov-rs1kq

[@jxucantab]

I did it this way, before actually watching the lengthy video. After binomial expansion and cancelling, it reduces to u^4+8u-20=0. then either u^2=2 or u^2=-10. For real solutions, t^2=2, thus t=+-sqrt(2). The real value for the (x,y) pair is (2+sqrt(2), 2-sqrt(2)). The complex pair would be (2+i*sqrt(10),2-i*sqrt(10)). Much quicker.

[@guyhoghton399]

Once you have their sum and product you can write down the equation for _x_ and _y_ immediately, since _t² - (x + y)t + xy = 0_ is satisfied when _t = x_ and when _t = y_ E.g. _x + y = 4, xy = 2_ _t² - 4t + 2 = 0_ with roots _2+√2, 2-√2_ which means that these are the respective values of _x_ and _y_ in either order.

[@ΑντώνιοςΤσακίρης]

This equation is aka Vieta's formula.

[@paulortega5317]

If f(n) = x^n + y^n then f(n+2) = f(1)*f(n+1) - xy*f(n) and f(0)=2 For this problem f(1) = 4, f(5) = 464, and we first need to solve for xy Let xy = b then f(n+2) = 4*f(n+1) - b*f(n) f(0) = 2 f(1) = 4 f(2) = 16 - 2b f(3) = 64 - 12b f(4) = 256 - 64b + 2b^2 f(5) = 1024 - 320b + 20b^2 = 464 20b^2 - 320b + 560 = 0 b^2 - 16b + 28 = 0 (b - 2)(b - 14) = 0 xy = 2 or xy = 14 each with 2 solutions

[@заряд-о3д]

Наиболее краткое решение с верным ответом, при условии применения более простых методов - является верным!.... Здесь такого нет - решение излишне усложнено, из за этого более длинное - значит решение некорректно.... Можно было сразу использовать формулу а^n + b^n , при n = 5 .... Более того решив систему двух уравнений и в последствии квадратное уравнение получилось бы по человечески!

[@user-kr1zj6lm2u]

I have been away from math for 25 years. I was really shocked at how it came flooding back. Old brain but it is not dead yet!

[@juliovasquezdiaz2432]

Hay que practicar matemáticas todos los días. Tengo 72 añitos jaja y sigo practicando. Saludos desde Chiclayo Norte del Perú

[@brianwilson2848]

I've worked in mathematics for decades, I have no idea what cancel is as an operation. Everybody that uses it seems to confuse it with dividing out, or subtracting out. However: X/X=1, X-X=0. The reason most people have trouble with algebra is they're using the same word for two completely different operations. There is no such operation as "cancel". Things either divide out, or they subtract out. You have to use the correct term if you're going to teach students what they're supposed to do. Many thanks to Mr Burt Fetters, my college algebra teacher, who drilled this into the class, and finally made algebra an understandable system.

[@knotwilg3596]

I agree with you that "cancel" can be confusing to beginner mathematicians who don't really know what's going on. So, when teaching elementary algebra, it should be avoided and instead the actual operations must be made explicit. As a shorthand for an elementary operation during a more elaborate exercise for algebra literates, I think it's fine.

[@prime423]

On the mark!!When in doubt, cancel out is NOT MATHEMATICS but gibberish.

[@Ярослав_Вдовин]

You could see that for x+y=4, xy=z, results will be symmetrical for each pair of rezults for both z.

[@宮前達也]

対称式なので、ｘｙを求めることが鍵だとは分かったが、その求め方が自分のやり方と違ってスマートなのに感心しました。あとｘｙを導き出した後の最後の解の求め方は２次方程式の解と係数の関係を自分は使った。

[@dinlendiricidrtv]

Thank you very much my dear friend

[@АлександрКомов-ъ2к]

Если уравнение содержит х и у в пятой степени, то это уравнение пятой степени, и тогда всех возможных корней (включая комплексные) должно быть пять. А тут четыре

[@domenicocacciatori]

Esatto. Il ragazzo è bocciato

[@CecilPonsaing]

Beautiful work.

[@learncommunolizer]

Thank you very much!!!

[@ТатьянаСтрокова-щ1р]

Так можно решать до бесконечности, а решить можно гораздо проще.

[@wes9627]

Study this much simpler solution method. Substitute x=z+2 and y=-z+2 into the given equation and rearrange to (z+2)^5-(z-2)^5-464=0 Pascal's Triangle: 1 5 10 10 5 1; 2[5*2z^4+10*2^3z^2+2^5]-464=0; 20z^4+160z^2-400=0; z^4+8z^2-20=0 z^2=(-8±12)/2=2 or -10 and z=±√2 or ±i√10 x=z+2=2±√2 or 2±i√10 and y=4-x=2∓√2 or 2∓i√10

[@charlesmitchell5841]

Nice problem. Nice job in solving it.

[@learncommunolizer]

Thank you very much!!!

[@SH331]

Using Powers And roots to equation have some rules And leads to reducing interval of usable numbers.

[@ThuNguyen---]

Bài toán khá phức tạp phải áp dụng nhiều kiến thức toán học, cùng vài trang giấy mời giải được !

[@dogethsamurai2390]

Thank you this a long and delay process that take a lot of step, and a lot memorization require. This video show how to solve this equation with a garanteen answer.

[@Miroslav_Bulgaria]

Good job! Well done! Otherwise I understand that may be in different regions and countries it simply use another approach to Quadratic equations as: ax2 + bx + c = 0 to standard solution…., but was shown it is easy for simple and small numbers but otherwise it will be really difficult to determine the answers….

[@medbou3136]

You must use relation betwen th summ and product of the roots for equation of 2nd degrees.

[@FRANKZHANG-un1ov]

1. \(x = 2 - \sqrt{2}, y = \sqrt{2} + 2\) 2. \(x = 2 + \sqrt{2}, y = 2 - \sqrt{2}\) 还有两个复数解，通常情况下我们只考虑实数解。

[@jyotishah9140]

It’s so long I think how to do in exam short period Thanks

[@babyghost4434]

nice solution i really love longer solutions it makes me feel great hahah

[@zainquadri1206]

I am as dumb as I started... If anything I am dumber...

[@sasisugu]

Explained well sir I have studied maths only upto 10th std. In +2 I've studied pure science This solution which taught by you is very difficult to me. So, so hard sir. How can I pick up it.🙄

[@MathForEverybody]

Nice!

[@learncommunolizer]

Thank you! Cheers!

[@kuuso1675]

When we meet x+y=a and xy=b, we better say x,y are the solutions of t^2-at-b=0. Though in the video (x-y)^2=8 is changed into x-y=±2√2, it should be |x-y|=2√2 and will need case treatment. We may omit them with proposed explanation.

[@kuuso1675]

t^2-at+b

[@ft7339]

Should be: t^2 - at + b = 0

[@vasileioszounarelis5794]

Μπράβο!!!

[@ЭдуардПлоткин-р3л]

Зачем же так сложно решать? Этот же ответ я получил через 5 минут.

[@sinisakarabatkovic-fq8qg]

While it is good for brain 🧠 maintenance to think or practice logic, it is crucial to learn how to put problems in mathematical form. If your problem is not correctly described by a mathematical equation then result is wrong. Garbage in garage out is what they teach you in computer classes.

[@phoebe543]

I think you'd get more positive feedback if you offered this as an alternative solution and do quadratic in method 1.

[@Ta-tamaan]

I like how you always say "bracket" but always put a "parenthesis"❤

[@pengliang151]

Why a degree 5 equation only have 4 sets of roots ? Is something missing ?

[@Dante-420]

If you solve x^5+y^5 = 464 for one variable, you get five solutions because it's a fifth degree polynomial equation. But this isn't a polynomial equation - it's a *system* of polynomial *equations*. Finding the number of solutions to a system of polynomial equations is much more difficult, because it's complicated by factors like the independence of equations and the possibility of infinitely many solutions (check out the wikipedia article for "system of polynomial equations"). In this particular case, we're lucky because the fifth degree polynomial x^5+y^5 is divisible by x+y so we just need to solve the fourth degree polynomial that results from the quotient, hence four solutions. (Also, normally we would need to prove the demoninator of our polynomial division is not zero, but here we are given x+y=4)

[@mohamedyassine6250]

Hello . Please I need the solution to this equation. x,y are reals such that: x²-xy+y²=76 What is the value of x+y=?

[@Caturiya]

I solved it I hoped nicely Allah Akbar 4 you kzbin.info/www/bejne/nYKld6ianbGXe9U

[@IEPYandPUYO]

Nice video! 👍 You should've written x1y1 and x2y2 at 11:51, because the way you wrote xy=14 and xy=2, it looks like you're saying that 14=2. 😁 Same thing for u. Should've been u1 and u2.

[@olayinkaadeleye3134]

Nice

[@learncommunolizer]

Thank you 👍👍👍

[@user-Aliceten]

x^5+y^5＝464なので 5乗の場合　解が5個存在しないとおかしいので 解が4個は不正解です ヒントだけ書いときますね もう一個の解は2を使いません　ｘかｙにマイナスがあるため　結構大きな数字が入ります

[@Dante-420]

I wrote this in another comment, but the reason we have 4 solutions is because this is a "system of polynomial equations" instead of a polynomial equation. In this particular system, we can divide x^5+y^5 by x+y to obtain a 4th degree equation, which is what is being solved. This division is valid because x+y=4≠0

[@Darisiabgal7573]

Of the solutions two were imaginary and two were trivial, but since one of the imaginary solutions was also trivial the is in only one non-imaginary non-trivial solution.

[@WildRhinocerosDriver]

А зачем так сложно то? ))) Пришла пора попробовать себя в физике или химии

[@АндрейКожевников-о8й]

Гораздо проще решать методом итераций: подставляем 2, первое выражение попадаем в точку а во втором при прямой подстановке получаем значение равное 64. Артнллеристы скажут, что до цели большой недолет. Подставляем в оба выражения Х=3, в обоих выражениях получаем большой перелет. Во втором выражении Х=648, поэтому взяв эти выражения в вилку(как говорят артеллериств) будем подбирать промедуточное число на отрезке 2 и 3.

[@ft7339]

So you will get only the real answers!

[@zohrealizadeh6827]

Thank you❤❤

[@thomasmunch2190]

Can you help me compute the flux integral of the surface S using the divergence theorem if the vector field F = yj and S is a closed vertical cylinder of height 2, with its base a circle of radius 1, on the xy-plane, centered at the origin ?

[@AlexMarkin-w6c]

The solution is straight forward for school students: y=4-x => x^5+(4-x)^5=464 (x^2-4x+2)(x^2-4x+14)=0. For real solutions , (x^2-4x+2)=0 (x-2)^2-2=0,. Alternatively, Vieta's formula and symmetry.

[@assat10]

Good job mate!

[@learncommunolizer]

Thank you very much!!!

[@сергейрождественский-ь9х]

❤❤❤❤

[@fei7954]

I more interested how many pages you used eventually

[@KingAntDaProphet]

Just divide by five and split it in half

[@reinymichel]

What a long ending !!! At the end, once you have xy=14 and x+4 = 4 x(4-x) = 14 x^2 - 4x + 14 = 0 x = (4 ± √-40)/2 , or x = 2 ± √10 i , same thing for the other case , so you your 2 x's , easy to get the y

[@ControlledDemolition]

x = ~.58 y = ~3.42

[@MrUtubePete]

Nice pen

[@louismendoza9795]

A good mathematician does not make a good teacher. I suggest using a logic line on the right of your equations so that you can explain their thought processes.

[@ВалентинГорелов-ф2э]

Учителю надо бы поучиться умножать целые числа

[@xopenmind]

too complicated .. xy=P and x+y=S you can solve by x^2 -Sx +P = 0 ...

[@estevan2065]

4 da matina, parei de assistir na hora que ela mudou de folha, poucas ideia pra Y menos ainda pra X

[@ajeethpandey]

A symple problem made complicated and much more the time 25 mins !

[@anakham79]

Despite this solution is very detailed, I think it is not entirely correct. Exponentiation of left and right equation parts, as well as multiplication of equation parts, is not equivalent transformation. x=-1 is not equivalent to x^2=(-1)^1, and even x=1 is not equivalent to x^3=1 in the field of complex numbers. If you added equations (3) and (4) to the initial system, it would be equivalent transformation of system as a whole. But in this case you should check every solution of your intermediate equation, whether it feats to both initial equations (1) and (2). And we all will see the practise of exponentiation to the power of 5 complex numbers and numbers with radicals. Or you should use fundamental theorem of algebra about polynomial complex root number, using transition to biquadratic equation, already mentioned in comments. But in this case may be it's worth just to use formula for biquadratic equation?

[@flyingfish2011]

完全看蒙了，如果是考试，估计半个小时做不完一道题呀😂

[@haothe1099]

X^5 + Y^5 = 464

[@fei7954]

How many pages you used 😊eventually

[@echodupeterwilliamgriefiel3832]

like your pen, whts the type name?

[@user-we5kx2tn6o]

Это всё равно что лететь из Берлина в Париже через Пекин.

[@user-tl8ib7dp4y]

😂😂

[@HoangVu-nt2uo]

Trực tiếp mũ 5 phương trình x+y=4 cho ra cái x^5 và y^5 luôn

[@rogelioroldan9527]

Siempre te complicás Ecotú querido. Sustitución x=4-y, lo sustituyes en la 2a ecuación. Mirá que fácil. No te compliques tanto Ecotú.

[@edwardtaschtschyan2707]

Х=1, у=3(х=3 у=1)

[@adefumiemax-macarthy950]

😊😊

[@rameshkmishra2517]

Why you are making it that much complicated

[@HiepThach-uw1ix]

Quá phức tạp

[@IvanPetrov-rs1kq]

Haven't you studied about a discriminant of a quadratic equation? 8th or 9th grade in the school. The problem is much easier to solve, not in 25 minutes.

[@SH331]

Ať time 8.36 u²-6u²!=5u² you forget minus

[@davereynolds739]

He's moving everything to the right side of the equation, so 5u^2 is correct.

[@alexandredutremblay7050]

Something bothers me. Intuitively, I would have thought that a 5 degrees équation should have 5 couples of solutions in C, no? Just like a 5 degrees equation of x has 5 solutions in C. I would guess that I'm wrong somewhere, but i dont know where, and if i'm not where ois the fifth solution ?

[@knotwilg3596]

There's an extra constraint x+y=4. This reduces the solution space with (at least) one degree.

[@DonaldKim33]

When I asked Chatgpt It is said that the x ≈ is 2.532 and the y ≈ is about 1.468.

[@maxlion3213]

...io vado al cinema..😢

[@waimanchin495]

一般聯立方程要的是實數解！

[@cachotrelles4715]

Buenas tardes, hermosa explicación. Y, como sostengo, las matemáticas, sus reglas y sus números SON EL IDIOMA...no importa la lengua en que se explique. Gracias

[@gLg359]

x5,y5=?

[@AlbaRosa-ui3xs]

Forse questi video sono rlivolti a chi non ha dimestichezza con la matematica e si sceglie un metodo più lungo,ma più semplice da comprendere...

[@BrandonCBlack-zn6qt]

why not try brute force method : X = 3.4142, : y = 0.585799. Thanks

[@markterribile6948]

Interesting problem, interesting solution technique. BUT you belabor and agonize over basic arithmetic, but leap over the error-prone operations such as collection of terms. Your presentation could be so much more helpful if you show the error-prone steps, but move quickly through the truly elementary stuff.

[@markterribile6948]

Followup: Each step you show should be as much as can be verified by quick inspection, neither more nor less. Anyone taking this problem on already has elementary algebra. The magic of this solution lies in generating formulas that allow the repeated substitution of known values. That's the magic, and the structure of that magic should be foremost in the presentation.

[@UshanaMakeba]

Can someone explain to me at 13:00 when he just puts in (x+y)^2 + (x-y)^2= 4xy, where do the (x-y)^2 and 4xy come from? It feels like he just adds them in out of nowhere, though he says it’s an algebraic rule.

[@catherine7304]

it is （x+y)^2 - (x-y)^2=4xy

[@sgenov]

If I was reviewing your work, I would give you a low grade for not using a simpler solution.

[@charliekollarovics5084]

Miért nem használod a másodfokú egyenlet képletét?

[@carlosdavila382]

12:45 es obvio pero como se te ocurrió

[@DuyGiap-z5k]

Những phép phân tích quá đơn giản mà diễn giải quá chi tiết làm mất thời gian và gây khó chịu quá

[@bathaoang5061]

X .y tinh bình phương lập phương trình vv. Quy đồng mẫu số và rút gọn phân số vv. OK bạn vvv.

[@lucksys]

IS a big fail cancel (x+y)² with the squere root , because this is the a absolute value definition. So is wrong !!!

[@thekennethofoz3594]

I find it really poor practice to use "x" for both a variable and a multiplication symbol in the same equations. So much opportunity for confusion, and it really slows down the scanning of a line.

[@afinstal]

Congratulations. You found the way to hate maths.

[@key_board_x]

(x + y)² = x² + y² + 2xy → given: x + y = 4 16 = x² + y² + 2xy (x + y)³ = (x + y)².(x + y) (x + y)³ = (x² + 2xy + y²).(x + y) (x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³ (x + y)³ = x³ + y³ + 3x²y + 3xy² x³ + y³ = (x + y)³ - 3x²y - 3xy² x³ + y³ = (x + y)³ - 3xy.(x + y) → given: x + y = 4 x³ + y³ = 64 - 12xy (x + y)⁵ = (x + y)².(x + y)².(x + y) (x + y)⁵ = (x² + 2xy + y²).(x² + 2xy + y²).(x + y) (x + y)⁵ = (x⁴ + 2x³y + x²y² + 2x³y + 4x²y² + 2xy³ + x²y² + 2xy³ + y⁴).(x + y) (x + y)⁵ = (x⁴ + 6x²y² + 4x³y + 4xy³ + y⁴).(x + y) (x + y)⁵ = x⁵ + x⁴y + 6x³y² + 6x²y³ + 4x⁴y + 4x³y² + 4x²y³ + 4xy⁴ + xy⁴ + y⁵ (x + y)⁵ = x⁵ + y⁵ + 5x⁴y + 5xy⁴ + 10x³y² + 10x²y³ x⁵ + y⁵ = (x + y)⁵ - 5x⁴y - 5xy⁴ - 10x³y² - 10x²y³ x⁵ + y⁵ = (x + y)⁵ - 5xy.(x³ + y³) - 10x²y².(x + y) → given: x + y = 4 x⁵ + y⁵ = 1024 - 5xy.(x³ + y³) - 40x²y² → recall: x³ + y³ = 64 - 12xy x⁵ + y⁵ = 1024 - 5xy.(64 - 12xy) - 40x²y² x⁵ + y⁵ = 1024 - 320xy + 60x²y² - 40x²y² x⁵ + y⁵ = 1024 - 320xy + 20x²y² → given: x⁵ + y⁵ = 464 464 = 1024 - 320xy + 20x²y² 20x²y² - 320xy + 560 = 0 x²y² - 16xy + 28 = 0 → let: z = xy z² - 16z + 28 = 0 Δ = (- 16)² - (4 * 28) = 256 - 112 = 144 = 12² z = (16 ± 12)/2 First case: z = 14 → recall: z = xy → xy = 14 Second case: z = 2 → recall: z = xy → xy = 2 Resume: x + y = 4 ← this is the sum S xy = 2 or xy = 14 ← this is the product P So x & y are the solution of the following equation: a² - Sa + P = 0 First case: xy = 14 and (x + y) = 4 a² - 4a + 14 = 0 Δ = (- 4)² - (4 * 14) = 16 - 56 = - 40 = 40i² a = (4 ± i√40)/2 a = (4 ± 2i√10)/2 a = 2 ± i√10 → x = 2 + i√10 Recall: y = 4 - x y = 4 - 2 - i√10 → y = 2 - i√10 → x = 2 - i√10 Recall: y = 4 - x y = 4 - 2 + i√10 → y = 2 + i√10 Second case: xy = 2 and (x + y) = 4 a² - 4a + 2 = 0 Δ = (- 4)² - (4 * 2) = 16 - 8 = 8 a = (4 ± √8)/2 a = (4 ± 2√2)/2 a = 2 ± √2 → x = 2 + √2 Recall: y = 4 - x y = 4 - 2 - √2 → y = 2 - √2 → x = 2 - √2 Recall: y = 4 - x y = 4 - 2 + √2 → y = 2 + √2

[@waimanchin495]

坐標內有i，是什麼意思？

[@waimanchin495]

是什麼意思？

[@吳泰樑]

將-1開根號之值，是個虛數，用i代表。

[@user-tl8ib7dp4y]

root -1 = i

[@arupsaha5967]

Fo u know math or u are writing mohabharat

[@ibrahimbakr8788]

Main key is the find XY ... after that we can solve this problem::))

[@NonyaBusiness-q6r]

It's 46÷23 =2 then 4 divided by 2 thats 2 that's 2, 2s so it's 2to the power or 2 2square idk how to articulate how you say it properly in mathematics

[@ashkannikzad1432]

140- its correct not140+

[@Nepomenik]

:)

[@秦劍鋒]

算完天都亮了，下一位

[@ivantdibere8020]

Подставлял кто-то эти решения в уравнение?У меня не выходит(выходит на первое).

[@ЮрийКочетов-р5п]

Да, подставлял. Обращаются в верные равенства. Все четыре пары являются корнями системы

[@ЮрийКочетов-р5п]

Кстати, у Вас ценное замечание на счёт проверки корней уравнений. 👍 При возведении частей уравнения в степень, при замене переменной (что в данном решении встречалось) могут появиться "посторонние корни", которые не являются корнями исходного уравнения. Так что проверка обязательна и без неё решение не является полным:-)

[@vahan6500]

Bruh . X=4-y …