Math Olympiad: How to Solve a System of Simultaneous Equations | Algebra Identities: Tricks to Know 

Appears to be a lengthy method. Here is one easy method. Let x^2 = a and y^2 = b. So that a - b = 24. Again ab = x^2 .y^2 = (xy)^2 = 35^2 = 1225. We now from algebraic formula (a + b)^2 = (a - b)^2 + 4ab. So that (a + b)^2 = 24^2 + 4.1225 = 5476 = 74^2. Now putting the values of a and b, we get (x^2 + y^2)^2 = 74^2. So x^2 + y^2 = 74 (we can not take -74 as the left side is sum of two squares term). Now x^2 + y^2 + 2xy = 74 + 2. 35 = 144. That is (x + y)^2 = 144. So our x + y = (+/ - ) 12

y = 35/x x² - y² = 24 x² - (35/x)² = 24 x⁴ - 24x² - 35² = 0 x⁴ - 24x² - 5²*7² = 0 (x² + 5²) (x² - 7²) = 0 x² = -5²(undefined), 7² 1. x = 7, y = 35/7 = 5 ==> x + y = 12 2. x = -7, y = 35/(-7) = -5 ==> x + y = -12 Ans: x + y = ±12

Because xy=35 the solutions must either both be +ve or both -ve, therefore there are no imaginary number solutions

Just by looking at the numbers, you can deduce 7 x 5 = 35. And by proving 7**2 - 5**2, you confirm. But it's good to see a complete answer by a didactic method to solve it.

We have X² -Y² = 24 (Eq. 1) and XY = 35 (Eq. 2). So, Y=35/X, we substitute in Eq. 1. Then X² - (35/X)² = 24. Solving we get X⁴ -24X² - 1225 = 0. We factor (X² -49)(X² +25)=0. Then X=±7 and Y=±5. So X + Y = ±12.

А в чем проблема просто из XY = 35 просто выразить Х, и получится Х = 35/Y? А потом поставить это выражение в 1 равенство, получая 1225/Y²-Y²=24. Домножаем на Y², чтобы избавиться от знаменателя, и получаем Y⁴+24Y²-1225=0... Отсюда найти Y - это уровень 8 класса. Почему так легко? У нас в школе в учебниках задачи намного сложнее задачи, а тут олимпиада. Может, этот способ просто нельзя здесь использовать, и об этом сказано в видео? Просто, я не очень хорош в английском

By inspection x = 7 and y = 5 (and their negatives since we're squaring). So x + y = 12 or -12. First clue: there's only so many ways to get to 35 with xy (assuming we're dealing with integers).

12. If this is a test, you can make some assumptions to answer quickly. Granted, you cannot do this in real life. My assumptions are that x and y are factors of 35 and x > y. Try x =7 and y = 5. 7^2 - 5^2 = 49 -25 = 24. x + y = 7 + 5 =12. It took me less than 5 seconds, which helps in a test where time is limited. Edit: x= -7 and y = -5 are also solutions.

この動画の解法よりも、2つの式から代入法でyのみの式を作った方が易しい。 5476が74の2乗はすぐに計算できないと思う。

35 is only can be found at 7x5 So the answer is tested after 49-25=24 The X + Y = 12

the task is Solve a System of Simultaneous Equations. in this case this task/statement is wrong. the solution is different in set of R and on set of C. who will decide which one is the good solution? in the first case you wrote X+Y=+-12. this is not exact solution because sum of X=1 Y=11 is +12 but wrong in the other equations. Well, I think in this case this video is misleading...

An elegant way to solve this system is to use complex numbers to avoid solving quadratic equation twice. Here we go. Considering x and y as real numbers. Set z=x+iy so z^2=x^2-y^2+2ixy=24+70i Then |z^2|=√(24^2+70^2)=74 Using z_=x-iy conjugate of z, We have z+iz_=x+iy+ix-i^2y So z+iz_=(x+y)(1+i) Then |z+iz_|^2=2|x+y|^2 Moreover, |z+iz_|^2=|(z+iz_)^2| So 2|x+y|^2=|z^2-z_^2+2izz_| =|z^2-z^2_+2i|z|^2| =|2i.Im(z^2)+2i|z|^2| =|2i.(Im(z^2)+|z|^2| =|2i|.|Im(z^2)+|z|^2| =2.|Im(z^2)+|z|^2| As Im(Z)= imaginary part and. |Z| module of complex Z. Then |x+y|^2=|Im(z^2)+|z^2|| So x+y=+/-√|Im(z^2)+|z^2|| Knowing that z^2=24+70i and |z^2|=74 then x+y=+/-√(70+74)=+/-√144 So x+y=+/-12 Greetings

This type of equation does not need the extensive work. The answer is 5+7 =12 7^2 - 5^2= 24

AN extremely complex procedure to solve a simple problem. !!!!

Did it in my head in 3 seconds

Translated from Japanese In Japan, integer problems are often encountered in high school. xy=35 ±35・±1 ±7・±5 (the opposite is also possible) x-y=±34, ±2 Since x+y is an integer x+y=24/(x-y) = 24/±2=±12 (answer)

SINCE GIVEN X TIMES Y EQUAL 35, AND AS 35 CAN ONLY BE 7 TIMES 5 (BOTH ARE UNIQUE NUMBERS), SO EITHER X=7 AND Y=5 OR THE OTHER WAY ROUND, I.E. X=5 AND Y=7. AND, AS REQUIRED BY THE FIRST EQUATION, X MUST BE 7 AND Y MUST BE 5, IN ORDER THE RESULT IS 24. SO, WHY SO LENGTHY APPROACH ?

Because 7+5=12

Took 2 secs 😂

Good work. But there was no need to make it long, or seem complicated. With application of basic principles you can guide a learner to the solution in less than 2 minutes.

It took me 5 seconds😅

X=12

•x+y = ? (3) •x²-y²= 24 (1) •xy=35 (2) So... if we take the 2nd equation [ xy = 35 ] : xy= 35 　⇒ x= 7∧ y= 5 Then : xy= 35 ⇒ (7)(5)= 35 ⇒ 35=35 Then if we proof these numbers in • x²-y² = 24 we'll realize that (7)²-(5)²= 24 ⇒49-25= 24 ⇒ 24=24 Then... we can proof that x+y = 12 ⇒ (7)+(5)= 12 ⇒ 12=12 Then 5 and 7 are solutions for these three equations... Atte: apologizes I Made a mistake...

Wouldn't it be much simpler to use unique prime factorization, which gives us x = +-7 and y = +-5 instantly ( xy = 35, so one is +-5 and the other is +-7, equation (1) tells us which is which). So we get x + y = +-12. Similar arguments hold for the two complex solutions x = +-5i and y = -+7i. Note the alternating + and - in this case. This gives x + y = +-2i.

wtf so easy for 12+ y.o. russian students express x=35/y substitute in first eq. just we solve the simplest biquadrate equation and getting y=+-5, and next x=+-7. 2 answers: 1)12 2)-12

errrr..... I saw the equation and the answer is 12 immediately as this generally do not need a sim eq xy times table is only if you use sub 7x5 or 1x35 and 7^2 -5^2 = 49 - 25 = 24.... so kinda obvious...

Now I know why I forgot all that algebra, none needed to solve the problem. Critical thinking will give the correct answer.

Чушь! Традиционное решение короче и проще!

All you had to do was: (x² - y²)² + 4x²y² = 576 + 4900 = 5476 = (x² + y²)² So you now have: x² - y² = 24 x² + y² = ±√5476 = ±74 …and from there it is trivial to solve. Alternatively, x² + y² + 2xy = (x + y)² = ±74 + 70 = -4 or 144 So you can directly get = x + y, without having to solve for x or y themselves.

The quickest and easiest way to solve is to factor the right-hand value. The only solution prime numbers 7 and 5! Assuming that the values for x equal 7 and Y equals 5, substituting these into the first equation confirm this!

(x+y)exp2-(x-y)exp2 = 4 xy let x+y =U then x-y= 24/U we find quickly U=12

y=(35/x) Therefore x^2- y^2 = x^2 - (35/x)^2 = 24 . x^2 - (1225/x^2) = 24. x^4 - 1225 = 24x^2. x^4 - 24x^2 - 1225 = 0 . Sub y for x^2 . y^2 - 24y - 1225. (y - 49) (y+25) = 0 . y=49 , y= - 25. y=x^2 x^2 = 49 x= + or - 7 xy=35 so x = 7 y=5 x+y=12. x= -7 y= -5 xy= -12. x= 5i y= 7i x+y=12i. x= -5i y= -7i x+y = - 12i.

multiply equation 1 by x^2 and substitute (xy)^2-> x^4-24x^2-35^2=0; solve quadratically for x^2= (24+/- sqrt(24^2+70^2))/2; x^2=12 +/- 37=49 or -25, x= +/-7 or +/- 5i; applying y=35/x you get all the results for x+y

(x-y)(x+y) = 24 case I : x-y = 1, x+y = 24 i.e x = 12+ 1/2 and y = 11+ 1/2 does not satisfy xy = 35 case II : x-y = 2, x+y = 12 i.e x = 7, y = 5 satisfies xy = 35 case III : x-y = 3, x+y = 8 i.e x = 5+1/2, y = 2+ 1/2 does not satisfy xy = 35 case III : x-y = 4, x+y = 6 i.e x = 5 , y = 1 does not satisfy xy = 35 Hereby x= 7, y = 5 is the only feasible solution

x^2-y^2=24 xy=35 (x+y)^2=x^2y-^2+4xy x^2-y^2=24 xy=35 (x+y )^2=x^2-y^2+4×35 (x+y)^2=24+140=144 (x+y)=+/-12

Xy=35 it is must be 7,5 Also x^2-y^2=24 So x=7 and y=5 X+y=12

lol you made it more complicated than it should be

35 = 7 X 5 So x times y = 7 times 5 x^2 - y ^2 = (x +y)(x-y) = (7+5)(7-5) = 12 times 2 = 24 so x + y = 7 + 5

Sorry..once again YOUR METHOD TO SOLVE ALGEBRA QUESTION ARE REALLY AWKWARD AND TOO COMPLICATED.. THERE ARE MUCH MORE EASIER METHODS..BUT YOU DON'T PAY ATTENTION TO THIS COMMENTS.. SORRY BUT WHERE DID YOU LEARN ALGEBRA???

答え12

We do not need tricks x+y=u we need find u x-y=v so 2x=u+v 2y=u-v x=(u+v)/2 y=(u-v)/2 After substitusion we have x^2-y^2=uv =24, x*y=((u+v)/2 ) * (u-v)/2 = (u^2-v^2)/4=35 so u^2-v^2=140 we are not interestet in v so: v=24/u u^2-(24/u)^2=140 (u^2)^2-24^2=140*u^2 (u^2)^2 -140*u^2 - 24^2= 0 u^2=144 or u^2=-4 u=12 or u=-12 or u = 2i or u=-2i

Hi I hope you are doing well. Could you please tell me the solution of this problem? X to the power of y equals y to the power of x . And x×y=8

35 is only divisible by 35, 1, 5, and 7. An 8 year old can do this.

Salut j'ai 13 ans et j'ai résolu cette équation en substituant y au carré par x au carré puis je l'ai remplacé dans l'équation donc je suis tombé sur le second dégre et j'ai trouvé comme résultat 48

The solution is easy and immediate. The least common multiple of 35 is 7 x 5. The sum a + b is 12. a^2 = 49 and b^2 = 25. The difference is 24. Very easy. Isn't ?

Had the answer in 10 seconds.

There aren't a lot of factors for 35. 7 and 5 are factors. 7^2 - 5^2 = 49 - 25 = 24.

Very labouriously solved ! We call this as Hamali !!

x == 7. y == 5, by inspection in about 2 minutes

At last!! I got the answer in seconds. I'm no genius; it was just an extremely easy problem. After all, how many factors of 35 are there? Right, only 7 and 5. Now that I think about it, it was insultingly easy. Shame on you.

Sound is inaudible

7+5=12

NO VOICE. CHECK ONCE

Volume is very low. Are you revealing any secret?

Interesting

I did this in my head in 2 seconds.

x is 7 and y is 5.

I had a sort of weird approach where I used vietas formula to create a quadratic equation with the given sum of "a" and product of 35, where x and y are the solutions to this. From this, the (x+y)(x-y) can be seen as (x+y) is "a" and (x-y) is sqrt(b^2-4c). b is just "a", and c is just 35. Set this all to 24 and it's a quadratic in disguise as a fourth degree. Factor it and you'll get (a^2-144)(a^2+4)=0 This gives all solutions for x+y aka "a"

x=7 and y=5

X= 7 Y= 5

let a = x^2 and b = y^2, so a-b = 24 and a*b = 35^2. By eliminating b, we have a(a-24) = 35*2, so a^2-24a-35*2 = 0. It is easily to know a = (24+/-sqrt (24^2+4*35^2))/2, so a = 49 or -25 (which can be excluded because a = x^2 cannot be negative). b = a-24 =25. It is easily to find either x=7 and y =5 or x=-7 and y = -5.

X+y=12

x^2 - y^2 =24, (x+y)(x - y)= 24......(1) xy = 35=7.5 x+y= 7+5=12, x - y=2 (1) holds good So x =7, y=5 Advice: for, IIT, Olympiad,any competition Intuition is very important. Reading invisible things properly.

X = 7 Y = 5 X+Y=12

Xy=35 and there is factors which gave this value is 5 and 7(which can be negative but won't satisfy x+y equation so they will be in positive and now put these factors in equation Ist.) 😋

Take all the factors of 24 i.e., 1x24 , 2x12, 3x8,4x6 or in opposite order like 24x1, 12x2 etc..and eqaute it to (x-y)(x+y) you'll find that when x+y is 12 only its will satisfy positive value of 35 for xy.

7x5=35, 7² = 49, 5² = 25, 49-25 = 24, x=7 y =5, x+y = 12... did it in my head at a glance of the thumbnail and never bothered to watch the video but I'm sure it's something super long that comes up with a different answer.. but when I plug my answer in it works..

It's obviously just the Pythagorean triple 7-24-25 leading very quickly to the values of x & y

Very OK, but for simple people living in the real world, everybody sees immediately that 7x5=35; the product of 2 primes!

Shouldn't that just be a +2i? You took the root of -4, so isn't that just 2i? Wait wait...no it isn't.Its an equation, so we have to consider the negative and positive root, though it comes to the same.

take the second equation: factors of 35 is 1*35 and 7*5.Firstly 1 * 35 doesn't satisfy the first equation but 7 *5 is a possibility. The answer to the first equation is a positive number. Since 7 is the bigger number that means x is 7 and y is 5. (7 + 5 = 12).

XY = 35 = 7*5 X+y=12

Решил задачу за секунду в уме. x=7 y=5 ответ 12.

答え　12

x=7, y=5=>x+y=7+5=12ans

This problem is not hard, you made it look hard with that unnecesary long procedure. If xy is not 0, you can say x=35/y, substitute and solve the biquadratic equation. Not just an easy and short way to go, but also you get the values of x and y, being x=±7 and y=±5. And I say this is not hard because every 2nd grade student knows how to deal with quadratic equations. Nice video anyway.

12. It's obvious. X and Y are 7 and 5 respectively.

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I solved IT just xy=35 And only thing (i think)that Will gave you 35 Is 7•5

For those of you commenting that you were able to get the answer 12, he is also providing complex number answer which you guys are not providing. Since the question did not specify explicitly whether the answer is in real or integer or complex, you have to provide all.

just by permutation & combination pick (5,7) as the answer. another set would be (-5,-7)

7+5=12

12

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12

too simple. x=+-7,y=+-5,....by watch only.

-12

good

Unnecessarily complicated :(

It is just a mental sum.I found out within 30 Seconds.

Let u = x^2 and v = (iy)^2 = -1 * y^2. Then u + v = 24 and u * v = x^2 * -1 * y^2 = - 1 * (x * y)^2 = -1 * 35^2 = -1225. u and v are conjugate solutions to t^2 - 24 * t - 1225 = 0. t = -25 or 49. If u = -25 and v = 49, then x = + or - 5i and y^2 = -v = -49 so y = + or - 7i. 5i * 7i = 35 * i^2 = -35. Similarly (-5i) * (-7i) = -35. So, x = 5i and y = -7i or x = -5i and y = 7i. Thus, x + y = + or - 2i. If we are restricting x and y to real numbers, then if u = 49 and v = -25. x = + or -7 and y^2 = -v = 25 so y = + or -5. While all 4 cases (7,5), (7,-5), (-7,5) and (-7,-5) satisfy x^2 - y^2 = 24, only (7,5) and (-7,-5) satisfy x * y = 35. So, x + y = + or - 12.