Olympiad Mathematics | Learn how to solve the system for X and Y quickly | Math Olympiad Training

[@mathbynisharsir5586]

Excellent question sir

[@PreMath]

Keep watching Glad you think so! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@guyhoghton399]

Once you get to x + y = ±5, xy = 6 a neat way to proceed is to use Vieta's formula which gives us the quadratic equation for any two roots x and y: z² - (x + y)z + xy = 0 Here z² ∓ 5z + 6 = 0 ∴ z = (±5 ±1)/2 (Note that the two ± symbols are independent so the above represents 2 sets of roots.) The roots are {3, 2} and {-2, -3}. Since x and y are the roots they can take either value respectively within each pair, so {(x, y)} = {(3, 2), (2, 3), (-2, -3), (-3, -2)}. Thus it was unnecessary to know in advance that x and y are integers.

[@eckhardfriauf]

At 3:00, I would have subtracted eq1 from eq2: (x^2 + y^2 + xy) - (x^2 + y^2 - xy) = 19 - 7, consequently: 2xy = 12, xy = 6. By the way, a very interesting question and training.

[@RSm12856]

X=3,y=2

[@timeonly1401]

It's so nifty that, instead of going straight for x & y, you solved for two other variables x^2+y^2 & xy. If we let u=x^2+y^2 & v=xy, then the system in x & y becomes: u-v=7 & u+v=19, resulting in u=13 & v=6. THEN, using those values of u & v, went after x & y. Very cool! Thx.

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!!!

[@HappyFamilyOnline]

Amazing presentation👍👍 Thanks for sharing✨✨

[@adebolaemmanuel9574]

Pls sir in terms of partial fraction denominator given (x^3 +4) how do I reduced it?

[@soniamariadasilveira7003]

Loved it

[@bigm383]

Thank you Professor for a very well explained solution!👍😀❤

[@waheisel]

Another great PreMath puzzle, Thank you. I don't think there are any non-integer solutions, or am I wrong? Substituting x=5-y into xy=6 you get y^2-5y+6=0 and the only solutions are y=3 and y=2 (and -3 and -2 for x=-5-y). i.e. only integers?

[@pramodsingh7898]

Thanks

[@mahalakshmiganapathy6455]

Really its useful

[@PreMath]

Glad to hear that! Thanks for your feedback! Cheers! You are awesome, Mahalakshmi. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@shrikrishnagokhale3557]

Good

[@shadmanhasan4205]

[1] (x^3+y^3)/(x+y) = (x^2-xy +y^2) = 7 [2] (x^3 - y^3)/(x-y) = (x^2+xy +y^2) = 19 These 2 systems can be remodeled as: (x^2+y^2) = 7 + xy, and (x^2+y^2) = 19 - xy Since LHS = RHS, This can go as: 7+xy = 19-xy -> 2xy = 12 -> xy=6

[@alster724]

Very easy

[@devondevon4366]

Answer x =3 and y=2 x^3 + y^3 = 7(x+y) [ multiply both sides by x+y] but since x^3 + y^3 = (x+y)(x^2+y^2 -xy) then 7(x+y) = (x+y)(x^2 + y^2-xy) 7 =x^2 + y^2 -xy [ divide both sides by x+y] equation M x^3 - y^3 = 19(x-y) [ multiply both sides by x-y] but since x^3-y^3 = (x-y)(x^2 + y^2 + xy), then 19(x-y)= (x-y)(x^2 + y^2 + xy) 19 = x^2 + y^2 + xy equation R -12 = -2xy [subtract equation R from equation M] 6 =xy; hence 2xy =12 Since 7=x^2 + y^2 -xy [see equation M], then 7 =x^2 + y^2-6 13 = x^2 + y^2 Since x^2 + y^2 + 2xy= (x+Y)^2 then 13 + 12 = (x+y)^2 25 = (x+y)^2 5= x+y equation P Since x^2 +y^2 -2xy = (x-y)^2 then 13 -12 = (x-y)^2 1 = (x-y)^2 1 = x-y equation Q 6= 2x (add equation P and Q) 3 =x Answer 5= 3+y [substitue the value of x=3 into equation P] 2= y Answer Answer x = 3 and y =3

[@alexandermorozov2248]

Можно решить в уме (если знать формулу суммы и разности кубов).

[@afifamyouni673]

excuse me, but your answer is missing an information, when you divided equation (1) by x+y and equation (2) by x-y, you should mention that x+y and x-y are both not equal to 0, that what you didn't do.

[@sahelibhunia7697]

2^3+3^3÷(2+3)=7 so it's 2and 3

[@Mbharwaj]

9th standerd question

[@giuseppemalaguti435]

x, y=2,3...3,2......x,y=-2,-3...-3,-2

[@JPTaquari]

two alternatives, there are more: X³ + Y³ = 7X + 7Y X³ - Y³ = 19X = 19Y ___________________ 2X³ = 26X - 12 Y ( / 2 ) X³ - 13X - 6Y = zero 1) X = 3 , Y = 2 2) X = 2 , Y = 3 Bingo!!!!

[@timeonly1401]

(1) How did you go from a cubic equation in x & y into specific values of x & y? (2) You solutions are missing the negative values of the ones you listed (ie., all components negative)...

[@JPTaquari]

I put two solutions, but a know there are more. I solve only looking, because it is very simple!

[@divyaraj140]

esy sir

[@devondevon4366]

X=3 AND Y=2

[@jackjill3259]

It's so lengthy ..,... baby... Make d 2d equation n solve it

[@RSm12856]

X=3 y=2

[@labbhattacharjee108]

To address a more generic problem (like if we had x^2+2y^2 in place of x^2+y^2 etc.), we should better replace x with 3/y to find a quadratic equation in y^2 and solve for y^2. Finally use x=3/y.

[@muneerabegum5362]

Copies of ayaans easy math channel

[@lk5367]

Дякую