Mathematical Olympiad | A Nice System of Equations

No video

Mathematical Olympiad | A Nice System of Equations

Рет қаралды 1,050

Күн бұрын

Mathematical Olympiad | A Nice System of Equations
Welcome back to infyGyan !! In this video, we'll be solving a challenging system of equations from the Mathematical Olympiad. This problem will test your algebraic skills and your ability to think critically. Join me as we solve this intriguing system step-by-step and uncover the secrets to mastering such problems. Make sure to watch till the end, try solving it yourself, and don't forget to like, comment, and subscribe for more math challenges!
🔍 In this video:
A detailed breakdown of a complex system of equations from the Mathematical Olympiad.
Techniques and strategies for solving systems of equations.
Tips to improve your problem-solving skills and algebraic thinking.
📌 About the Mathematical Olympiad:
The Mathematical Olympiad is a prestigious competition that challenges students with tough and creative problems. It requires deep understanding, practice, and a passion for mathematics.
📣 Call to Action:
Try solving the problem before watching the solution!
Share your solutions and thoughts in the comments below.
If you enjoyed this challenge, give it a thumbs up and subscribe for more math problems and solutions!
🔗 Useful Links:
• Two Victorious Ways to...
• Solving an Amazing Sys...
• Easier Than You Think ...
• A Nice System of Equat...
Time-stamps:
00:00 Introduction
00:44 Performing operations
03:40 Solving system of equations
07:26 Ordered triples
10:06 Quadratic equations
11:50 Solutions
#matholympiad #systemofequations #education #mathenthusiast #mathtutorial #mathematics #mathskills #problemsolving #algebra
👉 Don't forget to like, subscribe, and hit the notification bell to stay updated on more advanced math tutorials!
Thanks for watching!

Пікірлер: 6

@user-kp2rd5qv8g Ай бұрын

From the first two equations, we get (y-4)(x-z)=0. If y=4, the first eqn gives x+z=15 and the third, zx=44 > x=11,4, z=4,11. Since x,y,z are completely symmetric, (x,y,z) = (11,4,4), (4,11,4), (4,4,11). If z=x, we get x^3-76x+240=0 > x=z=6,4,-10. If x=z=6, y=6. If x=z=4, y=11 and if x=z=-10, y=-10. So, (x,y,z) = (11,4,4), (4,11,4), (4,4,11), (6,6,6), (-10,-10,-10).

@abcekkdo3749 Ай бұрын

（x,y,z)= (11,4,4) (4,4,11) (4,11,4) (6,6,6) (-10,-10,-10)

@user-ny6jf9is3t Ай бұрын

(χ,ψ,z) =(6,6,6) , (-10,-10,-10),(11,4,4), (4,11,4),(4,4,11)

@RajeshKumar-wu7ox Ай бұрын

X=11,y=4,z=4

@SidneiMV Ай бұрын

xyz + 4z² = 60z xyz + 4x² = 60x xyz + 4y² = 60y 3xyz + 4(x² + y² + z²) = 60(x + y + z) (x + y + z)² = x² + y² + z² + 2(xy + xz + yz) x² + y² + z² = (x + y + z)² - 2(xy + xz + yz)

@RealQinnMalloryu4 Ай бұрын

xy+4z= (56)+4z=60.(7^8)+4x 7^12^3+2^2 1^1^1^1 +1^2 .1^2 (x yz ➖ 2xyz+1) yz+4z =60.(56) +4z( 7^8)+2)+4x (7^12^3)+2^2x =60 1^1^1^1 +1^2x 1^2z (yzzlx ➖ 2yzx+1) zx +4y =60 (56)+4y (7^8)+4y (7^1^2^3)+2^2y 1^1^1^1 +1^2y= 1^2y (zxy ➖ 2zxy+1)