A Nice System of Equations | Math Olympiad | Algebra

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A Nice System of Equations | Math Olympiad | Algebra
Welcome to our Math Olympiad series! In this video, we dive into a captivating system of equations that's perfect for sharpening your problem-solving skills. Join us as we break down each step of the solution, providing clear explanations and helpful tips along the way. Whether you're gearing up for a Math Olympiad or just enjoy tackling challenging algebra problems, this video is for you. Can you solve it? Give it a try and let us know in the comments!
Don't forget to like, comment, and subscribe for more math challenges and solutions!
Topics covered:
System of equations
Algebra
Math Olympiad
Math Olympiad Training
Algebraic identities
Algebraic manipulations
Solving systems of equations
Solving cubic equation
Substitution
Problem Solving
Math tutorial
Math Olympiad Preparation
Additional Resources:
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Timestamps:
0:00 Introduction
0:34 Substitution
4:46 Cubic equation
6:45 Synthetic division method
7:45 Quadratic equation
8:12 Factorization
13:50 Solutions
#matholympiad #systemofequations #education #mathenthusiast #mathtutorial #mathematics #mathskills #problemsolving #algebra
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Опубликовано:

4 июл 2024

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Комментарии : 9

@mohammedsaysrashid3587 2 дня назад

It was a wonderful introducing thank you ,Sir 🙏

@ramunasstulga8264 2 дня назад

Is there faster way to find one solution for synthetic division than checking the factors of a constant of equation one by one?

@user-kp2rd5qv8g 3 дня назад

Let xyz=t. Then, x+xy=12-t > z+z(12-t) =30 > z=30/(13-t). Similarly, x=12/(22-t) and y = 21/(31-t). Thus, from the first equation, say, 12/(13-t) +(21)(12)/(13-t)^2 + t = 12 > t^3-65t^2+1306t-7560 = 0 > t = 10, 27, 28 > (x,y,z) = (1,1,10), (-12/5, 21/4, -15/7), (-2,7,-2).

@konroh2 2 дня назад

How did you get to the cubic equation? I'd like to see the work.

@tamirerez2547 День назад

Why did you write Let nyz=u Where did this n come from??!! This is how x looks like ❌ two straight lines crossing each other in the middle.

@RealQinnMalloryu4 2 дня назад

X+xy+xyz=12 (4) +(4)+ (4) (2^2) +(2^2) +(2^2).(1^1)+(1^1)+(1^2) (x^3^y^2z➖ x^3y^2z+1) y+yz+zyx=21 (1)+(10)+(10) (1^1) (2^5)+(2^5) .(1^1)+(2^1) 2^1 (z^3y^2x ➖ 2z^3y^2x+1)z +zx+zxy=30 (10)+(10)+(10) (2^5)+(2^5) +(2^5) (1^1)+(1^1)+(2^1) 2^1 (z^3x^2y ➖ 2z^3x^2y+1)