Mastering Systems of Equations for Success | Olympiads Prep!

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  • Опубліковано 19 чер 2024
  • Mastering Systems of Equations for Success | Olympiads Prep!
    🏆 Ready to elevate your math game to Olympiad-level? Dive into our comprehensive guide on mastering systems of equations! 📈 In this video, we unravel the secrets behind solving complex equations with ease, equipping you with the skills needed to ace Olympiad math competitions. From fundamental strategies to advanced techniques, we cover it all to ensure your success. Whether you're a seasoned competitor or just starting your Olympiad journey, this video is your ultimate preparation tool. Get ready to conquer the Olympiad arena with confidence! 🥇 #olympiadpreparation #mathsuccess #equations #systemsofequations #mathematics #problemsolving #competitionprep #matholympiad #olympiadmathematics #mathchallenges #algebra #maths
    Topics covered:
    System of equations
    Algebra
    Logarithms
    Properties of logarithm
    Cubic equation
    Algebraic identities
    Algebraic manipulations
    Vieta's Method
    Solving systems of equations
    Synthetic division method
    Rational root theorem
    Math tutorial
    Math Olympiad
    Math Olympiad Preparation
    Timestamps:
    0:00 Introduction
    0:32 Properties of logarithm
    2:48 Algebraic identities
    5:08 Cubic equation
    5:57 Synthetic division
    6:46 Quadratic equation
    8:50 Vieta's method
    9:55 Solutions
    👉 Don't forget to like, subscribe, and hit the notification bell to stay updated on more advanced math tutorials!
    Thanks for watching!
    @infyGyan

КОМЕНТАРІ • 5

  • @woobjun2582
    @woobjun2582 Місяць тому +1

    The given can be
    (1) x^3 +y^3 = 1/4
    (2) xy = 1/4
    By (1) and (2)
    4t^3 -3t -1 =0;
    (t -1)(2t +1)^2 = 0
    where t = x +y.
    Thus to solve
    2 system of eqns
    (i) x +y= 1 and xy =1/4
    (ii) x +y=-1/2 and xy =1/4.
    By (i),
    (x, y) = (1/2, 1/2)
    BY (ii)
    (x, y) = cmplx sols

  • @kassuskassus6263
    @kassuskassus6263 Місяць тому

    After some manipulations, we get x^3+y^3=1/4 and xy=1/4. we have to solve 64x^6-16x^2+1=0. Let u=x^3 and solve for u the new equation 64u^2-16u+1=0. We get twice u=1/8, then x=1/8 and y=1/8.

  • @tejpalsingh366
    @tejpalsingh366 Місяць тому +1

    X= Y= 1/ 2

  • @user-nd7th3hy4l
    @user-nd7th3hy4l Місяць тому +1

    X=1/2 et Y=1/2.

  • @sunil.shegaonkar1
    @sunil.shegaonkar1 Місяць тому +1

    If I solve it in this way, does it violate any principle of mathematics?
    Other View and please let me know, if anything is wrong :
    Substitute p for x and q for y. Then again substitute y for p & x for q. You arrive at the same set of two equations that we begin with.
    It means X and Y are both equal. So substitute Y=X in first equation. So you have X^3 = 1/8 => x = 1/2 = y.