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Mastering Systems of Equations: A Dual Approach Tutorial
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- čas přidán 17. 08. 2024
- Mastering Systems of Equations: A Dual Approach Tutorial
Ready to master systems of equations? 📊 In this tutorial, we'll explore the power of dual approaches to solving complex systems. Whether you prefer graphical methods or algebraic techniques, we've got you covered! Join us as we unravel the secrets of system solving and equip you with the skills you need to excel in math. Say goodbye to confusion and hello to confidence! Tune in now and elevate your equation game!
Time-Stamps:
0:00 Introduction
0:33 Method-1
1:55 Vieta's method
3:07 Quadratic formula
7:46 Method-2 (Substitution)
8:30 Solving systems of equations
10:33 Solutions
🔢✨#Equations #Mathematics #SystemOfEquations #ProblemSolving #Education #DualApproach #Tutorial #algebra #simultaneousequation #math #maths
🎯 This video is perfect for students, math enthusiasts, or anyone seeking to sharpen their problem-solving skills and gain confidence in dealing with system of equations. 🎓📈
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@infyGyan
a+b=11 and ab=30
(a-b)^2=(a+b)^2-4ab=121-120=1
a-b=+/-1
Now solve 2 sets of equations
a+b=11 and a-b=1 gives a=6 and b=5
a+b=11 and a-b =-1 gives a=5 and b=6
Use similar method to determine x and y.
Let x + y = a and xy = b. Then the given equations read a+b+1 = 12 > a+b = 11 and ab = 30. So, b = 30/a and therefore a + 30/a = 11 > a^2 - 11a = 30 > a = 6, b= 5 or a = 5, b =6. If a= 6, b= 5, x+y = 6 and y = 5/x > x^2 - 6x + 5 = 0 > (x,y) = (5,1), (1,5). If a= 5, b=6, x+y = 5 and y = 6/x > x^2 - 5x + 6 = 0 > (x,y) = (3,2), (2,3).
More reasonable is to choose linear dependence, substitute into nonlinear, anyways you receive the same equation
X.Y=3or2
x=2, y=3
Factorize instead of using formulas.
xy + (x + y) + 1 = 12 => xy + (x + y) = 11
xy(x + y) = 30 => x + y = 30/(xy)
xy + 30/(xy) = 11
(xy)² - 11xy + 30 = 0
xy = (11 ± 1)/2
xy = 6 or xy = 5
xy = 6 => y = 6/x
6 + x + 6/x = 11
x² - 5x + 6 = 0
x = (5 ± 1)/2
*x = 3 => y = 2*
*x = 2 => y = 3*
xy = 5 => y = 5/x
5 + x + 5/x = 11
x² - 6x + 5 = 0
x = (6 ± 4)/2
*x = 5 => y = 1*
*x = 1 => y = 5*