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Linear Equations System In Three Variables. The Methods are used : elimination and substitution. Example :. 4x + y + 3z =10 5x +3y +7z = 13 6x -5y -2z = 2 Find the value of x + y - z. Given that. Solution :. Back to eq. 4 2x +z = 4 2(3) +z = 4 6 + z = 4 z = -2.
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Linear Equations System In Three Variables The Methods are used : elimination and substitution
Example : 4x + y + 3z =10 5x +3y +7z = 13 6x -5y -2z = 2 Find the value of x + y - z Given that
Solution : Back to eq. 4 2x +z = 4 2(3) +z = 4 6 + z = 4 z = -2 4x + y + 3z =10 … eq 1 5x +3y +7z = 13 … eq 2 6x -5y -2z = 2 …eq 3 Eq.1 and eq.2 (elimination y) 4x + y +3z = 10 .3 12x + 3y +9z = 30 5x +3y +7z = 13 .1 5x +3y +7z = 13 7x + 2z = 17 … eq 4 Back to eq. 3 6x – 5y -2z = 2 6(3) -5y -2(-2) = 2 18 -5y +4 = 2 -5y = 2-4-18 = -20 y = 4 Eq.1 and eq.3 (elimination y) 4x + y +3z = 10 .5 20x + 5y +15z = 50 6x -5y -2z = 2 .1 6x -5y - 2z = 2 + 26x + 13z = 52 2x + z = 4 … eq 5 Set of solution is {(3,4,-2)} Thus x + y- z = 3+4-(-2) = 9 Eq.4 and eq.5 (elimination z) 7x + 2z = 17 .1 7x + 2z = 17 2x +z = 4 .2 4x + 2z = 8 - 3x = 9 x = 3
Simultaneous Equations Linear-Quadratic & Quadratic-Quadratic The Methods are used : substitution and elimination
Example : Solve the system : • 1. 2x + y = -1 • y = x2-4x • x + y + 1 = 0 • x2 +6xy +9y2 = 9 • y = x2 +x-2 • y = 2x2 -3x +1 • x2 – y2 = -3 • x2 + 2y2 = 9
PRACTICE:LKS, page 38, no : 1,2,5,8LKS, page 40, no : 1,7,9,10LKS, page 41, no : 2,4,7LKS, page 43, no : 1,2,4,6,8,9