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yg 3 dan 4 nyusul
1) (x-y)(x+2y) = 0 -->x = y atau x = - 2y
untuk x = y sub ke x² - xy + 2y² = 16
y² -y(y) + 2y²= 16
2y² = 16 --> y² = 8
y1 = √8, x1 =√8
y2 = -√8 , x2 = - √8
untuk x = - 2y sub ke x² - xy + 2y² = 16
(-2y)² -(-2y)(y) + 2y² = 16
4y² + 2y² + 2y²= 16
8y² = 16 --> y² = 2
y1 = √2, x1 = -2√2
y2 = -√2, x2 = 2√2
2. x² - 5xy + 6y² = 0
(x + y)(x - 6y) = 0
x = - y atau x = 6y sub ke x(x-2y) = 6 -6y²
untuk x = -y
-y (-y-2y) = 6 - 6y²
3y² +6y² = 6
y² = 6/9 = 3/2
y = √3/2
y₁= ¹/₂ √6 , x₁ = -¹/₂ √2
y₂ = - ¹/₂√2 , x₂ = ¹/₂√2
untuk x = 6y
6y(6y-2y) = 6 -6y²
24 y² + 6y² = 6
y² = 6/30 = 1/5
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y₁= ¹/₅ √5 , x₁ = ⁶/₅ √5
y₂ = - ¹/₅√5 ,x₂ = - ⁶/₅√5