Question

Berapakah nilai x dan y dari:
(3-akar2)(x+yakar2)=-akar2

Answer 1

(3 - √2)(x + y√2) = -√2
expanding ruas kiri, diperoleh
3x + 3y√2 - x√2 - 2y = -√2
(3x - 2y) + (3y - x)√2 = -√2
dengan mengidentikkan ruas kiri dan kanan, maka berlaku :
3x - 2y = 0
3x = 2y
x = (2/3)y
dan
3y - x = -1
substitusi nilai x,
3y - (2/3)y = -1
(7/3)y = -1
7y = -3
y = -3/7

substitusi nilai y ke persmaan x,
x = (2/3)y = (2/3)(-3/7) = -2/7

jadi, solusi buat (x,y) = (-2/7 , -3/7)

Answer 2

3x + 3y√2 - x√2 - 2y = -√2
3x + 3y√2 - x√2 - 2y = -√2
(3x-2y) - √2 ( x - 3y ) = -√2
3x - 2y = 0
3x = 2y
x = 2/3 y
x - 3y = 1
2/3 y - 3y = 1
2/3 y - 9/3 y = 1
-7/3 y = 1
y = -3/7
3x = 2 . -3/7
3x = -6/7
x = -6/7 . 1/3 = -2/7