Question

X+3y = 4
3x'2-4xy+5y'2 = 4
please jawab yaa bagi yang bisa

Answer 1

Dengan substitusi x + 3y = 4 --> x = 4- 3y
3 (4-3y)^2 - 4y(4-3y) + 5y^2 = 4
3(16 - 24y + 9y^2) - 16y +12y^2 + 5y^2 = 4
48 - 72 y + 27 y^2 - 16y+ 17y^2 = 4
44 y^2 - 88 y - 44 = 0
y^2 - 2y - 1 = 0
(y-1)^2 = 0 ---> y = 1 --> x = 4 - 3y --> x= 1
(x,y)= (1,1)