(i) Factorising 3x2 + 2xy – y2 we get
3x2 + 3xy – xy – y2 = 3x (x + y) – y (x + y)
= (3 x – y)(x + y)
So 3x2 + 2xy – y2 = 0 ⇒ (3x – y) (x + y) = 0
⇒ 3x – y = 0 and x + y = 0
(ii) 6 (x – 1)2 + 5 (x – 1)(y – 2) – 4(y – 2)2 = 0
⇒ 6(x2 – 2x +1) + 5(xy – 2x – y + 2) – 4(y2 – 4y + 4) = 0
(i.e) 6x2 – 12x + 6 + 5xy – 10x – 5y + 10 – 4y2 + 16y – 16 = 0
(i.e) 6x2 + 5xy – 4y2 – 22x + 11y = 0
Factorising 6x2 + 5xy – 4y2 we get
6x2 – 3xy + 8xy – 4y2 = 3x (2x – y) + 4y (2x – y)
= (3x + 4y)(2x – y)
So, 6x2 + 5xy – 4y2 – 22x + 11y = (3x + 4y + l )(2x – y + m)
Equating coefficient of x ⇒ 3m + 21 = -22 ... (1)
Equating coefficient of y ⇒ 4m – l = 11 ... (2)
Solving (1) and (2) we get l = -11, m = 0
So the separate equations are 3x + 4y – 11 = 0 and 2x – y = 0
(iii) 2x2 – xy – 3y2 – 6x + 19y – 20 = 0
Factorising 2x2 – xy – 3y2 we get
2x2 – xy – 3y2 = 2x2 + 2xy – 3xy – 3y2
= 2x(x + y) – 3y(x + y) = (2x – 3y) (x + y)
∴ 2x2 – xy – 3y2 – 6x + 19y – 20 = (2x – 3y + l)(x + y + m)
Equating coefficient of x 2m + l = -6 ... (1)
Equating coefficient of y -3m + l = 19 ... (2)
Constant term -20 = lm
Solving (1) and (2) we get l = 4 and m = – 5 where lm = – 20.
So the separate equations are 2x – 3y + 4 = 0 and x + y – 5 = 0