F.5 math binomial question!!!!

(1-ax)(1-bx)^6 = (1 - ax) (1 - (6C1)bx + (6C2)(b^2)x^2 - ... + (bx)^6)= (1 - ax) (1 - 6bx + 15(b^2)x^2 - ........... + (bx)^6)= 1 - 6bx + 15(b^2)x^2 - ax + 6abx^2 - 15a(b^2)x^3 + ...= 1 - (a+6b)x + (15b^2 + 6ab)x^2 - 15a(b^2)x^3 + ...So- (a+6b) = - 19a + 6b = 19.......(1)(15b^2 + 6ab) = 153b(5b + 2a) = 51.........(2)By (1) ,a = 19 - 6b , sub it to (2) :b(5b + 2(19 - 6b)) = 515b^2 + 38b - 12b^2 = 517b^2 - 38b + 51 = 0(7b - 17)(b - 3) = 0b = 17/7 or 3a = 19 - 6(17/7) = 31/7 or a = 19 - 6*3 = 1