StarPuppy SG need help for maths eq 3x - 2y = 1 [(x-2)(x-2)] + [(2y+3)(2y+3)] = 26 From (1): y= -1/2 + 1.5x anyone knows?
TYING SG I think you are the one who dono how to do. U cannot sub back to original eqn. Don believe you work out you will get 0.
papercut87 SG let 2y = 3x - 1 (also y = 1.5x - 0.5) sub into 2nd eqn to get: (x-2)^2 + (3x+2)^2 = 26 expand to get: 10x^2 + 8x - 18 = 0 solve and you'll have 2 sets of ans: x = 1, y = 1 x = -9/5, y = -16/5
snop_coolz SG could have help..but..scroll and saw ur help..so i dont need to rack my brain liao..anyway..applause..
TYING SG let y= - 1/2 + 1.5x be eqn 3 [(x-2)(x-2) + (2y+3)(2y+3)] = 26 [(x-2)(square) + (2y+3)(square)] = 26 [(2y+3)(sq.) - (x-2)(sq.)] = -26 sub y in. [2(-0.5+1.5x) + 3](sq) - (x-2)(sq) = -26 [(-1 + 3x) + 3](sq) - (x-2)(sq) = -26 (2+3x)(sq) - (x-2)(sq) = -26 (2+3x+x-2)(2+3x-x+2) = -26 (4x)(4+2x) = -26 (2x)(2+x) = -13 2x(sq) + 4x + 13 = 0 solve x den find y.
