Maths Qns

• StarPuppy

  4 Jan 07, 23:12

  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?
• StarPuppy

  4 Jan 07, 23:19

  bump
• dark_snowboy

  4 Jan 07, 23:19

  Originally posted by StarPuppy:

  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?

  just sub inside can le ma
• Classyfied

  4 Jan 07, 23:21

  Originally posted by StarPuppy:

  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?

  sub y=3/2x-1/2
  
  into (2) can liao le
• TYING

  4 Jan 07, 23:21

  Originally posted by StarPuppy:

  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?

  sub y into the second eqn and get x.
• MohamedF

  4 Jan 07, 23:22

  Originally posted by StarPuppy:

  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?

  3x - 2(-1/2 + 1.5x) = 1
  3x - 1 -3x = 1
  
  WHat the... I 4 years nvr do dis...
• jay_rocks

  4 Jan 07, 23:23

  sub sub sub
• dark_snowboy

  4 Jan 07, 23:24

  Originally posted by StarPuppy:

  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?

  wrong le ps
• MohamedF

  4 Jan 07, 23:28

  Originally posted by dark_snowboy:

  [ (x-2)^2) ] + [(2y+3)^2) = 26
  [ (x-2)^2) ] + [((2(-1/2 + 1.5x)+3)^2) = 26
  [ (x-2)^2) ] + [(4+3x)^2] = 26
  (x^2 -4x +4 ) + 16+9x^2+24x=26
  10x^2+20x+20=26
  10x^2+20x-6=0
  5x^2+10x-3=0
  (5x-1)(x-3)=0
  
  x=0.2, 3
  
  from (2),
  y=-0.5+1.5(0.2)
  y=-0.5+1.5(3)
  
  den sub into (1)
  
  

  Chey, no wonder... its one of those U curve answers... I hate those...
• TYING

  4 Jan 07, 23:30

  Originally posted by MohamedF:

  3x - 2(-1/2 + 1.5x) = 1
  3x - 1 -3x = 1
  
  WHat the... I 4 years nvr do dis...

  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

  4 Jan 07, 23:30

  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

  4 Jan 07, 23:33

  A maths..
• papercut87

  4 Jan 07, 23:34

  Originally posted by snop_coolz:

  A maths..

  so? btw, that wasnt very helpful in answering his question.
• StarPuppy

  4 Jan 07, 23:36

  Originally posted by papercut87:

  so? btw, that wasnt very helpful in answering his question.

  nevermind..its ok
  
  
  
  thanks for the help everybody
• MohamedF

  4 Jan 07, 23:37

  Requires 9 people to answer 1 questions... Should I be worried for Singapore...
• ifish

  4 Jan 07, 23:38

  Originally posted by papercut87:

  so? btw, that wasnt very helpful in answering his question.

  very helpful.. emaths ppl like me can just go away
• snop_coolz

  4 Jan 07, 23:39

  could have help..but..scroll and saw ur help..so i dont need to rack my brain liao..anyway..applause..
• TYING

  4 Jan 07, 23:43

  Originally posted by StarPuppy:

  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?

  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.
  
  
• Classyfied

  4 Jan 07, 23:51

  Originally posted by MohamedF:

  Requires 9 people to answer 1 questions... Should I be worried for Singapore...

  lol
  simultaneous somemore leh
• snop_coolz

  4 Jan 07, 23:55

  humm..deduction..TS sec 3...
• Twincat

  5 Jan 07, 00:12

  [quote]Originally posted by TYING:
  [b]
  
  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 <- this
  
  [(2y+3)(sq.) - (x-2)(sq.)] = -26 <- and this same meh?
  
  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.
  
  
• Twincat

  5 Jan 07, 00:19

  Originally posted by TYING:

  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.

  u also not too good...
  he can't solve for x and y, only prove LHS = RHS...
  
• Twincat

  5 Jan 07, 00:41

  Originally posted by papercut87:

  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

  sorry got to ask,
  for 10x^2 + 8x - 18 = 0
  beside solving quad. eqn. using trial and error is there any other method?
• lasboriquas

  5 Jan 07, 00:46

  kns i see this type of algebraic for my o level i rmb i instantly close the paper n went to slp...the onli thing i like in algebra is the word bra in it..
• papercut87

  5 Jan 07, 01:13

  Originally posted by Twincat:

  sorry got to ask,
  for 10x^2 + 8x - 18 = 0
  beside solving quad. eqn. using trial and error is there any other method?

  can use this eqn.. x = -b[+-]b^2-4ac / 2a where a = 10, b = 8, c = -18
  solve for x then.