Sec 2 Maths Problem

• kkhing

  1 May 11, 08:46

  Please help to solve these 2 Sec 2 Maths problem.

  1) Given that 3a + e = 7 and 4ae/3 = 2, find the value of e-3a.

   

  2) Given that m^2 + 8m = v^2 + 8v and that m is not equal to v, find the value of 5(m + v).

   

  Thanks.

   

   

   

• Darkness_hacker99

  1 May 11, 11:11

  --

• wee_ws

  1 May 11, 12:27

  Hi,

    What is the difficulty you are facing please? Thanks.

  Cheers,
  Wen Shih

• theawesomejo

  2 May 11, 15:16

  Not sure about this , maybe you can work it out ?

  1) Use simultaneous eqn method, then get the answer and then get the ans.

   

• Sakuragi86

  3 May 11, 18:41

  Originally posted by theawesomejo:

  Not sure about this , maybe you can work it out ?

  1) Use simultaneous eqn method, then get the answer and then get the ans.

   

  
  simultaneous eqn method cannot get any answer.

• wee_ws

  3 May 11, 19:36

  Hi,

    It seems to me that Q1 is not a reasonable question for a Sec 2 student. It is not worth solving. Thanks.

  Cheers,
  Wen Shih

• wee_ws

  3 May 11, 21:23

  Hi,

    Let e - 3a = k, then 3a = e - k -- (1).

    Given 3a + e = 7, we obtain

    e - k + e = 7, because of (1) 

    => 2e - k = 7

    => e = (k + 7)/2 -- (2).

    Given 4ae/3 = 2, we obtain ae = 3/2

    => 3ae = 9/2

    => (e - k)e = 9/2, because of (1)

    => [ (k + 7)/2 - k ] (k + 7)/2 = 9/2, because of (2)

    => (7 - k)(7 + k) / 4 = 9/2

    => 49 - k^2 = 18

    => k = +/- sqrt(31).

    The working is a little demanding at Sec 2 level, unless there is really a more elegant method that I have missed out?

    Thanks.

  Cheers,
  Wen Shih

• Chemfreak022

  3 May 11, 21:28

  3a + e = 7

  (3a + e)^2 = 49

  (3a + e)^2 = 9a^2  + 6ae + e^2

  4ae/3 = 2 ..... 2ae = 3 .... 6 ae = 9

  (e-3 a)^2 = e^2 - 6ae + 9a^2=  9a^2 + 6ae + e^2 - 12ae = 49 -18 = 31

  Therefore (e-3a) = sqrt(31)

   

   

• Chemfreak022

  3 May 11, 22:32

  I think mine is shorter, although i forgot the -ve root

• wee_ws

  4 May 11, 07:51

  Hi Chemfreak022,

    Thanks for sharing an elegant solution :)

    Students can rest assure that Cambridge will provide the suggestion to lead them to think in that direction. It may be phrased like this:

    It is given that e + 3a = 7 and ae = 3/2.

    Express (e - 3a)^2 in terms of (e + 3a) and ae. Hence find, exactly, the possible values of (e - 3a).

    Thanks.

  Cheers,
  Wen Shih

• wee_ws

  4 May 11, 08:08

  Hi,

    As a result of Q1 and the discussion that followed, it is worthwhile for the keen student to take note of the relationships between (a + b)^2 and (a - b)^2:

    1. (a - b)^2 = (a + b)^2 - 4ab,

    2. (a + b)^2 = (a - b)^2 + 4ab.

    Thanks.

  Cheers,
  Wen Shih

• syncopation_music

  5 May 11, 15:51

  agree with you .Master Wenshih , Cambridge do not wish to  slap themselves by writing ambiguous question in their papers.

  I read through some Examiner Report of IGCSE .They shared some insight in the type of question which will come out .

  What i worried is Cambridge could set english-based Mathematics question

  for example :subtract 3x-4 from x -2

  common folks will think that the working is (3x-4) - (x-2) = 2x+2

  however , the interpretation is wrong

  its (x-2) - (3x-4)

  which gives -2x +2

   

• wee_ws

  5 May 11, 23:15

  Hi syncopation_music,

    Thanks for the respectful address.

    We are learning from one another, so call me Wen Shih  :)

  Cheers,
  Wen Shih