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Question 18 GCSE higher tier

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  1. May 29, 2007 #1
    1. The problem statement, all variables and given/known data

    (a) [tex]\frac{3}{x}+\frac{3}{2x} = 2 [/tex]
    that is 4x = 9 so x = 2.25

    (b) I can do the first question (a) easy peasy , however the secnond question says "using your answer to part (a), or otherwise solve:
    [tex]\frac{3}{(y-1)^{2}}+\frac{3}{2(y-1)^{2}} = 2[/tex]

    What part of (a) do I use to solve part (b)

    3. The attempt at a solution

    for (b)
    1) Expand (y-1)² = y² - 2y + 1
    2) Expand 2(y-1)² = 2(y² - 2y + 1 ) = 2y² - 4y + 2
    3) Double 1'st fraction to get equal denominators then add fractions
    [tex]\frac{6}{2y^{2} - 4y + 2}[/tex]
    4) Crossmultiply so you get [tex]3 = 4y^{2} - 8y + 4[/tex]
    then complete the square
    (4y - 4)² - 16 + 4 = 3
    y = [tex]\frac{4 +- \sqrt{15}}{4}[/tex]

    but that GETS a bit complicated for NON CALCULATOR level so i tried using the formula which gave me y = 1 or 1 (when at the bottom of the question there is 2 answers). Have I made a mistake in my calculations, and what about using part (a) to help answer part (b), I havn't done that, how do they link in together?

    Thx
    :smile:
     
  2. jcsd
  3. May 29, 2007 #2

    cristo

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    You don't need to do anything like that. The clue is in the "using your answer to part (a)" part of the question. Do the two equations look similar? Can you spot a substitution of variables that will change the equation in (a) to that in (b)?
     
  4. May 29, 2007 #3

    Kurdt

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    If its a non-calculator paper I'd suggest leaving that 2.25 as a rational. It will be easier to simplify later on in the question. Just an extra bit of advice :smile:
     
  5. May 29, 2007 #4
    (b)l is like (a)... the first fraction is doubled to make equal denominators, both factions added together equal 2 and both have same nuerators.

    so

    [tex]\frac{6}{2(y-1)^{2}}+\frac{3}{2(y-1)^{2}} = 2[/tex]

    [tex]\frac{9}{2(y-1)^{2}} = 2 [/tex]

    yet im still not using my answer to part (a).... im missing somthing arn't I

    Thx
     
  6. May 29, 2007 #5

    Kurdt

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    How would you turn the fraction in part (b) into the fraction in part (a)?
     
  7. May 29, 2007 #6

    cristo

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    Suppose if you substituted x=(y-1)^2 into the second equation. What form does the equation turn into?
     
  8. May 29, 2007 #7
    it turns into the one from part (a). So where saying x is now the quadratic y² - 2y + 1?
     
  9. May 29, 2007 #8

    Kurdt

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    So if [tex] x=\frac{9}{4} [/tex] from part (a) and [tex] x=(y-1)^2[/tex] how will you solve for y?
     
  10. May 29, 2007 #9
    (y - 1)² = 9/4
    y² - 2y + 1 = 9/4
    then do i just complete square/use quadratic formula?
     
  11. May 29, 2007 #10

    Kurdt

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    So you have [tex](y-1)^2 = \frac{9}{4} [/tex]. You can simply take the square root. There is no need to factor this.
     
  12. May 29, 2007 #11
    [tex](y-1)^2 = \frac{9}{4} [/tex]

    so y-1 = 3/2
    y = 1/1 + 3/2 = 2/2 + 3/2 = 5/2
    so
    y = [tex]1\frac{1}{2}[/tex]
    cheerz
     
  13. May 29, 2007 #12

    cristo

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    You're missing a solution-- remember you said there was space for two solutions? Well, if [itex](y-1)^2=x[/itex] then [itex](y-1)=\pm\sqrt x[/itex].
     
  14. May 29, 2007 #13

    Kurdt

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    And I hope you understand the importance of keeping numbers as rationals on the non-calculator paper. You probably would have lost a mark for an answer like [tex] \pm\sqrt{2.25} +1[/tex].

    Good luck for next Monday!
     
  15. May 29, 2007 #14
    yeh I do, wouldn't of spotted 9/4 being good to root. super duper
    Thx
     
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