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    (Original post by hasan4life)
    your the salaan to my roti :flutter:
    hehe :flutter:
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    Why are you so jealous about biryani007?
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    (Original post by GnomeMage)
    is this a real question??
    Yes. from one of the STEP III papers
    (Original post by Mubariz)
    Loads of questions cannot be solved

    Posted from TSR Mobile
    Solve for x,y and z, this can be in terms of a and b (a and b are constants) so 3 main variables, 3 equations
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    (Original post by plusC)
    Yes. from one of the STEP III papers

    Solve for x,y and z, this can be in terms of a and b (a and b are constants) so 3 main variables, 3 equations
    Oh I thought it meant solve in terms of real numbers.

    Being a numpty me

    Anyway it's step 3 so I'm gonna back away

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    (Original post by Mubariz)
    Oh I thought it meant solve in terms of real numbers.

    Being a numpty me

    Anyway it's step 3 so I'm gonna back away

    Posted from TSR Mobile
    Don't let titles deceive you its pretty simple :teehee:
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    x+ay+z=2 (1)
    x+y+az=2 (2)
    2x+y+z=2b (3)
    (1)=(2) by default so x+ay+z=x+y+az
    subtract x and rearrange for a condition connecting y and z
    ay+z=y+az
    y(a-1)=z(a-1)

    Assuming a\not=1 we get the condition y=z so using this on (3) gives 2x+2y=2b \Rightarrow x+y=b label this (4)
    using this condition on (2) x+y+ay=2 \Rightarrow x+y+ay=2 (5)

    doing (5)-(4) yields ay=2-b \Rightarrow y=\dfrac{2-b}{a}=z
    using this on (4) yields x=b-y=b-\dfrac{2-b}{a}=\dfrac{ab+b-2}{a}

    and thats it (no solutions when a=1 or a=0 though)


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    (Original post by plusC)
    Yes. from one of the STEP III papers

    Solve for x,y and z, this can be in terms of a and b (a and b are constants) so 3 main variables, 3 equations
    bugger...i thought i need to provide a number for it.!
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    Ah so this is Simmy 0_0
 
 
 
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