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    (Original post by Ralfskini)
    lgx+lgy=lg3 (1)
    lg(3x+y)=1 (2)

    From (1): xy=3 ---> y=3/x

    lg(3x+y)=1, so 3x+y=10.

    3x+3/x=10
    3x^2-10x+3=0
    (3x-1)(x-3)=0
    x=1/3, x=3
    when x=1/3, y=9
    when x=3, y=1

    (and they worked when I checked them).
    yes thats what (1) should of been but they made a mistake in the book and made (1)lgx+lgy=1000.
    so the eqn would not work if that is the case.
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    (Original post by IntegralAnomaly)
    yes thats what (1) should of been but they made a mistake in the book and made (1)lgx+lgy=1000.
    so the eqn would not work if that is the case.
    agreed
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    (Original post by IntegralAnomaly)
    yes thats what (1) should of been but they made a mistake in the book and made (1)lgx+lgy=1000.
    so the eqn would not work if that is the case.
    well that mistake wasn't made in my book.
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    (Original post by Ralfskini)
    well that mistake wasn't made in my book.
    really?i thought these books were batched produced....
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    (Original post by TheWolf)
    from my book:

    (1) = lg x + lg y = 1000
    yeh same here...but ralfskini's is how it is supposed to be i reckon
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    (Original post by lgs98jonee)
    2cosxcos50-sinxsin50=sinxcos40+cosxsin40
    2cos50-tanxsin50=tanxcos40+sin40
    so 2cos50-sin40=tanx(cos40+sin50)
    so tan x= (2cos50-sin40)/(cos40+sin50)
    so x= tan-1 of all that
    this is wrong!

    2cos(x+50) = sin (x+40)
    2(cosxcos50-sinxsin50) = sinxcos40 + cosxsin40
    divide by cosx
    2(cos50 - tanxsin50) = tanxcos40 + sin40

    because cosx = sin(90 - x) => cos40 = sin50

    2cos50 - 2tansxsin50 = tanxsin50 + cos50
    2cos50 - cos50 = 3tanxsin50
    cos50 = 3tanxsin50
    tan50 = 3tanx

    tanx = tan50/3
    x = 21.665 and 201.665

    well, It's different from what the answer say but you must forgive me. It is two in the morning and i've had too many glases of wine.

    MB
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    (Original post by musicboy)
    this is wrong!

    2cos(x+50) = sin (x+40)
    2(cosxcos50-sinxsin50) = sinxcos40 + cosxsin40
    divide by cosx
    2(cos50 - tanxsin50) = tanxcos40 + sin40

    because cosx = sin(90 - x) => cos40 = sin50

    2cos50 - 2tansxsin50 = tanxsin50 + cos50
    2cos50 - cos50 = 3tanxsin50
    cos50 = 3tanxsin50
    tan50 = 3tanx

    tanx = tan50/3
    x = 21.665 and 201.665

    well, It's different from what the answer say but you must forgive me. It is two in the morning and i've had too many glases of wine.

    MB
    well i get the right answers anyway with mine



    (Original post by lgs98jonee)
    2cosxcos50-2sinxsin50=sinxcos40+cosxsin40
    2cos50-2tanxsin50=tanxcos40+sin40
    so 2cos50-sin40=tanx(cos40+2sin50)
    so tan x= (2cos50-sin40)/(cos40+2sin50)
    so x= tan-1 of all that

    I had missed out a 2 that is all... Surely you have overcomplicated it by miles musicboy
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    (Original post by musicboy)
    this is wrong!

    2cos(x+50) = sin (x+40)
    2(cosxcos50-sinxsin50) = sinxcos40 + cosxsin40
    divide by cosx
    2(cos50 - tanxsin50) = tanxcos40 + sin40

    because cosx = sin(90 - x) => cos40 = sin50

    2cos50 - 2tansxsin50 = tanxsin50 + cos50
    2cos50 - cos50 = 3tanxsin50
    cos50 = 3tanxsin50
    tan50 = 3tanx

    tanx = tan50/3
    x = 21.665 and 201.665

    well, It's different from what the answer say but you must forgive me. It is two in the morning and i've had too many glases of wine.

    MB
    I get 15.6 and 195.6

    Your answers don't work.
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    yeah mine do though
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    I am aware that it doesn't work. I was just wondering why.

    MB
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    (Original post by musicboy)
    cos50 = 3tanxsin50
    tan50 = 3tanx

    MB
    You have divided by sin50 to give tan50 - not correct as u know
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    (Original post by lgs98jonee)
    You have divided by sin50 to give tan50 - not correct as u know
    oops, yes. Damn, I should have spotted that. Admittedly when I posted it I had had 2 bottles of wine in the evening (I know it'ss excessive)


    MB
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    (Original post by musicboy)
    oops, yes. Damn, I should have spotted that. Admittedly when I posted it I had had 2 bottles of wine in the evening (I know it'ss excessive)


    MB

    ..still a good attempt considering what you'd had to drink.
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    (Original post by Hoofbeat)
    I asked about that question a while ago. There's a printing error in the question, and we were trying to work out what it should have said, but we couldn't. So I'm afraid you'll just have to leave that question. Normally, the methods involve getting rid of the logs and sovling simultaneously like normal.
    (Original post by TheWolf)
    can anyone do p2 examination style paper pg 189 heinemann book question 1?

    i got iT..... PIECE OF CAKE!

    lg x + lg y = lg 3

    lg ( 3x + y ) = 1


    Using laws of logs:

    lg x + lg y = lg 3 ----------> lg xy = lg 3

    so XY=3 {x= 3/y} and { Y= 3/x}

    Substritue in EQ.2:

    lg ( 3x+ 3/x) = 1

    Since base of logs is 10, then 3x+ 3/x = 10

    3x+ 3/x = 10 < Multiply by X?

    3x^2 + 3 + 10X=0
    Factorising:
    (3x-3)(x-1) X=1

    XY= 3 so Y=3

    I DID IT!!!!
    REDO to get the other set!
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    lg x + lg y = lg 3
    Can you not say lgx+lgy-lx3=0 Then use rules of log.
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    (Original post by KOH)
    i got iT..... PIECE OF CAKE!

    lg x + lg y = lg 3

    lg ( 3x + y ) = 1


    Using laws of logs:

    lg x + lg y = lg 3 ----------> lg xy = lg 3

    so XY=3 {x= 3/y} and { Y= 3/x}

    Substritue in EQ.2:

    lg ( 3x+ 3/x) = 1

    Since base of logs is 10, then 3x+ 3/x = 10

    3x+ 3/x = 10 < Multiply by X?

    3x^2 + 3 + 10X=0
    Factorising:
    (3x-3)(x-1) X=1

    XY= 3 so Y=3

    I DID IT!!!!
    REDO to get the other set!
    i repeat - there was a printing error with my book, but thanks anyways
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    (Original post by KOH)
    i got iT..... PIECE OF CAKE!

    lg x + lg y = lg 3
    1000 = 10^3
    1000 does not equal lg3

    MB
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    (Original post by musicboy)
    1000 = 10^3
    1000 does not equal lg3

    MB
    nah its a misprint. doing it with 1000 doesnt work
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    (Original post by lgs98jonee)
    2cosxcos50-2sinxsin50=sinxcos40+cosxsin40
    2cos50-2tanxsin50=tanxcos40+sin40
    so 2cos50-sin40=tanx(cos40+2sin50)
    so tan x= (2cos50-sin40)/(cos40+2sin50)
    so x= tan-1 of all that
    hot did u get from tan x= (2cos50-sin40)/(cos40+2sin50)
    to x= tan-1 of all that and is it tanx in the end?
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    (Original post by lexazver203)
    hot did u get from tan x= (2cos50-sin40)/(cos40+2sin50)
    to x= tan-1 of all that and is it tanx in the end?
    if u do tan-1 of both sides then tan-1(tanx)=x

    and x=tan-1(blah)

    try tan-1(tan60) or any number and u will get 60 back
 
 
 
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