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I Need Help On Algebraic Fractions!

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    Heya, im doing a past paper for the exam on Monday, it's a non calculator and I am stuck on a similar sort of question.

    \frac{3}{x-2}+\frac{8}{x-3}=2

    What I did was:

    \frac{3(x-3)+8(x-2)}{(x-2)(x-3)}=2

    I multiplied out the numerator and then cross multiplied the denominator onto the other side of the equals sign:

    11x-25=2\times(x-2)(x-3)
    11x-25=2(x^2-5x+6)
    11x-25=2x^2-10x+12
    2x^2-21x+37=0

    Here is where i got stuck..
    I had to options, either follow what Mathswatch had taught me and rearrange the whole equation to make 0 then factorise or use the quadratic formula..
    Since this is a non calculator i attempted the rearranging and factorising:

    2x^2-21x+37=0

    I tried the ABC method ( a=2 .. b= -21 .. c= 37)
    The rule is A\times C which would in this case  = 74
    The factors of 74 are : 1 and 74, 2 and 37 and neither of these add or subtract to make B which is -21..

    So I'm either lost, over complicating things or completely wrong!
    Please, can someone help me?
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    (Original post by Winch2012)
    Heya, im doing a past paper for the exam on Monday, it's a non calculator and I am stuck on a similar sort of question.

    \frac{3}{x-2}+\frac{8}{x-3}=2

    What I did was:

    \frac{3(x-3)+8(x-2)}{(x-2)(x-3)}=2

    I multiplied out the numerator and then cross multiplied the denominator onto the other side of the equals sign:

    11x-25=2\times(x-2)(x-3)
    11x-25=2(x^2-5x+6)
    11x-25=2x^2-10x+12
    2x^2-21x+37=0

    Here is where i got stuck..
    I had to options, either follow what Mathswatch had taught me and rearrange the whole equation to make 0 then factorise or use the quadratic formula..
    Since this is a non calculator i attempted the rearranging and factorising:

    2x^2-21x+37=0

    I tried the ABC method ( a=2 .. b= -21 .. c= 37)
    The rule is A\times C which would in this case  = 74
    The factors of 74 are : 1 and 74, 2 and 37 and neither of these add or subtract to make B which is -21..

    So I'm either lost, over complicating things or completely wrong!
    Please, can someone help me?
    I believe you're right - there's no simple solution, and you'd just have to give it in terms of the quadratic formula. http://www.wolframalpha.com/input/?i...%29%5E-1+%3D+2
    My calculator came out with a solution of 8.26.... so yeah, there's no simple solution

    Edit: Wolfram Alpha - the most amazing maths tool on the internet it's soo good
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    (Original post by Junaid96)
    I believe you're right - there's no simple solution, and you'd just have to give it in terms of the quadratic formula. http://www.wolframalpha.com/input/?i...%29%5E-1+%3D+2
    My calculator came out with a solution of 8.26.... so yeah, there's no simple solution

    Edit: Wolfram Alpha - the most amazing maths tool on the internet it's soo good
    The question was non-calculator.
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    Yeah however, the mark scheme says that X = 5 or X = -0.5

    The brief explanation is:
    M1 for common denominator on LHS or clearing fractions
    M1 for multiplying out brackets
    A1 for 2^2 – 9x + 5 = 0
    M1 for (2x ± 1)(x ± 5) or substitution into quadratic formula
    A1 for 5 and - 0.5
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    (Original post by ThatPerson)
    The question was non-calculator.
    True, but that doesn't stop you using the formula:
    (21+Sqrt137)/4 or (21-Sqrt137)/4 being the solutions as I get them
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    (Original post by ThatPerson)
    The question was non-calculator.
    Yes, I know. I just used a calculator to confirm my suspicions.

    (Original post by Icedstoat)
    True, but that doesn't stop you using the formula:
    (21+Sqrt137)/4 or (21-Sqrt137)/4 being the solutions as I get them
    Yep, and as I said in my OP, "you'd just have to give it in terms of the quadratic formula"
    (Original post by Winch2012)
    Yeah however, the mark scheme says that X = 5 or X = -0.5

    The brief explanation is:
    M1 for common denominator on LHS or clearing fractions
    M1 for multiplying out brackets
    A1 for 2^2 – 9x + 5 = 0
    M1 for (2x ± 1)(x ± 5) or substitution into quadratic formula
    A1 for 5 and - 0.5
    \frac{3}{x-2}+\frac{8}{x-3}=2 but.. 5 isn't a solution?
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    (Original post by Junaid96)
    ...
    (Original post by Icedstoat)
    ...
    (Original post by Winch2012)
    Yeah however, the mark scheme says that X = 5 or X = -0.5

    The brief explanation is:
    M1 for common denominator on LHS or clearing fractions
    M1 for multiplying out brackets
    A1 for 2^2 – 9x + 5 = 0
    M1 for (2x ± 1)(x ± 5) or substitution into quadratic formula
    A1 for 5 and - 0.5
    (Original post by ThatPerson)
    ...
    There is a typo.

    The question should be  \displaystyle \frac3{x-2} + \frac8{x+3} = 2

    The minus sign should be positive.
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    OP, if that's the mark scheme, then you've probably misread the question or something.
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    (Original post by Icedstoat)
    OP, if that's the mark scheme, then you've probably misread the question or something.
    See my previous post.
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    Yeah, I really need to learn to type faster before someone else says what I'm about to say
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    THANKS SO MUCH!
    I literally spent hours on this question !!

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