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    What is the answer to this? Can you show working please?!
    1/x-2 + 3/x+6 =1/2
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    (Original post by isabel12144)
    What is the answer to this? Can you show working please?!
    1/x-2 + 3/x+6 =1/2
    You should remove ambiguity by using brackets, so it would be, I'm assuming 1/(x-2) + 3(x+6) = 1/2, right, that is:

    \displaystyle \frac{1}{x-2} + \frac{3}{x+6} = \frac{1}{2}.

    Multiply throughout by (x-2)(x+6) and then expand brackets and solve the remaining quadratic:

    x+6 + 3(x-2) = \dfrac{(x+6)(x-2)}{2} - multiply both sides by two, re-arrange, etc... and then solve the quadratic.
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    (Original post by isabel12144)
    What is the answer to this? Can you show working please?!
    1/x-2 + 3/x+6 =1/2
    What do you think you should do? (hint: you've got some fractions there ...)
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    (Original post by SeanFM)
    What do you think you should do? (hint: you've got some fractions there ...)
    I did this:
    x+6/(x-2)(x+6) + 3(x-2)/(x-2)(x+6)

    x+6+3x-6/(x-2)(x+6) = 4x/(x-2)(x+6)

    4x/(x-2)(x+6)=1/2

    Then I didn't know what to do.
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    (Original post by isabel12144)
    I did this:
    x+6/(x-2)(x+6) + 3(x-2)/(x-2)(x+6)

    x+6+3x-6/(x-2)(x+6) = 4x/(x-2)(x+6)

    4x/(x-2)(x+6)=1/2

    Then I didn't know what to do.
    Good work :borat:

    Now it may help to 'get rid' of the fractions on each side so you can form a quadratic equation.. any ideas?
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    (Original post by SeanFM)
    Good work :borat:

    Now it may help to 'get rid' of the fractions on each side so you can form a quadratic equation.. any ideas?
    Do you do?

    4x=1/2*(x-2)(x-6)
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    (Original post by isabel12144)
    Do you do?

    4x=1/2*(x-2)(x-6)
    Careful with how you write things but yes, as long as that's  4x = \frac{(x-2)(x+6)}{2} that's fine, and you can also 'get rid' of the 1/2 on the right hand side...

    What next? :borat:
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    (Original post by SeanFM)
    Careful with how you write things but yes, as long as that's  4x = \frac{(x-2)(x+6)}{2} that's fine, and you can also 'get rid' of the 1/2 on the right hand side...

    What next? :borat:
    8x=(x-2)(x+6)
    ?
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    (Original post by isabel12144)
    8x=(x-2)(x+6)
    ?
    Correct now what can you do?
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    (Original post by isabel12144)
    8x=(x-2)(x+6)
    ?
    Keep going...

    Hint (only if you need it)
    Spoiler:
    Show
    You can expand the brackets
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    (Original post by SeanFM)
    Correct now what can you do?
    8x=x^2+4x-12
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    (Original post by isabel12144)
    8x=x^2+4x-12
    Yep, so it almost looks like a quadratic that you can solve. What's next?
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    (Original post by SeanFM)
    Yep, so it almost looks like a quadratic that you can solve. What's next?
    8x+12=x^2+4x
    12=x^2-4x
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    (Original post by isabel12144)
    8x+12=x^2+4x
    12=x^2-4x
    If you rearrange the equation so it equals 0, it might be easier to solve
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    (Original post by Andy98)
    If you rearrange the equation so it equals 0, it might be easier to solve
    is it this then?
    x^2-4x-12=0
    (x-6)(x+2)=0
    x=-6 or 2
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    (Original post by isabel12144)
    is it this then?
    x^2-4x-12=0
    (x-6)(x+2)=0
    x=-6 or 2
    You got the factorisation right, but the solutions incorrect
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    (Original post by isabel12144)
    is it this then?
    x^2-4x-12=0
    (x-6)(x+2)=0
    x=-6 or 2
    Almost - remember, you have to work out:
    x-6=0 and x+2=0
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    (Original post by isabel12144)
    is it this then?
    x^2-4x-12=0
    (x-6)(x+2)=0
    x=-6 or 2
    for (x-6)(x+2)=0
    either x-6=0 or x+2=0
    if you rearrange these you should get the correct values for x
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    (Original post by Andy98)
    You got the factorisation right, but the solutions incorrect
    x=-2 or 6
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    (Original post by isabel12144)
    x=-2 or 6
    correct.
 
 
 
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