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    (Original post by Andy98)
    correct.
    Thank you so much for your help!
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    (Original post by BlueBlueBells)
    Almost - remember, you have to work out:
    x-6=0 and x+2=0
    Thank you for your help! I really appreciate it!
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    (Original post by isabel12144)
    Thank you for your help! I really appreciate it!
    Glad to help
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    (Original post by SeanFM)
    Yep, so it almost looks like a quadratic that you can solve. What's next?
    Thank you sooo much for your help! I really appreciate it!!!!
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    (Original post by zooloowarrier)
    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
    Thank you so much for your help! I really appreciate it!!!
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    (Original post by isabel12144)
    Thank you so much for your help!
    Not problem :hat2:
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    Sorry to ask again but I just want to check i'm doing it right.

    1/x-5 +6/x =2
    x/x(x-5) + 6(x-5)/x(x-5)=2
    7x-30/x(x-5)=2
    7x-30/x^2-5x=2
    7x-30=2(x^2-5x)
    7x-30=2x^2-10x
    -30=2x^2-13x
    2x^2-13x+30=0
    Then you can't factorise so what do you do?
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    (Original post by isabel12144)
    Sorry to ask again but I just want to check i'm doing it right.

    1/x-5 +6/x =2
    x/x(x-5) + 6(x-5)/x(x-5)=2
    7x-30/x(x-5)=2
    7x-30/x^2-5x=2
    7x-30=2(x^2-5x)
    7x-30=2x^2-10x
    -30=2x^2-13x
    2x^2-13x+30=0
    Then you can't factorise so what do you do?
    There's an error here - "-30=2x^2-13x"
    it should be -30=2x^2-17x
    2x^2-17x+30=0 is easily solvable, it might also be worth looking up the 'quadratic formula' for quadratics that aren't so straight forward to factor.
 
 
 
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