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    Hey, I'm struggling to simplify a composite function. This is for question 3 part a.
    So far I have h(x)= (6/x+3) -2. How do I simplify this to give the answer -2x/x+3 (which is given to me)
    Thanks
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      (Original post by mrmigme)
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      Hey, I'm struggling to simplify a composite function. This is for question 3 part a.
      So far I have h(x)= (6/x+3) -2. How do I simplify this to give the answer -2x/x+3 (which is given to me)
      Thanks
      \displaystyle \frac{6}{x+3} - 2 = \frac{6}{x+3} - \frac{2(x+3)}{x+3} = \ ?
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      (Original post by EricPiphany)
      \displaystyle \frac{6}{x+3} - 2 = \frac{6}{x+3} - \frac{2(x+3)}{x+3} = \ ?
      I think I'm misunderstanding the basics of this topic lol. Why is it 2(x+3)/x+3 and not 2(6)/(x+3)
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        (Original post by mrmigme)
        I think I'm misunderstanding the basics of this topic lol. Why is it 2(x+3)/x+3 and not 2(6)/(x+3)
        Clearly,  2 = \dfrac{2(x+3)}{x+3}, x \not= -3.
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          (Original post by mrmigme)
          I think I'm misunderstanding the basics of this topic lol. Why is it 2(x+3)/x+3 and not 2(6)/(x+3)
          You have to multiply top and bottom of fraction by the same number, which is essentially multiplying by 1.
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          (Original post by mrmigme)
          I think I'm misunderstanding the basics of this topic lol. Why is it 2(x+3)/x+3 and not 2(6)/(x+3)
          Do you remember how to add fractions? Like in GCSE? You'd need to make common denominators?

          So to add \displaystyle \frac{1}{2} + \frac{1}{4} = \frac{1 \times 2}{2 \times 2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{2+1}{4}= \frac{3}{4}, right?

          Here it's the same thing: \displaystyle \frac{6}{x+3} -2 = \frac{6}{x+3} - \frac{2(x+3)}{(x+3)} = \frac{6 - 2(x+3)}{x+3}
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          (Original post by Zacken)
          Do you remember how to add fractions? Like in GCSE? You'd need to make common denominators?

          So to add \displaystyle \frac{1}{2} + \frac{1}{4} = \frac{1 \times 2}{2 \times 2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{2+1}{4}= \frac{3}{4}, right?

          Here it's the same thing: \displaystyle \frac{6}{x+3} -2 = \frac{6}{x+3} - \frac{2(x+3)}{(x+3)} = \frac{6 - 2(x+3)}{x+3}
          Bloody hell I can't believe I got caught up on how to do that. Thanks a lot!
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          (Original post by mrmigme)
          Bloody hell I can't believe I got caught up on how to do that. Thanks a lot!
          You're very welcome!
         
         
         
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