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Math Help - hard fraction

  1. #31
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    hard fraction-maths2.jpg
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  2. #32
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    You're not getting the common denominator correctly. You multiply each fraction, top and bottom, by what it's missing in the denominator in order for that fraction to have the common denominator. So, what would you multiply the first fraction by (it must be equal to 1, or you'll change the fraction!) in order for the first fraction to have the common denominator of 2(x+2)?
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  3. #33
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    6?
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  4. #34
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    No, you're guessing. Think! What do you need to multiply 2 by to get 2(x+2)? There's less here than meets the eye, trust me. There's no trick questions.
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  5. #35
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    x+2
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  6. #36
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    That's right. So now multiply the first fraction by (x+2)/(x+2). You need to employ the same reasoning on the second fraction. What do you need to multiply x+2 by to get 2(x+2)?
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  7. #37
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    could you show me how i would do this please. i know how to do it but im not sure how to set it out.
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  8. #38
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    Multiplying fractions is the more straight-forward part. You need to do this:

    \dfrac{x-1}{2}\left(\dfrac{x+2}{x+2}\right)-\dfrac{4}{x+2}\left(\dfrac{2}{2}\right).

    Do you see how I got this expression? Could you duplicate this effort on your own?
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  9. #39
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    i don't understand how you got that! could you show me what we have worked out from the start.
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  10. #40
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    I'd rather see you try to pull together what we've done so far. I'll give you a hint: what I did in post # 38 is, mathematically, precisely what I did in post # 24. If it helps, think of x+2 as a single symbol. Substitute, for example, y = x+2, if it helps. If it doesn't help, ignore that idea.

    So, go back through the posts, and think about what we're trying to do. Draw heavily on your experience in adding fractions that are just numbers, like in post # 24. We are doing exactly the same thing, mathematically.
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  11. #41
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    Here's a little analogy to consider also...

    Multiplying a fraction by a form of 1 expressed as a fraction
    does not change the fraction but allows it to be combined with another with the same denominator.
    Then you can think in terms of apples+apples as we have halves+halves, quarters+quarters, sixths+sixths etc...

    All we need do is use the denominator from the other fraction to write 1.

    \displaystyle\frac{1}{2}+\frac{5}{3}=\left(\frac{3  }{3}\right)\frac{1}{2}+\left(\frac{2}{2}\right)\fr  ac{5}{3}=\frac{3}{6}+\frac{10}{6}=\frac{3+10}{6}=\  frac{13}{6}

    We took the 3 from the second fraction and wrote \frac{3}{3}
    and 2 from the first fraction to write \frac{2}{2}

    Now, suppose we split 3 into 3=x+2 so that x=1

    but we write \displaystyle\frac{1}{2}+\frac{5}{x+2}=\frac{13}{6  }

    so that we must deduce x from here.

    All we do is the same thing we did when we had no x.

    We multiply the first fraction by \frac{x+2}{x+2} and the second one by \frac{2}{2}

    This will result in both LHS fractions having a common denominator,
    so we are adding like fractions again...
    sixths+sixths, quarters+quarters etc...but in this case 2(x+2)'s+2(x+2)'s
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  12. #42
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    right then, would i do:

    hard fraction-1.jpg

    after this im not sure what to do, so could you point me in the right direction please?
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  13. #43
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    I'm afraid that's incorrect. Do you understand how I got the expressions in post # 38? That's the point from where you need to continue. The mathematics I used to get that are exactly what Archie Meade said, and what I've said before. I think you understand what a common denominator is, but you're not actually getting one yet. Compare post # 38 with, say, Archie's post, until you understand why post 38 is the right thing to do. Then go from there.

    Here's another possible track (but it will still require adding fractions!):

    \dfrac{x-1}{2}-\dfrac{4}{x+2}=1

    \dfrac{x-1}{2}=1+\dfrac{4}{x+2} (moved fraction to the RHS)

    \dfrac{x-1}{2}=\dfrac{x+2}{x+2}+\dfrac{4}{x+2}. Note that 1=\dfrac{x+2}{x+2}, so I just substituted it in.

    Then continue.
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  14. #44
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    i know how you got it on post 38 but how is what i put wrong on post 42? where am i going wrong? which part is wrong?
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  15. #45
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    Try this:

    \left(\dfrac{x-1}{2}\right)\left(\dfrac{x+2}{x+2}\right)-\left(\dfrac{4}{x+2}\right)\left(\dfrac{2}{2}\righ  t)=\dfrac{(x-1)(x+2)}{2(x+2)}-\dfrac{(4)(2)}{2(x+2)}.

    Foil out the LHS.

    What was wrong with your last attempt was that most of the numerators disappeared! You divided by what was needed to get the common denominator, but you didn't also multiply. If I take 2, and divide by 3, I get 2/3, which is not equal to 2. If, on the other hand, I take 2, and then multiply and divide by 3, I get 6/3, which is equal to 2. That is, because I multiplied by 3/3, I didn't change the fraction. You changed the fraction because you only divided by what you needed to get the common denominator.
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