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Thread: Algebraic fractions: Question e: 5/[x (x + 1)] - 2/x + 3/(x + 1)

  1. #1

    Question Algebraic fractions: Question e: 5/[x (x + 1)] - 2/x + 3/(x + 1)

    Question:

    . . . . .[tex]\mbox{(e) }\, \dfrac{5}{x\, (x\, +\, 1)}\, -\, \dfrac{2}{x}\, +\, \dfrac{3}{x\, +\, 1}[/tex]

    I'm getting myself confused with this one.

    1) I multiplied all the denominators, so they all became x(x+1)x(x+1).
    2) Then 5x(x+1) - 2x(x+1)(x+1) + 3x(x+1)x
    3) Becomes, 5x2 + 5x - 2x3 - 4x2 - 2x + 3x3 + 3x2
    4) Simplified, 4x2 + 3x + x3 / x(x+1)x(x+1).

    The answer is x + 3 / x(x+1). I don't know how to get this.

    Appreciate your help all.

    Thxs.
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    Last edited by stapel; 02-10-2018 at 06:53 PM. Reason: Typing out the text in the graphic; creating useful subject line.

  2. #2
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by radnorgardens View Post
    Question:

    . . . . .[tex]\mbox{(e) }\, \dfrac{5}{x\, (x\, +\, 1)}\, -\, \dfrac{2}{x}\, +\, \dfrac{3}{x\, +\, 1}[/tex]

    I'm getting myself confused with this one.

    Simplified, [4x2 + 3x + x3]/[x(x+1)x(x+1)]

    The answer is [x + 3]/[x(x+1)]

    I don't know how to get this.
    Notice that x^3 + 4x^2 + 3x can be factored (to start, every term contains at least one factor of x). If you completely factor this numerator, you can then cancel common factors, to get the known answer.

    Also, it's true that x(x+1)x(x+1) is a common denominator, but it's not the easiest one to use.

    In the given expression, the three denominators are x(x+1), x, and x+1.

    Can you see that x(x+1) is also a common denominator? Using that one (at the beginning) is less work.
    Last edited by stapel; 02-10-2018 at 06:52 PM. Reason: Copying typed-out graphical content into reply.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  3. #3
    Quote Originally Posted by mmm4444bot View Post
    Notice that x^3 + 4x^2 + 3x can be factored (to start, every term contains at least one factor of x). If you completely factor this numerator, you can then cancel common factors, to get the known answer.

    Also, it's true that x(x+1)x(x+1) is a common denominator, but it's not the easiest one to use.

    In the given expression, the three denominators are x(x+1), x, and x+1.

    Can you see that x(x+1) is also a common denominator? Using that one (at the beginning) is less work.
    Thanks mmm4444bot. So much easier with x(x+1). Don't know why I didn't see that.

    5/x(x+1) - 2(x+1)/x(x+1) + 3x/x(x+1)

    5 - 2x - 2 + 3x = x+3

    Thus, x+3/x(x+1)

    Many thanks.

  4. #4
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by radnorgardens View Post
    5/[x(x+1)] - 2(x+1)/[x(x+1)] + 3x/[x(x+1)]

    5 - 2x - 2 + 3x = x+3

    Thus, [x+3]/[x(x+1)]
    As Denis noted, we need to put grouping symbols around some numerators and denominators, when typing algebraic ratios with a keyboard. This is to ensure that people reading these texted expressions know what's on top and what's on bottom. Let us know, if you're not sure about this.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  5. #5
    Quote Originally Posted by mmm4444bot View Post
    As Denis noted, we need to put grouping symbols around some numerators and denominators, when typing algebraic ratios with a keyboard. This is to ensure that people reading these texted expressions know what's on top and what's on bottom. Let us know, if you're not sure about this.
    Ah yes. I understand I should have done that. Thanks for point this out.

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