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Math Help - Adding fractions in algebra

  1. #1
    Newbie Grich's Avatar
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    Adding fractions in algebra

    Okay, I have the expression:
    <br />
\frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}<br />
    now I need help to understand how these fractions are added together.
    I have the solution to it, but upon analysing it I am a bit confused:
    <br />
\frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}<br />
    <br />
= \frac{6}{2(x - 1)(x + 1)} \times \frac{3(x - 1)}{3(x - 1)} <br />
+ \frac{1}{6(x - 1)^2} \times \frac{x + 1}{x + 1} \\<br />
    <br />
= \frac{18(x - 1)}{6(x - 1)^2(x + 1)} + <br />
\frac{x + 1}{6(x - 1)^2(x + 1)}<br />
    <br />
= \frac{18(x - 1) + x + 1}{6(x - 1)^2(x + 1)}<br />

    1. Why are we multiplying \frac{6}{2(x - 1)(x + 1)} by \frac{3(x - 1)}{3(x - 1)} and how did we get the conclusion to use 3(x - 1) in the first place.
    2. Same with \frac{1}{6(x - 1)^2} and x + 1.

    I get the rest of it but those two parts of the solution don't make sense to me!? Are they using an Least Common Denominator (LCD) or something?

    Thanks in advance.
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  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
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    Quote Originally Posted by Grich View Post
    Okay, I have the expression:
    <br />
\frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}<br />
    now I need help to understand how these fractions are added together.
    I have the solution to it, but upon analysing it I am a bit confused:
    <br />
\frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}<br />
    <br />
= \frac{6}{2(x - 1)(x + 1)} \times \frac{3(x - 1)}{3(x - 1)} <br />
+ \frac{1}{6(x - 1)^2} \times \frac{x + 1}{x + 1} \\<br />
    <br />
= \frac{18(x - 1)}{6(x - 1)^2(x + 1)} + <br />
\frac{x + 1}{6(x - 1)^2(x + 1)}<br />
    <br />
= \frac{18(x - 1) + x + 1}{6(x - 1)^2(x + 1)}<br />

    1. Why are we multiplying \frac{6}{2(x - 1)(x + 1)} by \frac{3(x - 1)}{3(x - 1)} and how did we get the conclusion to use 3(x - 1) in the first place.
    2. Same with \frac{1}{6(x - 1)^2} and x + 1.
    I get the rest of it but those two parts of the solution don't make sense to me!? Are they using an Least Common Denominator (LCD) or something?

    Thanks in advance.
    Hello Grich,

    I wouldn't do it that way. Let me show you my approach and see if you can follow it.

    \frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}

    First, simplify the first term of your sum by dividing 2 into 6.

    \frac{3}{(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}

    Find the LCD: 6(x-1)^2(x+1)

    Now multiply each term of your sum by the LCD and make appropriate cancellations (divisions)

    6(x-1)^2(x+1)\left(\frac{3}{(x-1)(x+1)}\right)+6(x-1)^2(x+1)\left(\frac{1}{6(x-1)^2}\right)

    Now, put the simplified numerator over your LCD and finish up

    \frac{18(x-1)+x+1}{6(x-1)^2(x+1)}
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  3. #3
    Newbie Grich's Avatar
    Joined
    Aug 2008
    From
    Sydney
    Posts
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    Thanks masters, that was great help. I knew it was something to do with the LCD!
    Here is how I solved the solution:
    Find the LCD, which is:
    6(x - 1)^2(x + 1)
    The working out:
    = \left(\frac{6}{2(x - 1)(x + 1)} \times \frac{3(x - 1)}{3(x - 1)}\right)<br />
+ \left(\frac{1}{6(x - 1)^2} \times \frac{(x + 1)}{(x + 1)}\right)
    <br />
= \frac{18(x - 1)}{6(x - 1)^2(x + 1)} + \frac{(x + 1)}{6(x - 1)^2(x + 1)}<br />
    Therefore it will equal:
    <br />
\frac{18(x - 1) + (x + 1)}{6(x - 1)^2(x + 1)}<br />
    I have to remember to simplify the prackets though!
    <br />
\frac{18(x - 1) + x + 1}{6(x - 1)^2(x + 1)}<br />
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