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How to add/subtract two algebraic fractions with common denominators? Watch

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    Such as:

    8/(x-5)(x+7) - 1/(x-5)(x-2)

    Title is worded slightly stupidly.

    How to add/subtract two algebraic fractions with common factors
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    The denominator remains the same, you just subtract/add the top numbers. So in this case it's 8 - 7.

    Just think if I had 3/4 and took 1/4, I'd have 2/4 left.

    EDIT: Ignore that, just realised the second bracket was different. Do cross multiplication to get rid of the thing that isn't common.
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    (Original post by BTAnonymous)
    Such as:

    8/(x-5)(x+7) - 1/(x-5)(x-2)
    Same way as you would for a normal number one.

    Like:

    \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{3-1}{12} .

    Here it's just:

    \displaystyle

\begin{align*} \frac{8}{(x-5)(x+7)} - \frac{1}{(x-5)(x-2)} &= \frac{8(x-2)}{(x-2)(x-5)(x+7)} - \frac{(x+7)}{(x-2)(x-5)(x+7)}\\ & = \frac{8(x-2) - (x+7)}{(x-2)(x-5)(x+7)} \end{align*}
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    (Original post by Zacken)
    Same way as you would for a normal number one.

    Like:

    \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{3-1}{12} .

    Here it's just:

    \displaystyle

\begin{align*} \frac{8}{(x-5)(x+7)} - \frac{1}{(x-5)(x-2)} &= \frac{8(x-2)}{(x-2)(x-5)(x+7)} - \frac{(x+7)}{(x-2)(x-5)(x+7)}\\ & = \frac{8(x-2) - (x+7)}{(x-2)(x-5)(x+7)} \end{align*}
    Excellent! Thank you again!
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    (Original post by BTAnonymous)
    Excellent! Thank you again!
    No problem!
 
 
 
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