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Math Help - basic question

  1. #1
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    basic question

    The expression 4/n+1 + 3/n-1 is equal to....

    Is the answer 7n-1/n^2-1 ????

    Thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Stevo_Evo_22 View Post
    The expression 4/n+1 + 3/n-1 is equal to....

    Is the answer 7n-1/n^2-1 ????

    Thanks
    \frac{4}{n+1}+\frac{3}{n-1}=?

    We want to combine these two fractions. However, their denominators are different. Multiply both top and bottom of the first fraction by n-1 and multiply both top and bottom of the second fraction by n+1:

    \frac{4}{n+1}\cdot{\color{red}\frac{n-1}{n-1}}+\frac{3}{n-1}\cdot{\color{red}\frac{n+1}{n+1}}

    Multiply through and simplify. Take note that (n+1)(n-1)=n^2+n-n-1=\color{red}n^2-1.

    \frac{4(n-1)}{n^2-1}+\frac{3(n+1)}{n^2-1}=\frac{4n-4+3n+3}{n^2-1}=\color{red}\boxed{\frac{7n-1}{n^2-1}}

    Hope this makes sense!
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  3. #3
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    Lol actually the question was extremely easy (even for me, maths tragic). I got the same answer you did, but the problem is it's a multiple choice question and neither of them are what I had.

    The options are:

    A=7n-1/1-n^2
    B=1-7n/1-n^2
    C=7n-1/n^2-1
    D=7/n^2-1
    E=7/n

    Is the answer that we got the same as any of these???
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Stevo_Evo_22 View Post
    Lol actually the question was extremely easy (even for me, maths tragic). I got the same answer you did, but the problem is it's a multiple choice question and neither of them are what I had.

    The options are:

    A=7n-1/1-n^2
    B=1-7n/1-n^2
    C=7n-1/n^2-1
    D=7/n^2-1
    E=7/n

    Is the answer that we got the same as any of these???
    Yes.

    \frac{7n-1}{n^2-1}=\frac{-(1-7n)}{-(1-n^2)}=\color{red}\boxed{\frac{1-7n}{1-n^2}}.

    B would be the best answer.
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