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toriebabii Group Title

3/x+13/x-11

  • 2 years ago
  • 2 years ago

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  1. toriebabii Group Title
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    \[\left[\begin{matrix}3 & 13 \\ x & x-11\end{matrix}\right]\]

    • 2 years ago
  2. SWAG Group Title
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    • 2 years ago
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  3. andriod09 Group Title
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    are they fractions or are they dividends?

    • 2 years ago
  4. toriebabii Group Title
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    there fractions being added.

    • 2 years ago
  5. toriebabii Group Title
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    sorry forgot to write that

    • 2 years ago
  6. satellite73 Group Title
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    i think it is \[\frac{3}{x}+\frac{13}{x-11}\] but i could be wrong

    • 2 years ago
  7. satellite73 Group Title
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    in any case add fraction the same way you always do \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bc}\]

    • 2 years ago
  8. toriebabii Group Title
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    Thats exactly how you write it

    • 2 years ago
  9. satellite73 Group Title
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    if i am right in the question, you should get \[\frac{3(x-11)+13x}{x(x-11)}\]which you can clean up with some algebra

    • 2 years ago
  10. andriod09 Group Title
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    \[\frac{3}{x}+\frac{13}{x-11}\]\[\frac{3(x-11)+13x}{x^2-11}\] would that be right @satellite73

    • 2 years ago
  11. satellite73 Group Title
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    if you multiply out in the denominator it will be \(x^2-11x\)

    • 2 years ago
  12. toriebabii Group Title
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    Im really confused can you do it step by step?

    • 2 years ago
  13. satellite73 Group Title
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    btw i made a typo above, it should be \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]

    • 2 years ago
  14. satellite73 Group Title
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    @toriebabii did you see what i wrote above about adding fractions? it is the same way with variables as with numbers \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]

    • 2 years ago
  15. satellite73 Group Title
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    in your case \(a=3,b=x,c=13,d=x-11\)

    • 2 years ago
  16. satellite73 Group Title
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    put them directly in to the right hand side of the method for addition \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\] with careful use of parentheses, and you will get the answer

    • 2 years ago
  17. andriod09 Group Title
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    so then like mine said it would be: 3x+13x−11 \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]\[\frac{3}{x}\frac{13}{x-11}\]\[\frac{3(x-11)+13x}{x(x-11)}

    • 2 years ago
  18. andriod09 Group Title
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    \[*\frac{3(x-11)+13x}{x(x-11)}*\]

    • 2 years ago
  19. toriebabii Group Title
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    and thats the answer?

    • 2 years ago
  20. andriod09 Group Title
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    no, now you simplify. \[\frac{3(x-11)+13x}{x(x-11)}\]\[\frac{3x-33+13x}{x^2-11}\]\[\frac{16x-33}{x^2-11}\] do you see anything wlse to do in this equation?

    • 2 years ago
  21. andriod09 Group Title
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    i believe that would be the answer.

    • 2 years ago
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