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Hollywood_chrissy Group Title

Can somebody please explain th3 attached to me. is there a proof or something???

  • 2 years ago
  • 2 years ago

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  1. Hollywood_chrissy Group Title
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    • 2 years ago
  2. koalamon Group Title
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    just think about it, it will come to you.

    • 2 years ago
  3. Hollywood_chrissy Group Title
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    @koalamon now about a hint

    • 2 years ago
  4. asnaseer Group Title
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    hint: in this equation:\[\frac{a+b}{a-b}=\frac{c+d}{c-d}\]divide the numerator and denominator of the left-hand-side by 'b' and divide the numerator and denominator of the right-hand-side by 'd'.

    • 2 years ago
  5. Hollywood_chrissy Group Title
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    @asnaseer, even after your hint, i am still running into difficulty

    • 2 years ago
  6. DLS Group Title
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    Isn't this Componendo & Dividendo? ;)

    • 2 years ago
  7. Hollywood_chrissy Group Title
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    what is componedo and dividendo???

    • 2 years ago
  8. DLS Group Title
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    It is a theorem in which the denominator is added to the numerator & numerator is added with negative denominator e. g. if \[\frac{x}{y}=\frac{a}{b}\] then by componendo & dividendo we have \[\frac{( x + y)}{( x- y ) }\]= \[\frac{( a +b )}{( a - b )}\] (Proof) \[\frac{3}{2}\] =\[\frac{ 6}{4}\] then by componendo & dividendo \[\frac{3+ 2}{3-2}\] = \[\frac{6 +4}{6- 4}\] OR 5/ 1 = 10/ 2 that is true

    • 2 years ago
  9. Hollywood_chrissy Group Title
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    @DLS do you understand asnaseer's hint???

    • 2 years ago
  10. DLS Group Title
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    I think the theorem is better, he just said if u do the same thing/operation with a digit on one side,u do that on other too so overall effect nullifies

    • 2 years ago
  11. Hollywood_chrissy Group Title
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    you helped me greatly thanks

    • 2 years ago
  12. asnaseer Group Title
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    This is what I meant - starting with this:\[\frac{a+b}{a-b}=\frac{c+d}{c-d}\]lets first divide the numerator and denominator of the left-hand-side by 'b' to give:\[\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\frac{c+d}{c-d}\]then lets divide the numerator and denominator of the right-hand-side by 'd' to give:\[\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\frac{\frac{c}{d}+1}{\frac{c}{d}-1}\]now note that you are given that:\[\frac{a}{b}=\frac{c}{d}\]so the equality of both sides follows from this relation.

    • 2 years ago
  13. asnaseer Group Title
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    I don't think you can just assume the "Componendo and dividendo" theorem directly as this question is effectively asking you to prove it.

    • 2 years ago
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