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zzr0ck3r

  • 3 years ago

If ai+b = c+d and aj+b = c+e prove that x = (jd-ie)/(d-e) is a solution to ax+b=c

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  1. zzr0ck3r
    • 3 years ago
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    this has nothing to do with complex, that is just some variable i.

  2. Directrix
    • 3 years ago
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    What is the title of the course in which you were assigned this problem?

  3. zzr0ck3r
    • 3 years ago
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    um, its from the nine chapters on chinese mathematics, its a 400 level math history class, but this math should not be hard... I just have no idea how to do it for some reason. I try and expand everything and it gets very very very ugly..and I get lost in it.

  4. zzr0ck3r
    • 3 years ago
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    its from chapter 7 in the nine chapters

  5. zzr0ck3r
    • 3 years ago
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    http://www-history.mcs.st-and.ac.uk/HistTopics/Nine_chapters.html scroll down to chapter 7

  6. Directrix
    • 3 years ago
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    When I see i, I think the imaginary number. When I see e I think of another constant. I would have to change the variables to have a chance of working the problem.

  7. zzr0ck3r
    • 3 years ago
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    lol I hear ya, yeah they are all just arbitrary variables.

  8. zzr0ck3r
    • 3 years ago
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    well you know what i mean..

  9. Directrix
    • 3 years ago
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    I do know. Hold up just few seconds while I wrap up two short problems in progress.

  10. zzr0ck3r
    • 3 years ago
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    Ill be here all night:)

  11. Directrix
    • 3 years ago
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    You can move this question to the closed section and type up a new question in a new thread, if there is another question.

  12. zzr0ck3r
    • 3 years ago
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    ah this is all I got, but thanks.

  13. Directrix
    • 3 years ago
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    I am not at all familiar with the "double false position," at least not by that name. There are helpers here who would be.

  14. zzr0ck3r
    • 3 years ago
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    its very strange because my teacher left a note saying just plug in x and varify...lol

  15. ghazi
    • 3 years ago
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    \[x=\frac{ c-b }{ a }\] now find \[\frac{ c-b }{ a } \] from \[ai+b=c+d, ai+b=c+e\]

  16. ghazi
    • 3 years ago
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    sorry aj +b=c+e

  17. ghazi
    • 3 years ago
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    ai-d=c-b and aj-e= c-b now add above equations a(i+j) -(d+e)=2(c-b) now divide both sides by a \[(i+j)-\frac{ d+e }{ a }=2\frac{ c-b }{ a }=2x\] i think now you can do it :)

  18. ghazi
    • 3 years ago
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    also if you subtract ai-d=c-b from aj-e=c-b you will have a(j-i)=e-d therefore\[a=\frac{ e-d }{ j-i }\] now plug in a and rationalise

  19. ghazi
    • 3 years ago
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    and i am sorry if there are sign mistakes :(

  20. zzr0ck3r
    • 3 years ago
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    sweet thnx

  21. zzr0ck3r
    • 3 years ago
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    What am i pluging a into?

  22. zzr0ck3r
    • 3 years ago
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    @ghazi

  23. zzr0ck3r
    • 3 years ago
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    Nm

  24. ghazi
    • 3 years ago
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    well i am sorry i wasnt here , you have to put value of a in to the fraction\[(i+j)-\frac{ d+e }{ a }\] the a in the denominator so that you can rearrange everything and have answer in terms of d and e , i and j only

  25. zzr0ck3r
    • 3 years ago
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    Yeah i got it, tyvm

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