zzr0ck3r
  • zzr0ck3r
If ai+b = c+d and aj+b = c+e prove that x = (jd-ie)/(d-e) is a solution to ax+b=c
Mathematics
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SOLVED
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katieb
  • katieb
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zzr0ck3r
  • zzr0ck3r
this has nothing to do with complex, that is just some variable i.
Directrix
  • Directrix
What is the title of the course in which you were assigned this problem?
zzr0ck3r
  • zzr0ck3r
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.

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zzr0ck3r
  • zzr0ck3r
its from chapter 7 in the nine chapters
zzr0ck3r
  • zzr0ck3r
http://www-history.mcs.st-and.ac.uk/HistTopics/Nine_chapters.html scroll down to chapter 7
Directrix
  • Directrix
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.
zzr0ck3r
  • zzr0ck3r
lol I hear ya, yeah they are all just arbitrary variables.
zzr0ck3r
  • zzr0ck3r
well you know what i mean..
Directrix
  • Directrix
I do know. Hold up just few seconds while I wrap up two short problems in progress.
zzr0ck3r
  • zzr0ck3r
Ill be here all night:)
Directrix
  • Directrix
You can move this question to the closed section and type up a new question in a new thread, if there is another question.
zzr0ck3r
  • zzr0ck3r
ah this is all I got, but thanks.
Directrix
  • Directrix
I am not at all familiar with the "double false position," at least not by that name. There are helpers here who would be.
zzr0ck3r
  • zzr0ck3r
its very strange because my teacher left a note saying just plug in x and varify...lol
ghazi
  • ghazi
\[x=\frac{ c-b }{ a }\] now find \[\frac{ c-b }{ a } \] from \[ai+b=c+d, ai+b=c+e\]
ghazi
  • ghazi
sorry aj +b=c+e
ghazi
  • ghazi
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 :)
ghazi
  • ghazi
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
ghazi
  • ghazi
and i am sorry if there are sign mistakes :(
zzr0ck3r
  • zzr0ck3r
sweet thnx
zzr0ck3r
  • zzr0ck3r
What am i pluging a into?
zzr0ck3r
  • zzr0ck3r
@ghazi
zzr0ck3r
  • zzr0ck3r
Nm
ghazi
  • ghazi
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
zzr0ck3r
  • zzr0ck3r
Yeah i got it, tyvm

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