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zzr0ck3r
 2 years ago
If ai+b = c+d and aj+b = c+e prove that x = (jdie)/(de) is a solution to ax+b=c
zzr0ck3r
 2 years ago
If ai+b = c+d and aj+b = c+e prove that x = (jdie)/(de) is a solution to ax+b=c

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zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1this has nothing to do with complex, that is just some variable i.

Directrix
 2 years ago
Best ResponseYou've already chosen the best response.0What is the title of the course in which you were assigned this problem?

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1um, 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.

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1its from chapter 7 in the nine chapters

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1http://wwwhistory.mcs.stand.ac.uk/HistTopics/Nine_chapters.html scroll down to chapter 7

Directrix
 2 years ago
Best ResponseYou've already chosen the best response.0When 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
 2 years ago
Best ResponseYou've already chosen the best response.1lol I hear ya, yeah they are all just arbitrary variables.

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1well you know what i mean..

Directrix
 2 years ago
Best ResponseYou've already chosen the best response.0I do know. Hold up just few seconds while I wrap up two short problems in progress.

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1Ill be here all night:)

Directrix
 2 years ago
Best ResponseYou've already chosen the best response.0You can move this question to the closed section and type up a new question in a new thread, if there is another question.

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1ah this is all I got, but thanks.

Directrix
 2 years ago
Best ResponseYou've already chosen the best response.0I am not at all familiar with the "double false position," at least not by that name. There are helpers here who would be.

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1its very strange because my teacher left a note saying just plug in x and varify...lol

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.3\[x=\frac{ cb }{ a }\] now find \[\frac{ cb }{ a } \] from \[ai+b=c+d, ai+b=c+e\]

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.3aid=cb and aje= cb now add above equations a(i+j) (d+e)=2(cb) now divide both sides by a \[(i+j)\frac{ d+e }{ a }=2\frac{ cb }{ a }=2x\] i think now you can do it :)

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.3also if you subtract aid=cb from aje=cb you will have a(ji)=ed therefore\[a=\frac{ ed }{ ji }\] now plug in a and rationalise

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.3and i am sorry if there are sign mistakes :(

zzr0ck3r
 2 years ago
Best ResponseYou've already chosen the best response.1What am i pluging a into?

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.3well 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
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