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

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zzr0ck3rBest ResponseYou've already chosen the best response.1
this has nothing to do with complex, that is just some variable i.
 one year ago

DirectrixBest ResponseYou've already chosen the best response.0
What is the title of the course in which you were assigned this problem?
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
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.
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
its from chapter 7 in the nine chapters
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
http://wwwhistory.mcs.stand.ac.uk/HistTopics/Nine_chapters.html scroll down to chapter 7
 one year ago

DirectrixBest ResponseYou've already chosen the best response.0
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.
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
lol I hear ya, yeah they are all just arbitrary variables.
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
well you know what i mean..
 one year ago

DirectrixBest ResponseYou've already chosen the best response.0
I do know. Hold up just few seconds while I wrap up two short problems in progress.
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
Ill be here all night:)
 one year ago

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

zzr0ck3rBest ResponseYou've already chosen the best response.1
ah this is all I got, but thanks.
 one year ago

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

zzr0ck3rBest ResponseYou've already chosen the best response.1
its very strange because my teacher left a note saying just plug in x and varify...lol
 one year ago

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

ghaziBest ResponseYou've already chosen the best response.3
aid=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 :)
 one year ago

ghaziBest ResponseYou've already chosen the best response.3
also 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
 one year ago

ghaziBest ResponseYou've already chosen the best response.3
and i am sorry if there are sign mistakes :(
 one year ago

zzr0ck3rBest ResponseYou've already chosen the best response.1
What am i pluging a into?
 one year ago

ghaziBest ResponseYou've already chosen the best response.3
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
 one year ago
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