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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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this has nothing to do with complex, that is just some variable i.
What is the title of the course in which you were assigned this problem?
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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its from chapter 7 in the nine chapters scroll down to chapter 7
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.
lol I hear ya, yeah they are all just arbitrary variables.
well you know what i mean..
I do know. Hold up just few seconds while I wrap up two short problems in progress.
Ill be here all night:)
You can move this question to the closed section and type up a new question in a new thread, if there is another question.
ah this is all I got, but thanks.
I am not at all familiar with the "double false position," at least not by that name. There are helpers here who would be.
its very strange because my teacher left a note saying just plug in x and
\[x=\frac{ c-b }{ a }\] now find \[\frac{ c-b }{ a } \] from \[ai+b=c+d, ai+b=c+e\]
sorry aj +b=c+e
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 :)
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
and i am sorry if there are sign mistakes :(
sweet thnx
What am i pluging a into?
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
Yeah i got it, tyvm

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