## zzr0ck3r 2 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

1. zzr0ck3r

this has nothing to do with complex, that is just some variable i.

2. Directrix

What is the title of the course in which you were assigned this problem?

3. 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.

4. zzr0ck3r

its from chapter 7 in the nine chapters

5. zzr0ck3r

http://www-history.mcs.st-and.ac.uk/HistTopics/Nine_chapters.html scroll down to chapter 7

6. 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.

7. zzr0ck3r

lol I hear ya, yeah they are all just arbitrary variables.

8. zzr0ck3r

well you know what i mean..

9. Directrix

I do know. Hold up just few seconds while I wrap up two short problems in progress.

10. zzr0ck3r

Ill be here all night:)

11. 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.

12. zzr0ck3r

ah this is all I got, but thanks.

13. 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.

14. zzr0ck3r

its very strange because my teacher left a note saying just plug in x and varify...lol

15. ghazi

$x=\frac{ c-b }{ a }$ now find $\frac{ c-b }{ a }$ from $ai+b=c+d, ai+b=c+e$

16. ghazi

sorry aj +b=c+e

17. 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 :)

18. 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

19. ghazi

and i am sorry if there are sign mistakes :(

20. zzr0ck3r

sweet thnx

21. zzr0ck3r

What am i pluging a into?

22. zzr0ck3r

@ghazi

23. zzr0ck3r

Nm

24. 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

25. zzr0ck3r

Yeah i got it, tyvm