Question

If ai+b = c+d and aj+b = c+e prove that x = (jd-ie)/(d-e) is a solution to ax+b=c

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

its from chapter 7 in the nine chapters

Answer 5

http://www-history.mcs.st-and.ac.uk/HistTopics/Nine_chapters.html

scroll down to chapter 7

Answer 6

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.

Answer 7

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

Answer 8

well you know what i mean..

Answer 9

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

Answer 10

Ill be here all night:)

Answer 11

You can move this question to the closed section and type up a new question in a new thread, if there is another question.

Answer 12

ah this is all I got, but thanks.

Answer 13

I am not at all familiar with the "double false position," at least not by that name. There are helpers here who would be.

Answer 14

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

Answer 15

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

Answer 16

sorry aj +b=c+e

Answer 17

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 :)

Answer 18

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

Answer 19

and i am sorry if there are sign mistakes :(

Answer 20

sweet thnx

Answer 21

What am i pluging a into?

Answer 22

@ghazi

Answer 23

Nm

Answer 24

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

Answer 25

Yeah i got it, tyvm