Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

its from chapter 7 in the nine chapters
http://www-history.mcs.st-and.ac.uk/HistTopics/Nine_chapters.html 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 varify...lol
\[x=\frac{ c-b }{ a }\] 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?
Nm
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question