[RESOLVED] Reaaranging expression

[.paul.]

Can someone have a look through my workings and tell me whether it's correct?



Thanks for your help

[jemidiah]

You're correct, and Wolfram Alpha agrees.

[.paul.]

Wolfram Alpha?

It's a question from my End of (Maths) Module exam. I submitted it last month and I don't get feedback. I was proud of that and I'm glad I was right to be proud of it...

edit: I found the link...

[.paul.]

Thanks for checking it.

[.paul.]

Actually Jemidiah, there's a difference between my solution and the solution in Wolfram Alpha.

I make it b = -((12c - 6a) / (3 - c))

As opposed to b = -((6a - 12c) / (c - 3))

They give equal results?

[jemidiah]

They're not symbol-for-symbol the same, but yes, they're equal, since 3-c is -(c-3), and 12c-6a is -(6a-12c). I had checked to make sure. You can also have Wolfram Alpha check for symbolic equality, see here.

[krishnanayak]

Prewriting is hard. Good work. Way to go!