#1

Can you help? I'm working on a computer project, and it's been a while since algebra.

A,B, and C are constants, so I don't want to calculate this every time x changes. I want to get something like x*Q, where Q is a constant that won't eat up the processor.

(x(A-1)-B(A-1))/C + 1

A,B, and C are constants, so I don't want to calculate this every time x changes. I want to get something like x*Q, where Q is a constant that won't eat up the processor.

(x(A-1)-B(A-1))/C + 1

#2

If you need help it should be called LAWLGEBRA Honestly though. You know what you're doing apply it

#3

Calculate out the constants could be...

x * ((A-1)/C) - ((BA-B)/C) + 1

x * ((A-1)/C) - ((BA-B)/C) + 1

#4

x = (-C + B(A-1)) / (A-1)

Voila.

So how exactly does this help you with algebra?

Voila.

So how exactly does this help you with algebra?

#5

Hey, thanks alot!

#6

I have an algebra exam tomorow .. been studying for it all day

now I'm on UG, forgetting every single thing I learned

now I'm on UG, forgetting every single thing I learned

#7

x = (-C + B(A-1)) / (A-1)

Voila.

So how exactly does this help you with algebra?

Er, where'd you get that = from?

(x(A-1)-B(A-1))/C + 1

(x(A-1)/C) - (B(A-1)/C) + 1

x * ((A-1)/C) - (B(A-1)/C) + 1

Not quite there yet. This won't work in your x*Q plan because then you'd end up with

Q = ((A-1)/C) - (B(A-1)/C) + 1

x*Q = x * (((A-1)/C) - (B(A-1)/C) + 1)

Which is something other than x * ((A-1)/C) - (B(A-1)/C) + 1.

x * ((A-1)/C) - (B(A-1)/C) + 1

x * ((A-1)/C) - ((B(A-1)/C) - 1) - divide by ((A-1)/C)

x - (((B(A-1)/C) - 1))/((A-1)/C)

So you can do x-Q if Q = (((B(A-1)/C) - 1))/((A-1)/C). I hope.

#8

Dang, anyone with a guitar and a problem can ask a question about anything here and get help, thanks!

So I got that same answer and then tested it. The program didn't work, so I simplified the answer to x-Q=B+C/(A-1)... wait, scratch everything. I even said in the beginning "xQ=(blah blah)". Why were we saying Q=(junk) when xQ=(junk). Thanks for helping me see that... Ok, got it... thanks again...

So I got that same answer and then tested it. The program didn't work, so I simplified the answer to x-Q=B+C/(A-1)... wait, scratch everything. I even said in the beginning "xQ=(blah blah)". Why were we saying Q=(junk) when xQ=(junk). Thanks for helping me see that... Ok, got it... thanks again...

#9

Wait, I spoke too soon. Now it's set up right, and I "absorbed" that one 1 into A. So, can you solve for Q and eliminate the x? If it's not possible with that 1 their, you can eliminate it.

xQ=A(x-B)/C+1.

xQ=A(x-B)/C+1.