geeenkie i hate algebra

could someone help me!

heres the problem

Solve the given formula for the variable shown in bold. State the rstrictions if any, for the formula obtained to be meaningful.

1. S = n/2(a+1);a

HELP PLEASE!

come on! help me! or at least tell me if you know how to do it or not!

err don't double post please. kthx.

is n a variable too?

Half of n times (a + 1) or n over 2 times (a + 1)? And what kinda math are you in?

this seems like precalc stuff haha i passed algebra 2 with an a and have never seen any of this

Hmm, is that the formula for the sum of numbers in an arithmetic sequence? Yeah, that looks like precal stuff. Hmm, I don't really get what you're asking though.

all you gotta do is get a by itself by doing the opposite to both sides.. (does that even make sense?)
here:

S = n/2(a+1) multiply by two on the right sides to get rid of the two.

2s=n(a+1) then you divide by n to get rid of the n...

2s/n=a+1 then subtract 1

2s-1/n=a <----and that's your a.

i hope that's what you're asking for. if not, then sorry.

its alegbra 1.....

also thanks ^^ for helping!

the asymtope is at a=-1

taking calculus AP
and look at 1. S = n/2(a+1);a
want to said uhm...how u do it?

whis is there 2 variable n and a?

yeeaaaaaah ive taken alg1 and am currently in alg2, yet i have no idea what youre trying to do there

S = n/2(a+1) [like someone mentioned above...multiply both sides by 2 to cancel it out]
2s = n(a+1) now u have to do the parentheses [n x a + n x 1 (n)]
2s = na + n

i think that's it....

i'm assuming that the (a+1) is in the denominator along with the 2.. if i misinterpreted the problem, please correct me.

Part One: Solving for 'a'
really, this is quite simple, people.

1. Distribute the denominator:
S = n/(2a+2)

2. Multiply both sides by the denominator (2a+2)
S(2a+2) = n

3. Divide both sides by S.
2a+2 = n/S

4. Subtract 2 from both sides.
2a = (n/S) - 2

5. Divide both sides by 2.
a = (n/2s) - 1

Thus... you solved for the variable 'a,' right? this doesn't involve calculus at all, does it? i'm very confused. anyway...

Part Two: Restrictions
Since 2S is the denominator on the right side and the denominator cannot be zero, the set of 'S' would be all real numbers except zero.

S =/= 0

ok...i got my grade back..perplexism was right..thanks guys

topic closed