Math Is Fun Forum

benmilesbc

Member

Registered: 2007-05-11

Posts: 2

Tricky equation!

Hello guys, first post here.

I am trying to solve a tricky equation for a friend's business and have got so far with the problem.

The problem as given to me is as follows: Make S the subject of:

S = ( C+(CJ-C)+R+(RJ-R)+( L(S+Q+U+K)) ) – ((SJ-S)+U+((CJ-C) – (SJ-S)))

Tidying this up

S = (C + R + (CJ-C) + (RJ-R) + L(S+Q+U+K)) - (U + (SJ-S) + ((CJ-C) - (SJ-S)))

This is where I start to stuggle and my Maths A - level starts to betray me - getting rid of the brackets. I think a muinus outside the bracket changes the signs inside? So:

S = (C + R + CJ - C + RJ - R + LS + LQ + LU + LK) - (U + SJ - S + CJ - C - SJ + S)

S = C + R + CJ - C + RJ - R + LS + LQ+ LU + LK - U - SJ + S -CJ +C +SJ -S

Simplifying (+C removes -C and +R removes - R, -S removes +S?)

S = CJ + RJ + LS + LQ + LU + LK - U - SJ -CJ + C + SJ

SJs also cancel so

S = CJ + RJ + LS + LQ + LU + LK - U - CJ + C

S / LS = CJ + RJ + LQ + LU + LK - U - CJ + C

The question is - am I right so far, and how do I finish this and make S the subject. Any Maths genius able to help me?

Many Thanks

Ben

Last edited by benmilesbc (2007-05-11 03:37:42)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Tricky equation!

You're right so far, apart from the very last step. Instead of dividing by LS, you take it away.

You had:
S = CJ + RJ + LS + LQ + LU + LK - U - CJ + C

Cancel the CJs and subtract LS:
S - LS = RJ + LQ + LU + LK - U + C

Factorise the left side:
S(1-L) = RJ + LQ + LU + LK - U + C

Divide by (1-L):
S = (RJ + LQ + LU + LK - U + C)/(1-L)

Done!

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Why did the vector cross the road?
It wanted to be normal.

benmilesbc

Member

Registered: 2007-05-11

Posts: 2

Re: Tricky equation!

Thanks for that but urg, the problem is a lot more complicated than i Thought, and I had screwed up previously making the above totally wrong lol. Ill post it in its total:

S = (C+V+R+T+A) – (Q+U+K)

But

A = (L(S+Q+U) – (S+Q+U))
V = (CJ-C)
T = (RJ-R)
Q = (SJ-S)
K = (CJ-C) – (SJ-S)

So

S = ( C+V+R+T+A ) – (Q+U+K)

S = ( C+(CJ-C)+R+(RJ-R)+ (L(S+Q+U) – (S+Q+U)) ) – ((SJ-S)+U+((CJ-C) – (SJ-S)))

S = ( C+(CJ-C)+R+(RJ-R)+ (L(S+(SJ-S)+U) – (S+(SJ-S)+U))) – ((SJ-S)+U+((CJ-C) – (SJ-S)))

Tidying this up:

S = ( C+ R+(CJ-C)+(RJ-R)+ (L(S+(SJ-S)+U) – (S+(SJ-S)+U))) – (U+(SJ-S)+ ((CJ-C) – (SJ-S)))

Removing Brackets ( – outside the brackets changes the signs inside)

S = ( C+ R+CJ-C+RJ-R+(LS+L(SJ-S)+LU) – (S+SJ-S+U))) – (U+SJ-S+CJ-C–SJ+S)
S = ( C+ R+CJ-C+RJ-R+(LS+LSJ-LS+LU) – S-(SJ-S)-U)) – (U+SJ-S+CJ-C–SJ+S)
S = ( C+ R+CJ-C+RJ-R+LS+LSJ-LS+LU – S-SJ+S-U) – (U+SJ-S+CJ-C–SJ+S)
S = C+ R+CJ-C+RJ-R+LS+LSJ-LS+LU – S-SJ+S-U – U-SJ+S-CJ+C+SJ-S

Simplify This ( i think this is maybe where i screwed up if not before)

S = RJ+LSJ+LU-SJ-2U+C

Subtract LSJ and add SJ to each side

S –LSJ+SJ = RJ+LU-2U+C

Factorise the left side

S( –LJ+J) = RJ+LU-2U+C

Divide by ( –LJ+J)

S = (RJ+LU-2U+C) / ( –LJ+J)

Trouble is, it nowhere near works when using actual figures, so my equation must be wrong

Any help would be much appreciated!

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Tricky equation!

You're very nearly right. The mistake is when you factorised, just before the end.

S - LSJ + SJ = S(1 - LJ + J). You just forgot to put that 1 in there.

Following that through gives you a final equation of S = (RJ+LU-2U+C) / (1–LJ+J).

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Why did the vector cross the road?
It wanted to be normal.