# Simple algebra question - pulling my hair out

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• Jun 17th 2010, 06:32 PM
Siddy
Simple algebra question - pulling my hair out
Hi,
I have a formula i have to use, depending on the data give in the exam i will have to re-arrange it before using it. Problem is i just cant get it work. I my notes i didnt use a lot of detail and i cant remember how i did it in the class, i thought it was so simple at the time, but now i just cant do it! My lecture says its too easy to need worked solutions, so all i have is the answers, not worked solutions.

f = c ( s - a ) / s ( c - a )

If the question ask for "f", obviously i can do it, but sometimes it asked for "s".
When i try to rearrange, i end up getting "s/s" at some point and the formula explodes in my face.

eg:
c= 0.98
a= 0.89
f= 0.45
s= ? (answer sheet says 0.928)

if you are interested the forumla is for calculating the fraction crystallinity in a polymer sample, from densities, its simple linear relationship.
• Jun 17th 2010, 06:45 PM
undefined
Quote:

Originally Posted by Siddy
Hi,
I have a formula i have to use, depending on the data give in the exam i will have to re-arrange it before using it. Problem is i just cant get it work. I my notes i didnt use a lot of detail and i cant remember how i did it in the class, i thought it was so simple at the time, but now i just cant do it! My lecture says its too easy to need worked solutions, so all i have is the answers, not worked solutions.

f = c ( s - a ) / s ( c - a )

If the question ask for "f", obviously i can do it, but sometimes it asked for "s".
When i try to rearrange, i end up getting "s/s" at some point and the formula explodes in my face.

eg:
c= 0.98
a= 0.89
f= 0.45
s= ? (answer sheet says 0.928)

if you are interested the forumla is for calculating the fraction crystallinity in a polymer sample, from densities, its simple linear relationship.

Attachment 17893

The LaTeX:

Code:

\\ f=\frac{c(s-a)}{s(c-a)}\\ \\ fs(c-a)=c(s-a)\\ \\ s(fc-fa)=cs-ca\\ \\ s(fc-fa-c)=-ca\\ \\ s=\frac{-ca}{fc-fa-c}
• Jun 17th 2010, 07:17 PM
Siddy
ありがとう ございます!
• Jun 17th 2010, 09:17 PM
Wilmer
Quote:

Originally Posted by Siddy
f = c ( s - a ) / s ( c - a )

You need another set of brackets! f = c ( s - a ) / (s ( c - a ))

f = (cs - ca) / (cs - as)
f(cs - as) = cs - ca
Can you wrap it up now?
• Jun 18th 2010, 02:09 AM
Siddy

1/T = (x /a) + (1-x/b)

eg.
1/288 = (x/373) + (1-x/203)

from trial and error, i get x = 0.65 (and 1-x 0.35). But how do i do it algebraically?
• Jun 18th 2010, 02:33 AM
sa-ri-ga-ma
Quote:

Originally Posted by Siddy

1/T = (x /a) + (1-x/b)

eg.
1/288 = (x/373) + (1-x/203)

from trial and error, i get x = 0.65 (and 1-x 0.35). But how do i do it algebraically?

1/288 = (x/373) + 1/203 - x/203

x/203 - x/373 = 1/203 - 1/288

x*( 1/203 - 1/373 ) = 1/203 - 1/288

Now can you solve for x?
• Jun 18th 2010, 02:44 AM
Siddy
Thank you very much, i can do it now!

so i have to separate the x and then factorise, thanks!
• Jun 18th 2010, 06:05 AM
undefined
Sorry about the broken link. Fixed my post above. (The site is here.)
• Jun 20th 2010, 05:54 PM
Siddy
http://latex.codecogs.com/gif.latex?...{-ca}{fc-fa-c}

I dont understand how to get from step 3 to step 4.
You subtract "cs" from both sides?
Shouldnt the left side;
http://latex.codecogs.com/gif.latex?...%20cs%20=-ca\\
• Jun 20th 2010, 06:01 PM
undefined
Quote:

Originally Posted by Siddy
http://latex.codecogs.com/gif.latex?...D%7Bfc-fa-c%7D

I dont understand how to get from step 3 to step 4.
You subtract "cs" from both sides?
Shouldnt the left side;
http://latex.codecogs.com/gif.latex?...s%20=-ca%5C%5C

LaTeX is working on the forum again by the way, using  tags.

You could say I skipped a step. Notice that

$\displaystyle s(fc-fa)-cs = s(fc-fa-c)$
• Jun 20th 2010, 06:30 PM
Siddy
I thought it would be this:

$\displaystyle s(fc - fa) - cs = s(fc - fa - c - 1)$
• Jun 20th 2010, 06:36 PM
undefined
Quote:

Originally Posted by Siddy
I thought it would be this:

$\displaystyle s(fc - fa) - cs = s(fc - fa - c - 1)$

Well no, what you have on the RHS corresponds with $\displaystyle s(fc-fa)-cs-s$.

Maybe this will clarify:

$\displaystyle s(fc - fa) - cs = s\bigg((fc - fa)-c\bigg) = s(fc-fa-c)$
• Jun 20th 2010, 06:42 PM
Siddy

I don't understand where the "s" in "cs" go?
• Jun 20th 2010, 06:44 PM
undefined
Quote:

Originally Posted by Siddy
$\displaystyle s(fc - fa) - cs = sfc -sfa -sc = s(fc-fa-c)$