Please simplify

[sharpcolorado]

c = (1+(a/(1-a))) / (1-(a/(1-a)))

c is a constant I know. I am trying to simplify and solve for a.

 

[JeffM]

c = (1+(a/(1-a))) / (1-(a/(1-a)))

c is a constant I know. I am trying to simplify and solve for a.

Please start by reading and responding to the site with the following clickable url:

http://www.freemathhelp.com/forum/forums/34-Administration-Issues

 

[tkhunny]

Common denominators in both the numerator portion and the denominator portion. A simplification should present itself.

 

[HallsofIvy]

It may be easier to see as \(\displaystyle \frac{1+ \frac{a}{1- a}}{1- \frac{a}{1- a}}\)

Now get "common denominators" as tkhunny suggested.

 

[Bob Brown MSEE]

Trick

Set x = a/(1-a)
and solve for x in original
and solve for a in my equation

 

[soroban]

Hello, sharpcolorado!

\(\displaystyle \text{Solve for }a\!:\;c \:=\:\dfrac{1+\frac{a}{1-a}}{1-\frac{a}{1-a}}\)

We have: .\(\displaystyle \dfrac{1+\frac{a}{1-a}}{1-\frac{a}{1-a}} \;=\;c\)

Multiply by \(\displaystyle \frac{1-a}{1-a}:\;\dfrac{(1-a)(1 + \frac{a}{1-a})}{(1-a)(1-\frac{a}{1-a})} \;=\;c \quad\Rightarrow\quad \dfrac{1-a + a}{1-a-a} \:=\:c \)

. . . . . . . . . \(\displaystyle \dfrac{1}{1-2a} \:=\:c \quad\Rightarrow\quad 1 \:=\:c - 2ac \quad\Rightarrow\quad 2ac \:=\:c-1\)

Therefore: .\(\displaystyle a \:=\:\dfrac{c-1}{2c}\)

 

[sharpcolorado]

Hello, sharpcolorado!


We have: .\(\displaystyle \dfrac{1+\frac{a}{1-a}}{1-\frac{a}{1-a}} \;=\;c\)

Multiply by \(\displaystyle \frac{1-a}{1-a}:\;\dfrac{(1-a)(1 + \frac{a}{1-a})}{(1-a)(1-\frac{a}{1-a})} \;=\;c \quad\Rightarrow\quad \dfrac{1-a + a}{1-a-a} \:=\:c \)

. . . . . . . . . \(\displaystyle \dfrac{1}{1-2a} \:=\:c \quad\Rightarrow\quad 1 \:=\:c - 2ac \quad\Rightarrow\quad 2ac \:=\:c-1\)

Therefore: .\(\displaystyle a \:=\:\dfrac{c-1}{2c}\)

Thank you all very much. This was for a geological engineering observation, expressing critical parameters of rock tensile strength to its brittleness.