solving a proportion

To "b" or not to "b"......that is the question......is it nobler in the minds of men to expandeth answers to infinity through the Queen's and heureka's employment of LaTex, or, should other - more compact - answers serve our  purposes just as well ????

Do you mean

$$\begin{array}{rll} \frac{3}{3b+4}&=&\frac{2}{b-4}\\\\ (b-4)(3b+4)\times\frac{3}{3b+4}&=&\frac{2}{b-4}\times(b-4)(3b+4)\\\\ (b-4)\times\frac{3}{1}&=&\frac{2}{1}\times(3b+4)\\\\ 3(b-4)&=&2(3b+4)\\\\ 3b-12&=&6b+8\\\\ 3b-12-3b&=&6b+8-3b\\\\ -12&=&8+3b\\\\ -12-8&=&8+3b-8\\\\ -20&=&3b\\\\ b&=&\frac{-20}{3}\\\\ b&=&-6\frac{2}{3}\\\\ \end{array}$$

check

LHS =     $${\frac{{\mathtt{3}}}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{20}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}} = {\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{16}}}} = -{\mathtt{0.187\: \!5}}$$

RHS =       $${\frac{{\mathtt{2}}}{\left({\mathtt{\,-\,}}\left({\frac{{\mathtt{20}}}{{\mathtt{3}}}}\right){\mathtt{\,-\,}}{\mathtt{4}}\right)}} = {\mathtt{\,-\,}}{\frac{{\mathtt{3}}}{{\mathtt{16}}}} = -{\mathtt{0.187\: \!5}}$$

what is b in

  3            2

'''''''''''''=''''''''''

3b+4      b - 4

$$\small{ \begin{array}{rll} \dfrac{3}{3b+4} &=& \dfrac{2}{b-4} \\\\ \dfrac{3b+4}{3} &=& \dfrac{b-4}{2} \\\\ b + \dfrac43 &=& \dfrac{b}{2} - 2 \\\\ \dfrac{b}{2} &=& - 2 - \dfrac43 \\\\ \dfrac{b}{2} &=& -\dfrac{10}3 \\\\ b &=& -\dfrac{20}{3}\\\\ b &=& -6\dfrac{2}{3}\\\\ \end{array} }$$

Thanks Heureka    

Yep - My turn to make it look difficult.      LOL 

No.. Heureka and I have got that one nailed!

You should try learning Latex.  It is like everything, the more you know the easier it is to add to your knowledge.  

Getting started is the hardest bit but then it becomes easy.  And you can start with really trivial things here so even getting started is not so hard. :)

It is much easier to makes sense of well presented material - and that usually means LaTex.