Melody

Thanks Heureka    

Yep - My turn to make it look difficult.      LOL 

@@ End of Day Wrap  Thurs 2/7/15   Sydney, Australia Time   9:20pm  ♪ ♫

Good evening,     

Today our exceptional answers were provided by Sir-Emo-Chappington CPhill, Alan Radix, WizBoy, Mellie, Heureka, Oscarcampos02, MathsGod1 and TitaniumRome.   Thank you  

$$\begin{array}{rll}
\frac{2}{x^2-5x+6}+\frac{5}{x^2-4x+3}&=&\frac{6}{x^2-3x+2}\\\\
\frac{2}{(x-3)(x-2)}+\frac{5}{(x-3)(x-1)}&=&\frac{6}{(x-2)(x-1)}\\\\
(x-3)(x-2)(x-1)\left(\frac{2}{(x-3)(x-2)}+\frac{5}{(x-3)(x-1)}\right)&=&(x-3)(x-2)(x-1)\left(\frac{6}{(x-2)(x-1)}\right)\\\\
2(x-1)+5(x-2)&=&6(x-3)\\\\
2x-2+5x-10&=&6x-18\\\\
7x-12&=&6x-18\\\\
1x&=&-6\\\\
x&=&-6
\end{array}$$

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

Fri 3/7/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts:

FTJ means 'For the juniors' 

Your question does not make sense.

Perhaps you could rephrase it?

You are on a roll today Chris.  Your logic is irrefutable      

(That's what I think anyway)      LOL

That is great logic Chris.  

It is so simple and so logical but I would not have thought of it   

That is really good logic I think Chris    

I forgot the 'divide by 49' one when i did it.                

Thanks Chris,

I am going to do it too    

$$\\40^8\times75^{18}\\\\
=10^8\times 4^8\times (\frac{3}{4}\times 100)^{18}\\\\
=10^8\times 4^8\times \frac{3^{18}}{4^{18}}\times 100^{18}\\\\
=10^8\times \frac{3^{18}}{4^{10}}\times 100^{10}\times 100^{8}\\\\
=10^8\times 100^{8}\times 3^{18}\times 25^{10}\\\\
=10^8\times 10^{16}\times 3^{18}\times 25^{8}\\\\
=10^{24}\times 3^{18}\times 25^{8}\\\\
$25 to the 8 will end in a 5$\\
$3 to the power of anything does not end in a 0 $\\
$The last digits of powers of 3 are 3,9,7,1,3,9,$\\
$There are 4 numbers in the pattern. 18=2mod4$\\
$Hence 3 to the 18 ends in a 9$\\
so\\
3^{18}\times 25^{8} $ will end in a 5$\\\\$$

So there will be 24 trailing zeros - just like CPhill said