(5y^2)/(y^2-16)=10/(y-4)+(6y)/(y+4) how would i go about solving this equation?
+118723 $$\begin{array}{rll}
\frac{5y^2}{y^2-16}&=&\frac{10}{y-4}+\frac{6y}{y+4}\\\\
\frac{5y^2}{(y-4)(y+4)}&=&\frac{10}{y-4}+\frac{6y}{y+4}\qquad
\mbox{You should notice that the left denomonator is the diff of 2
squares}\\\\
\end{array}
\\\mbox{The easiest thing to do is to multiply both sides by the lowest common denominator which is (x-4)(x+4)}\\\\
\begin{array}{rll}
\frac{5y^2(y-4)(y+4)}{(y-4)(y+4)}&=&\frac{10(y-4)(y+4)}{y-4}+\frac{6y(y-4)(y+4)}{y+4}\qquad \\\\
5y^2&=&10(y+4)+6y(y-4)\qquad \\\\
\end{array}$$
And you should be able to finish it from there.
If you don't understand or you need more help let us know what your problem is and we will help you more. 
+118723 $$\begin{array}{rll}
\frac{5y^2}{y^2-16}&=&\frac{10}{y-4}+\frac{6y}{y+4}\\\\
\frac{5y^2}{(y-4)(y+4)}&=&\frac{10}{y-4}+\frac{6y}{y+4}\qquad
\mbox{You should notice that the left denomonator is the diff of 2
squares}\\\\
\end{array}
\\\mbox{The easiest thing to do is to multiply both sides by the lowest common denominator which is (x-4)(x+4)}\\\\
\begin{array}{rll}
\frac{5y^2(y-4)(y+4)}{(y-4)(y+4)}&=&\frac{10(y-4)(y+4)}{y-4}+\frac{6y(y-4)(y+4)}{y+4}\qquad \\\\
5y^2&=&10(y+4)+6y(y-4)\qquad \\\\
\end{array}$$
And you should be able to finish it from there.
If you don't understand or you need more help let us know what your problem is and we will help you more.
+118723 A LaTex question. Heureka could you help me please.
I just split that answer into 2 arrays so that I could put the full line comment in.
For this answer that is fine but sometimes it would mean that the equal signs will not line up.
Do you know how to do this without starting a new array?
Thank you.
Melody.
