Algebra equation? My question is how to solve this algebra equation.
$$\frac8{y-2}-\frac{13}2=\frac3{2y-4}$$
Can anyone help me solve this.The issue is that I am stuck. 
 A: Assuming that the equation is as I interpreted it in the edit, you can start by putting everything over a common denominator. Since $2y-4=2(y-2)$, the righthand side already has the least common denominator of the lefthand side. Now $$\frac8{y-2}-\frac{13}2=\frac8{y-2}\cdot\frac22-\frac{13}2\cdot\frac{y-2}{y-2}=\frac{16}{2(y-2)}-\frac{13(y-2)}{2(y-2)}=\frac{16-13(y-2)}{2y-4}\;,$$ so you’re really solving $$\frac{16-13(y-2)}{2y-4}=\frac3{2y-4}\;.$$ Assuming that $y\ne 2$, you can multiply both sides by $2y-4$ to get $$16-13(y-2)=3\;,\tag{1}$$ which I expect you can solve pretty easily.
Note that if you were to get $y=2$ as a solution to $(1)$, you’d have to throw it out: it’s excluded by the denominators $y-2$ and $2y-4$ in the original equation, since they can’t be $0$. The step of multiplying through by $2y-4$ is legitimate only if you’re not multiplying by $0$.
Note also that if you can see right away that each of the three denominators is a factor of $2y-4$, you can simply multiply both sides by $2y-4$ directly. I inserted the extra steps at the beginning because they explain just why you would choose that multiplier.
A: I think your problem is how to get the varialbe $y$, "on top", right? 
EDIT:As is mentioned in the comments below, notice that $y\not = 2$ since $y-2$ appears in the denominator. $y\not =2$ because otherwise you would be dividing by $0$ which is not defined. Also, $2y-4$ is also in the denominator, so this implies $2y\not = 4$. Dividing both sides by $2$ and we again get that $y\not =2$.
HINT: factor $2y-4$ as $2(y-2)$. Then you can multiply both sides by $(y-2)$
$$\frac8{y-2}-\frac{13}2=\frac3{2y-4}$$
$$\frac{8(y-2)}{y-2}-\frac{13(y-2)}2=\frac{3(y-2)}{2(y-2)}$$
after cancelling terms you get
$$8-\frac{13(y-2)}2=\frac{3}{2}$$
Ill let you do the rest
