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.

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I am sure how to format it but this is what I mean (8/y-2)-(13/2)=(3/2y-4) –  James Del Rio Apr 28 '12 at 18:41
Do you mean the way I edited it, or $$\frac8y-2-\frac{13}2=\text{something}\;?$$ –  Brian M. Scott Apr 28 '12 at 18:44
Hint: Multiply both sides by $(y-2)$. Or, if the problem is correctly written the way Brian put it in the comments instead of in the post above, multiply both sides by $y$. –  Brett Frankel Apr 28 '12 at 18:46
I like how Brian edit it the first time on my original post. –  James Del Rio Apr 28 '12 at 18:46
I did that the problem is on 13/2 part I get y^2 –  James Del Rio Apr 28 '12 at 18:46

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.

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thanks for an excellent reponse –  James Del Rio Apr 28 '12 at 18:57

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

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Note that the condition $y\neq 2$ should be stated after multyplying by $y-2$. –  Pedro Tamaroff Apr 28 '12 at 18:52
@PeterTamaroff Since $y-2$ appears in the denominator of the original problem, it is implicit that $y\neq 2$ from the beginning. –  Brett Frankel Apr 28 '12 at 18:55
@Brett: It’s nevertheless good practice for beginners to be aware of restrictions on the manipulations that they perform, and this particular restriction is often forgotten. –  Brian M. Scott Apr 28 '12 at 18:57
@PeterTamaroff, thanks for the comment, I should be more careful with this problem. –  Edison Apr 28 '12 at 19:17