Posted by **Mike** on Tuesday, May 29, 2007 at 12:33pm.

Can someone please show me how to solve these for me?

5/(y-2)=y+2

(x/3)-(2/3)=(1/y-5)

(y+2)/7=1(y-5)

(x^2)/(x-4)-7/(x-4)=0

x^2/(x+3)-5/(x+3)=0

I assume that your second equation has y as a typo. I will use that as an example.

(x/3)-(2/3)=(1/x-5)

Multiply both sides by 3. I will also assume that (1/x-5) is the same as (1/x)-5 rather than 1/(x-5).

x-2=3/x-15

Get the x terms on one side and the numerical terms on the other by adding 2 and subtracting 3/x from both sides.

x-3/x=17

Multiply both sides by x.

x^2-3=17x

x^2-17x-3=0

Unfortunately, I cannot carry the process further. However, it should make the process clear.

I hope this helps. Thanks for asking.

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