<PRE><B>Question 1040622</B>
(x-8)/(x-5)+30/(x^2-25)=0
x^2-25=(x+5)(x-5)
common denominator for the left side is (x+5)(x-5).
Multiply the first term by (x+5); so (x-8)(x+5)=x^2-3x-40
The second term is unchanged, and their sum is x^-3x-40+30=x^2-3x-10.
That is all over (x+5)(x-5)
x^2-3x-10=(x-5)(x+2)
Therefore (x+5)(x+2)/(x+5)(x-5)= (x+2)/x-5), since (x+5) cancels.
(x+2)/(x-5)=0
That occurs when x=-2. You can multiply both sides of the equation by the denominator (x-5). That gets rid of the denominator on the left, and 0*(x-5)=0
-----------------------
Check
(-10/-7)+(30/-21)=0
(10/7)-(10/7)=0, after reducing the second term.
</PRE>