Algebra
posted by Bretagne .
Solve.
x/x+3  3/x3 = x^2+9/x^29
Answer: x = 0?

Plug your 0 into the x in your equation and you get 1 = 1, which is false, which mean: no, the answer is not x = 0.
Since you didn't show your work, I have no idea what you did wrong on your process... or maybe you just guess 0 and didn't do any work at all. But I'll tell you a few steps to approach this problem.
First, you find a common denominator for the 3 fractions (Hint: factor out the x^29). Now they all have the same denominator, you can basically ignore the denominator part and combine all the liketerms on the numerator and then bring "x" on one side and solve for x. Plug the number back into your equation to check and if it works, you know you got the right answer.
*I did it and I found the answer quite tricky, you sure you typed the problem correctly? Anyway, there's still an answer, just a tricky one. 
I did the problem again and still got zero. I give up

Without parentheses your equation is ambiguous. I will assume that what is intended is
x/(x+3)  3(x3) = (x^2+9)/(x^29)
Multiply both sides by (x+3)(x3), which is the same as x^29, and you get
x(x3) 3(x+3) = x^2 +9
x^2 6x 9 = x^2 + 9
6x = 18
x = 3
x=0 is not a solution. You can verify that by substituing x=0 into the original equation. You get 1 = 1, which cannot be true
x= 3 result causes zeros to appear in denominators on both sides of the original equation, so technically speaking, the equation is meaningless there. Nevertheless, the equation tends toward validity as x approaches 3