# How to solve $y=-x^3/(x^2-9)$ for $x$

This is not a homework question. I am doing a independent study refresher on precalc prior to taking calculus. Wolfram gives an unbelievably long series of steps with techniques I have not even heard of. Yet this is a problem out of a precalc book. Here is what I have so far:

1) Multiply both sides by $(x^2 - 9)$- (factoring the denominator doesn't seem to help): $y(x^2-9)= -x^3$

2) Distribute the $y$ on the left side: $x^2y-9y=-x^3$

3) Move $y$'s to right-side and $x$'s to left, by adding $x^2$ to both sides and subtracting $9y$ from both sides: $x^2y+x^3=9y$

4) Now factor $x^2$ out of the terms on left side: $x^2(y+x)=9y$

And this is as far as I can get. Can someone show the steps and if I am correct up to this point? This is my first post in Mathematics so please point out any errors in my question. Thank You!

• The first step is wrong. Should be $y(x^2-9) = x^3$. But fixing that doesn't help much -- you're still left with a difficult problem. – bubba Mar 2 '14 at 13:53
• Thanks to whoever formatted my question so it's readable. I have taken note of the edits and will format properly with any future questions. – wkyniston Mar 2 '14 at 13:54
• bubba, I must have left the - sign out of the original equation. fixed it. – wkyniston Mar 2 '14 at 13:58

After some algebra, you get $$x^3 + yx^2 - 9y = 0$$ This is a cubic equation in $x$ (which just means that its highest power of $x$ is $x$-cubed). These kinds of equations can be solved, but the process is difficult, in general, which is why Wolfram gave you a huge mess. See here for details.
If you are at the pre-calculus level, I'd say that this problem is much too hard for you. My guess is that book has a typo. If the $x^3$ were $x^2$ it would be a lot more reasonable.