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!

  • $\begingroup$ 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. $\endgroup$ – bubba Mar 2 '14 at 13:53
  • 1
    $\begingroup$ 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. $\endgroup$ – wkyniston Mar 2 '14 at 13:54
  • $\begingroup$ bubba, I must have left the - sign out of the original equation. fixed it. $\endgroup$ – 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.

In some cases, there is a clever easy solution that avoids the messy general approach, but I don't see one here.

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.

  • $\begingroup$ I have taken up to and including precalc. This is merely a refresher, but I admit, I was thrown by seeing this problem and others like it in a precalc textbook. In fact its in the second chapter with no examples for solving. Would it make a difference to say I just want to isolate the x on one side (solve for x)? $\endgroup$ – wkyniston Mar 2 '14 at 14:12
  • $\begingroup$ Thanks for the link on cubic equations bubba. At least I understand why I haven't been able to make any progress on this equation. I agree with you it is currently above my level. So I accepted your answer. I'm going to assume this is a typo in the text. The other equations I referred to ended up not being as hard as I thought given they were not cubes. Thanks! $\endgroup$ – wkyniston Mar 2 '14 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.