Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to sove the system of equations $$\begin{cases} x^3 y-y^4=7,\\ x^2 y+2 xy^2+y^3=9. \end{cases} $$ I tried divide these two equations we obtain $$\dfrac{x^3 - y^3}{(x+y)^2 } = \dfrac{7}{9}$$ From here, I don't know how to solve.

share|cite|improve this question
Here's one try: $x^3-y^3=(x-y)(x^2+xy+y^2)$. – Gyu Eun Lee Dec 20 '12 at 4:40
up vote 6 down vote accepted

We'll prove that $(2,1)$ is the only real solution. First note that if $y = 0$, the second equation is unsolvable, so $y\neq 0$.

Solving the second equation for $x$ via the quadratic formula gives \begin{align*} x &= \frac{-2y^2 \pm \sqrt{4y^4-4y(y^3-9)}}{2y}\\ &= \frac{-y^2\pm 3\sqrt{y}}{y} \\ &= -y \pm y^{-\frac{1}{2}}.\end{align*}

In particular, $y \geq 0$. Subbing this into the first equation gives \begin{align*}7 &= \left(-y \pm 3y^{-\frac{1}{2}} \right)^3 y - y^4 \\ &= -2y^4\pm 9y^{\frac{5}{2}}-27y\pm 27y^{-\frac{1}{2}}.\end{align*}

This is equivalent to $$ 0 = -2y^{\frac{9}{2}} \pm 9y^{\frac{6}{2}}-27y^{\frac{3}{2}} -7y^\frac{1}{2} \pm 27.$$

Now, substitute in $z = \sqrt{y}$, which is possible since we already know $y\geq 0$. This gives $$0 = -2z^9 \pm 9z^6 - 27z^3 -7z \pm 27$$

If we choose the $-$ sign, then clearly there is no solution with $z$ real, so we must have $$0 = -2z^9 + 9z^6 - 27z^3-7z+27.$$

Now, using the rational roots theorem, maple, direct inspection, etc, we see $z=1$ is a solution.

I claim that $f(z)-2z^9 + 9z^6 - 27z^3 - 7z + 27$ is always decreasing, so $z=1$ is the only solution.

To see this, compute the derivative, getting $f'(z)=-18z^8 + 54z^5-81z^2 - 7 = 9(-2z^8 + 6z^5-9z^2 - \frac{7}{9})$. I claim this is always negative. The $9$ out in front doesn't effect this, so we'll ignore it in intermediate computations.

Why is $f'(z)$ always negative? Lets find its maximum. So, take another derivative, getting $f''(z)-16z^7+30z^4-18z = 2z(-8z^6 + 15z^3 - 9)$. Of course, this is $0$ when $z=0$, but the other part is quadratic in $z^3$ and so we see that it's never $0$. Since $f'(z)$ goes to $-\infty$ as $x\rightarrow \pm \infty$, it follows that if it only has one critical point, this must be a max. So, the max of $f'(z)$ occurs when $z = 0$, where $f'(0) = -7 < 0$.

Since $f'(z) < 0$, $f(z)$ is always decreasing, so $z= 1$ must be its only solution.

When $z=1$, $\sqrt{y} = 1$ so $y=1$, and subbing back in shows that $x=2$.

share|cite|improve this answer
Please check your solution at "Solving the second equation for x via the quadratic formula gives..." – minthao_2011 Dec 20 '12 at 9:34
@I'm not sure what you're referring to, it still looks ok to me. – Jason DeVito Dec 20 '12 at 14:47
Note that, $$\sqrt{4y^4 -4y(y^3 -9)}=6\sqrt{y}$$. – minthao_2011 Dec 20 '12 at 15:07
You're right!! Let me see how this messes with everything... – Jason DeVito Dec 20 '12 at 15:17
It took forever, but it's fixed (I think). – Jason DeVito Dec 20 '12 at 16:30

Guess and check works: $x=2$, $y=1$. (I tried forcing $x+y$ to be 3 and that's what I got.)

For a more general method for finding all solutions, I suppose applying the appropriate root-finding algorithm to approximate the solutions could work.

share|cite|improve this answer
Well, that's one solution. Maybe there are others? – Gerry Myerson Dec 20 '12 at 4:37
Alpha finds only that real root. – Ross Millikan Dec 20 '12 at 4:48
Huh. Guess I got pretty lucky then. – Gyu Eun Lee Dec 20 '12 at 4:50
Depends on where the problem comes from. Class problems often have solutions you can find that way. The rational root theorem is a great example-it works better in class than in real life. This looks like one that was supposed to be solved the way you did. Looking for small integer solutions is often productive. – Ross Millikan Dec 20 '12 at 5:47

$$\begin{cases} x^3 y-y^4=7,\\ x^2 y+2 xy^2+y^3=9. \end{cases}$$

$x^3y-y^4=y(x^3-y^3)$, while $x^2y+2xy^2+y^3=y(x+y)^3$.

We try the transformation $x+y=z$.

Then $y(x^3-y^3)=y(x-y)(x^2+xy+y^2)=y(z-2y)(z^2-xy)=y(z-2y)(z^2-(z-y)y)=$ $-2y^4+3y^3z-3y^2z^2+yz^3=7$ On the other hand, the second equation gives $yz^2=9$, so $y=\frac{9}{z^2}$. We expand and then notice that it is actually equal to $9(z^3-18)(z^6-9z^3+81)=7z^8$. Then we get $z=3$ as a root.

The problem is that I do not know how do I show that there are no other roots, but probably someone who knows calculus can be able to resolve this? or probably by a weird factoring?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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