# How can I solve the system of equations?

How can I solve the system of equations? $$\begin{cases} x^2 y^2+12 x y^3-18 x y-18y^4-4 y^2+27=0,&\\ x^2 y^2-3 x y^3-3 x y+5 y^2=0. \end{cases}$$ I have not any idea to solve.

## 3 Answers

Hint

Subtract the second equation from the first. Use the resulting expression to express $x$ as a function of $y$. Replace $x$ by this expression in the second equation and simplify.

I am sure that you can take from here.

$$\begin{cases} x^2 y^2+12 x y^3-18 x y-18y^4-4 y^2+27=0,&\\ x^2 y^2-3 x y^3-3 x y+5 y^2=0. \end{cases}$$

We subtract the second equation from the first to get $15xy^3-15xy-18y^4-9y^2+27=0$. But $15xy^3-15xy=15xy(y^2-1)$, and $-18y^4-9y^2+27=-9(2y^4+y^2-3)=9(-2y^2-3)(y^2-1)$.

Hence $15xy^3-15xy-18y^4-9y^2+27$$=(y^2-1)(15xy-18y^2-27)=3(y^2-1)(5xy-6y^2-9)$. Thus either $y^2=1$ (hence $y=\pm 1$, or $5xy-6y^2-9=0$.

Case 1: $y=\pm 1$.

If $y=1$, Substitute to the second equation to get $x^2-6x+5=0$, so $x=1$ or $x=5$. If $y=-1$, Substitute to the second equation to get $x^2+6x+5=0$, so $x=-1$ or $x=-5$. We can get these solutions by factoring.

Case 2: $5xy-6y^2-9=0$

So now we proceed to write $x^2 y^2-3 x y^3-3 x y+5 y^2=0$ in terms of $y$, by changing all $xy$ terms into $\frac{6y^2-9}{5}$.

Thus $x^2y^2-3xy^3-3xy+5y^2=(\frac{6y^2-9}{5})^2-(\frac{6y^2-9}{5})y^2-3(\frac{6y^2-9}{5})+5y^2=0$. This is just a quadratic in $y^2$. But then this quadratic is equal to $3(y^2)^2-14(y^2)+108=0$ (after expanding and regrouping), and it has no real roots as its discriminant, $14^2-4(3)(108)$, is negative.

Thus our only solutions are $\boxed{(x,y)=(\pm 1, \pm 5), (\pm 5, \pm1)}$.

After subtracting the second equation from the first equation, factor out the GCF of first two terms and factor the last three terms. You will then be able to factor further, leaving only a quadratic to be solved.