Let $x$ and $y$ be real numbers satisfying $\frac{x^2y^2 - 1}{2y-1}=3x.$...

Question

Let $x$ and $y$ be real numbers satisfying $\frac{x^2y^2 - 1}{2y-1}=3x.$ Find the largest possible value of $x.$

I'm not sure how to do this question. Any help is greatly appreciated!

Answer 1

Since $\frac{x^2y^2 - 1}{2y-1}=3x$,

\begin{align*} x^2y^2-1&=6xy-3x\\ x^2y^2-6xy+(3x-1)&=0 \end{align*}

As $y$ is real, the discriminant of the above equation (in $y$) is non-negative

\begin{align*} (-6x)^2-4(x^2)(3x-1)&\ge0\\ 4x^2(9-3x+1)&\ge0\\ x&\le\frac{10}{3} \end{align*}

The largest possible value of $x$ is $\frac{10}{3}$.