# Largest possible value of $x$

Problem:

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

How would I do this? Would I multiply each side by $2y-1$?

• That might be a good start. Do you know about Lagrange multipliers? – Harald Hanche-Olsen May 7 '17 at 19:52
• Never heard of them – JenkinsMa May 7 '17 at 19:54
• Okay, then a direct method is called for. The answer by dxiv looks good. – Harald Hanche-Olsen May 7 '17 at 19:58

Hint: eliminate the denominator and consider it as a quadratic equation in $y\,$: $$x^2 \cdot y^2 - 6x \cdot y +3x -1 = 0$$
$$\frac{1}{4} \Delta = 9 x^2 -x^2(3x -1) = x^2(10 - 3x) \ge 0$$
• indeed, it is also difficult to tell when $x$ can be written as a function of $y$. Thanks! – qbert May 7 '17 at 20:05