Largest value

For the first equation, we have $3x^2 - 13x + 14 =0$ .

By Al-Khawarizmi Formula, we have $x_{1,2} = \frac{13\pm1}{6}$ .

For the second equation, we have $6y^2 - 13y + 6 =0$ .

Again, by Al-Khawarizmi Formula, we have $y_{1,2} = \frac{13\pm5}{12}$ .

Now, we required to find the largest possible value of $2xy$ .

Since $\frac{13+1}{6} \space > \space \frac{13-1}{6}$ and $\frac{13+5}{12} \space > \space \frac{13-5}{12}$ , we have

$2xy = 2 \times \frac{14}{6} \times \frac{18}{12}$ ,

Which leaves us, $\boxed{7}$ .

Solve for x and y only. That's the practical approach.