$$\eqalign{ & 11C_{2x}^x = 6C_{2x + 1}^{x + 1} \Leftrightarrow 11 \cdot \frac{{\left( {2x} \right)!}}{{x!x!}} = 6 \cdot \frac{{\left( {2x + 1} \right)!}}{{x!\left( {x + 1} \right)!}} \Leftrightarrow 11 \cdot \frac{{\left( {2x} \right)!}}{{x!}} = 6 \cdot \frac{{\left( {2x + 1} \right)!}}{{\left( {x + 1} \right)!}} \Leftrightarrow \cr & 11 \cdot \frac{1}{{x!}} = 6 \cdot \frac{{\left( {2x + 1} \right)}}{{\left( {x + 1} \right)!}} \Leftrightarrow 11 \cdot \frac{{\left( {x + 1} \right)!}}{{x!}} = 6 \cdot \left( {2x + 1} \right) \Leftrightarrow \cr & 11 \cdot \left( {x + 1} \right) = 6 \cdot \left( {2x + 1} \right) \Leftrightarrow 11x + 11 = 12x + 6 \Leftrightarrow x = 5 \cr} $$

$%x = 5$%