(6) \( x^{2}-5 x y+6 y^{2}=0 \)

This equation resembles a quadratic in \( x \), which can be analyzed using the quadratic formula. Rearranging gives \( x^{2}-5xy+6y^{2}=0 \). In this case, \( a = 1 \), \( b = -5y \), and \( c = 6y^2 \). By employing the quadratic formula, the values of \( x \) can be derived as \( x = \frac{5y \pm \sqrt{(-5y)^2 - 4(1)(6y^2)}}{2(1)} \), which simplifies to \( x = \frac{5y \pm y}{2} \). Thus, the solutions can be expressed as \( x = 3y \) and \( x = 2y \). Understanding the nature of this equation not only highlights the relationship between \( x \) and \( y \) but also provides a real-world application in optimization problems, where such equations can model constraints in areas like economics or engineering. By knowing how to solve for \( x \) in terms of \( y \), one can optimize resources effectively, such as maximizing output while minimizing costs.