Given the conditions $x > y > 0$ and the equation:

$6x^2 - 5xy - 4y^2 = 0,$

we need to find the value of the fraction:

$\frac{x - y}{x + y}.$

Step-by-Step Solution

1. Rewrite the quadratic equation in terms of a single variable:

   The given equation is:

   $6x^2 - 5xy - 4y^2 = 0.$

   Let's treat this as a quadratic equation in $x$:

   $6x^2 - 5xy - 4y^2 = 0.$

   The standard form of a quadratic equation $ax^2 + bx + c = 0$ has solutions given by:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

   Here, $a = 6$, $b = -5y$, and $c = -4y^2$. Substituting these values gives:

   $x = \frac{-(-5y) \pm \sqrt{(-5y)^2 - 4 \cdot 6 \cdot (-4y^2)}}{2 \cdot 6}.$

   Simplify inside the square root:

   $x = \frac{5y \pm \sqrt{25y^2 + 96y^2}}{12}.$

   Combine like terms:

   $x = \frac{5y \pm \sqrt{121y^2}}{12}.$

   Since $\sqrt{121y^2} = 11y$, we have:

   $x = \frac{5y \pm 11y}{12}.$

2. Find the possible values of $x$:

   There are two possible values for $x$:

   $x_1 = \frac{5y + 11y}{12} = \frac{16y}{12} = \frac{4y}{3},$ $x_2 = \frac{5y - 11y}{12} = \frac{-6y}{12} = -\frac{y}{2}.$

   Since $x > y > 0$, $x$ must be positive. Therefore, we discard $x_2 = -\frac{y}{2}$ because it is negative.

   So, we have:

   $x = \frac{4y}{3}.$

3. Compute the fraction $\frac{x - y}{x + y}$:

   Substitute $x = \frac{4y}{3}$:

   $\frac{x - y}{x + y} = \frac{\frac{4y}{3} - y}{\frac{4y}{3} + y}.$

   Simplify the numerator and the denominator:

   $x - y = \frac{4y}{3} - y = \frac{4y - 3y}{3} = \frac{y}{3},$ $x + y = \frac{4y}{3} + y = \frac{4y + 3y}{3} = \frac{7y}{3}.$

   Thus:

   $\frac{x - y}{x + y} = \frac{\frac{y}{3}}{\frac{7y}{3}} = \frac{y}{3} \cdot \frac{3}{7y} = \frac{1}{7}.$

Final Answer

The value of the fraction $\frac{x - y}{x + y}$ is:

$\frac{x - y}{x + y} = \frac{1}{7}.$

Would you like further details or have any questions?

Here are some related questions to consider:

1. How would the solution change if the equation was $6x^2 - 5xy + 4y^2 = 0$?
2. What are the possible values of $x$ if $x < y$?
3. How do we find the roots of a quadratic equation using the discriminant?
4. Can you explain why we discarded the negative value for $x$?
5. How can this problem be solved using a different method, such as factoring?

Tip: Always check the conditions of a problem before deciding which solutions are valid.