If x > y > 0, 6x²-5xy-4y² = 0, find the value of the fraction x-y÷x+y

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.

If x > y > 0, 6x² - 5xy - 4y² = 0, find the value of the fraction x - y ÷ x + y.

Step-by-step solution to solving the quadratic equation 6x² - 5xy - 4y² = 0, and finding the value of the fraction x - y / x + y. This problem covers key algebraic concepts including quadratic equations and simplifying fractions. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation