If `(5x^(2)-3y^(2)):xy=11:2` then what is the positive value of `(x)/(y)` ?

Options

A. `(5)/(2)`
B. `(3)/(2)`
C. `(5)/(3)`
D. `(7)/(2)`

Answer

To solve the problem, we start with the given equation:

\[ \frac{5x^2 - 3y^2}{xy} = \frac{11}{2} \]

### Step 1: Cross-multiply the equation
We can cross-multiply to eliminate the fraction:

\[ 2(5x^2 - 3y^2) = 11xy \]

### Step 2: Expand the equation
Expanding the left side gives us:

\[ 10x^2 - 6y^2 = 11xy \]

### Step 3: Rearrange the equation
Rearranging the equation to one side results in:

\[ 10x^2 - 11xy - 6y^2 = 0 \]

### Step 4: Treat this as a quadratic in terms of \(x\)
We can treat this equation as a quadratic equation in \(x\):

\[ 10x^2 - 11xy - 6y^2 = 0 \]

### Step 5: Use the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 10\), \(b = -11y\), and \(c = -6y^2\):

\[ b^2 - 4ac = (-11y)^2 - 4(10)(-6y^2) = 121y^2 + 240y^2 = 361y^2 \]

### Step 6: Substitute back into the quadratic formula
Substituting back into the formula gives:

\[ x = \frac{11y \pm \sqrt{361y^2}}{2 \cdot 10} \]

This simplifies to:

\[ x = \frac{11y \pm 19y}{20} \]

### Step 7: Calculate the two possible values for \(x\)
This gives us two possible values for \(x\):

1. \(x = \frac{30y}{20} = \frac{3y}{2}\)
2. \(x = \frac{-8y}{20} = -\frac{2y}{5}\) (we discard this as we want the positive value)

### Step 8: Find the ratio \(\frac{x}{y}\)
Now, we find the ratio \(\frac{x}{y}\):

\[ \frac{x}{y} = \frac{\frac{3y}{2}}{y} = \frac{3}{2} \]

### Final Answer
Thus, the positive value of \(\frac{x}{y}\) is:

\[ \frac{3}{2} \]