Solve for x and y :
`(a)/(x)-(b)/(y) = 0` and
`(ab^(2))/(x)+(a^(2)b)/(y) = a^(2) + b^(2)`, Where `x, y ne 0`.

Options

A. `x =(a(a^(2)+b^(2)))/(a^(2)+b^(2)),y=(a^(2)-b^(3))/(-(a^(2)+b^(2)))`
B. `x= (a(a^(2)-b^(2)))/(a^(2)+b^(2)),y=(a^(2)b-b^(3))/(-(a^(2)+b^(2)))`
C. `x=(a^(2)+b^(2))/(a(a^(2)-b^(2))),y=(-(a^(2)+b^(2)))/(a^(2)b-b)`
D. x = a, y = b

Answer

To solve the given equations for \( x \) and \( y \):

1. **Write down the equations:**
\[ \frac{a}{x} - \frac{b}{y} = 0 \quad \text{(1)} \]
\[ \frac{ab^2}{x} + \frac{a^2b}{y} = a^2 + b^2 \quad \text{(2)} \]

2. **From equation (1), express \( \frac{a}{x} \) in terms of \( \frac{b}{y} \):**
\[ \frac{a}{x} = \frac{b}{y} \]
Cross-multiplying gives:
\[ ay = bx \quad \text{(3)} \]

3. **Substitute \( y \) from equation (3) into equation (2):**
From equation (3), we can express \( y \) as:
\[ y = \frac{bx}{a} \]
Substitute this into equation (2):
\[ \frac{ab^2}{x} + \frac{a^2b}{\frac{bx}{a}} = a^2 + b^2 \]
Simplifying the second term:
\[ \frac{ab^2}{x} + \frac{a^2b \cdot a}{bx} = a^2 + b^2 \]
\[ \frac{ab^2}{x} + \frac{a^3b}{bx} = a^2 + b^2 \]
Combine the fractions:
\[ \frac{ab^2 + a^3b}{x} = a^2 + b^2 \]

4. **Factor out \( b \) from the numerator:**
\[ \frac{b(a^2 + ab)}{x} = a^2 + b^2 \]

5. **Cross-multiply to eliminate the fraction:**
\[ b(a^2 + ab) = x(a^2 + b^2) \]

6. **Solve for \( x \):**
\[ x = \frac{b(a^2 + ab)}{a^2 + b^2} \]

7. **Now substitute \( x \) back into equation (3) to find \( y \):**
From equation (3):
\[ y = \frac{bx}{a} \]
Substitute \( x \):
\[ y = \frac{b \cdot \frac{b(a^2 + ab)}{a^2 + b^2}}{a} \]
Simplifying gives:
\[ y = \frac{b^2(a^2 + ab)}{a(a^2 + b^2)} \]

8. **Final results:**
Thus, the solutions for \( x \) and \( y \) are:
\[ x = \frac{b(a^2 + ab)}{a^2 + b^2}, \quad y = \frac{b^2(a^2 + ab)}{a(a^2 + b^2)} \]