Solve for x and y in the following question.
`(2)/(x+2y) + (1)/(2x-y)+ (5)/(9)= 0, (9)/(x+2y)+ (6)/(2x-y)+4= 0`

Options

A. `x= 1, y= 2`
B. `x= 2, y= 1`
C. `x= 2, y= (1)/(2)`
D. `x= (1)/(2), y= 2`

Answer

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

1. **Write the equations**:
\[ \frac{2}{x + 2y} + \frac{1}{2x - y} + \frac{5}{9} = 0 \]
\[ \frac{9}{x + 2y} + \frac{6}{2x - y} + 4 = 0 \]

2. **Rearrange the equations**:
For the first equation:
\[ \frac{2}{x + 2y} + \frac{1}{2x - y} = -\frac{5}{9} \]
For the second equation:
\[ \frac{9}{x + 2y} + \frac{6}{2x - y} = -4 \]

3. **Substitute variables**:
Let:
\[ A = \frac{1}{x + 2y}, \quad B = \frac{1}{2x - y} \]
Then the equations become:
\[ 2A + B = -\frac{5}{9} \quad \text{(1)} \]
\[ 9A + 6B = -4 \quad \text{(2)} \]

4. **Eliminate one variable**:
To eliminate \( B \), we can multiply equation (1) by 6:
\[ 12A + 6B = -\frac{30}{9} \quad \text{(which simplifies to } -\frac{10}{3}\text{)} \]
Now we have:
\[ 12A + 6B = -\frac{10}{3} \quad \text{(3)} \]
\[ 9A + 6B = -4 \quad \text{(2)} \]

5. **Subtract equation (2) from equation (3)**:
\[ (12A + 6B) - (9A + 6B) = -\frac{10}{3} + 4 \]
This simplifies to:
\[ 3A = -\frac{10}{3} + \frac{12}{3} \]
\[ 3A = \frac{2}{3} \]
\[ A = \frac{2}{9} \]

6. **Substitute \( A \) back to find \( B \)**:
Substitute \( A \) into equation (1):
\[ 2\left(\frac{2}{9}\right) + B = -\frac{5}{9} \]
\[ \frac{4}{9} + B = -\frac{5}{9} \]
\[ B = -\frac{5}{9} - \frac{4}{9} = -\frac{9}{9} = -1 \]

7. **Substitute back to find \( x \) and \( y \)**:
Recall:
\[ A = \frac{1}{x + 2y} = \frac{2}{9} \quad \Rightarrow \quad x + 2y = \frac{9}{2} \quad \text{(4)} \]
\[ B = \frac{1}{2x - y} = -1 \quad \Rightarrow \quad 2x - y = -1 \quad \text{(5)} \]

8. **Solve the system of equations (4) and (5)**:
From equation (5):
\[ y = 2x + 1 \quad \text{(6)} \]
Substitute (6) into (4):
\[ x + 2(2x + 1) = \frac{9}{2} \]
\[ x + 4x + 2 = \frac{9}{2} \]
\[ 5x + 2 = \frac{9}{2} \]
Multiply through by 2 to eliminate the fraction:
\[ 10x + 4 = 9 \]
\[ 10x = 5 \quad \Rightarrow \quad x = \frac{1}{2} \]

9. **Find \( y \)**:
Substitute \( x = \frac{1}{2} \) back into equation (6):
\[ y = 2\left(\frac{1}{2}\right) + 1 = 1 + 1 = 2 \]

10. **Final solution**:
The solution is:
\[ x = \frac{1}{2}, \quad y = 2 \]