Find x and y: ` (x)/(2) - ( y )/(9) = 6, `
` ( x) /( 7) + ( y ) /(3) = 5 `.

Options

A. ` x = 14 , y = 9 `
B. ` x = 15 , y = 9 `
C. ` x = 14 , y = 3 `
D. ` x = 16 , y = 9 `

Answer

To solve the equations:

1. **Write down the equations**:
\[ \frac{x}{2} - \frac{y}{9} = 6 \quad \text{(1)} \]
\[ \frac{x}{7} + \frac{y}{3} = 5 \quad \text{(2)} \]

2. **Clear the fractions by finding the LCM**:
- For equation (1), the LCM of 2 and 9 is 18.
- Multiply the entire equation by 18:
\[ 18 \left(\frac{x}{2}\right) - 18 \left(\frac{y}{9}\right) = 18 \cdot 6 \]
This simplifies to:
\[ 9x - 2y = 108 \quad \text{(3)} \]

- For equation (2), the LCM of 7 and 3 is 21.
- Multiply the entire equation by 21:
\[ 21 \left(\frac{x}{7}\right) + 21 \left(\frac{y}{3}\right) = 21 \cdot 5 \]
This simplifies to:
\[ 3x + 7y = 105 \quad \text{(4)} \]

3. **Now we have the system of equations**:
\[ 9x - 2y = 108 \quad \text{(3)} \]
\[ 3x + 7y = 105 \quad \text{(4)} \]

4. **Use the elimination method**:
- To eliminate \(x\), we can multiply equation (4) by 3:
\[ 3(3x + 7y) = 3(105) \]
This gives us:
\[ 9x + 21y = 315 \quad \text{(5)} \]

5. **Now subtract equation (3) from equation (5)**:
\[ (9x + 21y) - (9x - 2y) = 315 - 108 \]
This simplifies to:
\[ 23y = 207 \]

6. **Solve for \(y\)**:
\[ y = \frac{207}{23} = 9 \]

7. **Substitute \(y\) back into one of the original equations to find \(x\)**. We'll use equation (3):
\[ 9x - 2(9) = 108 \]
Simplifying gives:
\[ 9x - 18 = 108 \]
\[ 9x = 108 + 18 \]
\[ 9x = 126 \]
\[ x = \frac{126}{9} = 14 \]

8. **Final solution**:
\[ x = 14, \quad y = 9 \]