Solve the following system of equations: `x/7+y/3=5,\ \ \ x/2-y/9=6`

Answer

To solve the system of equations:

1. **Equations Given:**
\[ \frac{x}{7} + \frac{y}{3} = 5 \quad \text{(Equation 1)} \]
\[ \frac{x}{2} - \frac{y}{9} = 6 \quad \text{(Equation 2)} \]

2. **Clear the Fractions in Equation 1:**
- The least common multiple (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:
\[ 3x + 7y = 105 \quad \text{(Equation 3)} \]

3. **Clear the Fractions in Equation 2:**
- 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:
\[ 9x - 2y = 108 \quad \text{(Equation 4)} \]

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

5. **Use the Elimination Method:**
- We want to eliminate one variable. Let's eliminate \(x\).
- To do this, we can multiply Equation 3 by 3 to match the coefficient of \(x\) in Equation 4:
\[ 3(3x + 7y) = 3(105) \]
- This gives us:
\[ 9x + 21y = 315 \quad \text{(Equation 5)} \]

6. **Subtract Equation 4 from Equation 5:**
\[ (9x + 21y) - (9x - 2y) = 315 - 108 \]
- This simplifies to:
\[ 21y + 2y = 207 \]
- Combining like terms gives:
\[ 23y = 207 \]
- Now, solve for \(y\):
\[ y = \frac{207}{23} = 9 \]

7. **Substitute \(y\) back into Equation 3 to find \(x\):**
\[ 3x + 7(9) = 105 \]
- This simplifies to:
\[ 3x + 63 = 105 \]
- Subtract 63 from both sides:
\[ 3x = 105 - 63 \]
- This gives:
\[ 3x = 42 \]
- Now, solve for \(x\):
\[ x = \frac{42}{3} = 14 \]

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